Calculus with Analytic Geometry 3–Fall 2018

Practice Exercises for the Final Exam–Solutions

1. The points P (1, 2, 3), Q(1, 0, 6), and R(4, 2, 4) are 3 vertices of the parallelogram spanned by the vectors~PQ, ~PR.

(a) Find the coordinates of the fourth vertex.

(b) A bug is crawling along the line of equation r(t) = (1 + t)i + (2 + t)j + (3 + 2t)k. Suppose time ismeasured in seconds, distance in meters. At time t = 0 it is at the point of coordinates (1, 2, 3). It thenmoves in the positive t direction for a distance of exactly 1 meter along the line, and stops. Determinethe coordinates of the point Z at which the bug has stopped.

(c) Find the volume of the parallelepiped having the parallelogram of before as a base, and Z as a vertex.

The fourth vertex of the parallelogram, call it S, will satisfy

~PS = ~PQ+ ~PR = 〈0,−2, 3〉+ 〈3, 0, 1〉 = 〈3,−2, 4〈

. Since ~PS is obtained by subtracting the coordinates of P from those of S; to get the coordinates of S weadd those of P to the components of ~PS. The answer to part (a) is (4, 0, 7).For part (b) we have to figure out first for which t > 0 is |r(t)−(1, 2, 3)| = 1. This works out to

√t2 + t2 + 4t2 =

1, thus t = 1/√

6. The answer is thus Z =

(1 +

1√6, 2 +

1√6, 3 +

2√6

).

For (c) we have that the parallelepiped is spanned by the vectors ~PQ, ~PR, andvecPZ = (1/

√6, 1,√

6, 2/√

6). The volume is then the absolute value of the triple product

( ~PQ× ~PR) · ~PZ =

∣∣∣∣∣∣∣∣∣∣0 −2 3

3 0 1

1/√

6 1/√

6 2/√

6

∣∣∣∣∣∣∣∣∣∣=

19√6

The answer is19√

6.

2. Consider the curve given parametrically by x = a cos t, y = b sin t, z = c sin t, where a, b, c are positive realnumbers. Show that it lies in a plane. Find the equation of the plane.

One can solve this by some educated guessing or in a more systematic way. I begin with the systematicway. A plane is determined by three points. So if we find three non-collinear points on the curve they willdetermine a plane and if the curve lies in a plane, that should be the plane. So for the three points let ustake some values of t where it is easy to compute the curve coordinates for that t. I think the best valuesare t = 0, π/4, and π/2. With t = 0 we get the point (a, 0, 0) (take as base point), with t = π/4, the point(a/√

2, b/√

2, c/√

2). With t = π/2 we get (0, b, c). A vector perpendicular to the plane through these threepoints is thus given by ∣∣∣∣∣∣∣∣∣∣

i j k

a√2− a b√

2c√2

−a b c

∣∣∣∣∣∣∣∣∣∣= −(

√2− 1)acj + (

√2− 1)abk.

Any vector parallel to this one would do; one such vector is −cj+bk. The equation of the plane perpendicularto this vector and through (a, 0, 0) is −cy + bz = 0. Plugging the expression for the curve into the equationof the plane we get

−cb sin t+ bc sin t = 0

2

which clearly holds for all t. The answer is −cy + bz = 0.

There is an alternative approach. The vector 〈0,−c, b〉 is easy to guess as being perpendicular to all values

of the curve, meaning the curve lies on the plane through the origin perpendicular to this vector.

3. A curve is described by the vector function r(t) = t2i + t3j + 1tk (t > 0). Find the parametric equations of

the tangent line at the point where t = 2.

We have

r′(t) = 2ti + 3t2j− 1

t2k.

For t = 2, the curve is at r(2) = 4i+8j+ 12k and a tangent vector is r′(2) = 4i+12j− 1

4k. Thus, the followingare parametric equations for this line.

x = 4 + 4t, y = 8 + 12t, z =1

2− 1

4t.

4. A curve is described by the vector function r(t) = t2i + t3j + 1tk (t > 0). Find the equations of the normal

plane at the point where t = 2.

In the previous exercise we found that the point where t = 2 is (4, 8, 1/2) and that direction vector of thetangent line is 〈4, 12,−1/4〉. This allows one to write out at once the equation of the plane as

4(x− 4) + 12(y − 8)− 1

4(z − 1

2) = 0.

One can work on this equation and rewrite it in the form 16x+ 48y − z = 447.

5. A curve is given parametrically by x = t2, y = t− t3, z = 0. Show that it self-intersects (goes twice throughthe same point) and find the equations of the two tangent lines at that point.

For the curve to self intersect there must be two parameter values, say s < t at which all three coordinatesare equal. This gives the following equations for s, t:

s2 = t2, s− s3 = t− t3, 0 = 0.

The last equation is, of course, always true and can be ignored. From the first one we get s = ±t thust = −s. Plugging this into the second equation gives s − s3 = −s + s3, thus 2(s + s3) = 0. We get threesolutions: s = −1, 0, 1. If s = −1 then t = −s = 1. If s = 0 then t = 0; but t > s. Similarly, if s = 1, thent = −1 violating s < t. The solutions are s = −1, t = 1. The point in question is the end point of the vectorr(−1) = r(1) = 〈1, 0, 0〉; i.e., the point (1, 0, 0). Now r′(t) = 〈2t, 1 − 3t2, 0〉 so two tangent vectors directingthe two lines are

r′(−1) = 〈−2,−2, 0〉, r′(1) = 〈2,−2, 0〉.From all of this information we get at once that the two lines have equations given parametrically by

x = 1− 2t, y = −2t, z = 0,

andx = 1 + 2t, y = −2t, z = 0.

6. (a) Find the length of the curve described by the position vector r(t) = 12ti + 8t3/2j + 3t2k for 0 ≤ t ≤ 1.It might help to realize that 144 = 4 · 36 and that (t+ 2)2 = t2 + 4t+ 4.

r′(t) = 12i + 12t1/2j + 6tk, thus

|r′(t)| =√

144 + 144t+ 36t2 =√

36(4 + 4t+ t2 =√

36(t+ 2)2 = 6(t+ 2).

Thus

L = 6

∫ 1

0

(t+ 2) dt = 15 .

3

(b) Suppose that the curve of part (a); that is, the curve of position vector r(t) = 12ti+ 8t3/2j+ 3t2k, givesthe position of a particle at time t. Determine the time th at which the particle has gone through halfthe length of the curve. That is, find th such that the length of r(t), 0 ≤ t ≤ th equals the length ofr(t), th ≤ t ≤ 1.

Since the total length of the curve is 15 (length units) we must find th such that the length of the segmentfor 0 ≤ t ≤ th is 15/2 units. This means that we should have∫ th

0

|r′(t)| dt =15

2.

Using the expression for |r′(t)| found in part (a), we get

6

∫ th

0

(t+ 2) dt =15

2; that is, 3t2h + 12th =

15

2.

Multiplying by 2, we get 6t2h + 24th − 15 = 0. Solving we get

th =−24±

√936

12= 2± 1

2

√78.

Since 0 < th < 1 we have to choose the plus sign, getting

th =1

2

√26− 2 ≈ 0.5495.

7. Find the vectors T, N, and B for the curve given by r(t) = 〈t2, 23 t3, t〉 at the point where t = 2.

Note: Computations can be a bit nasty, so work carefully. There is a point where it could help to know that(2t+ 1)2 = 4t2 + 4t+ 1.

We have r′(t) = 〈2t, 2t2, 1〉, thus

|r′(t)| =√

4t2 + 4t4 + 1 =√

(2t2 + 1)2 = 2t2 + 1.

Thus

T(t) = 〈 2t

2t2 + 1,

2t2

2t2 + 1,

1

2t2 + 1〉.

Setting t = 2 we get that

T = 〈49,

8

9,

1

9〉.

Next, being careful!,

T′(t) = 〈2(2t2 + 1)− 8t2

(2t2 + 1)2,

4t(2t2 + 1)− 8t3

(2t2 + 1)2,− 4t

(2t2 + 1)2〉

= 〈 2− 4t2

(2t2 + 1)2,

4t

(2t2 + 1)2,−4t

(2t2 + 1)2〉.

Since we do not have to differentiate any more, we could set t = 2 right now. Makes life easier. But to makethings as challenging as possible, we won’t do this yet. Then

|T′(t)| =1

(2t2 + 1)2

√(2− 4t2)2 + 16t2 + 16t2

=1

(2t2 + 1)2

√16t4 + 16t2 + 4 = 2(2t2 + 1).

Thus

N(t) = 〈1− 2t2

2t2 + 1,

2t

2t2 + 1,−2t

2t2 + 1〉.

4

Setting t = 2 we get

N = 〈−7

9,

4

9,−4

9〉.

Finally

B = T×N = 〈−4

9,

1

9,

7

9〉.

8. Find the velocity, acceleration, and speed (as functions of time t) of a particle with the position vectorr(t) = 〈t sin t, t cos t, t2〉.

v(t) = r′(t)〈sin t+ t cos t, cos t− t sin t, 2t〉,

speed = |v(t)| =√

(sin t+ t cos t)2 + (cos t− t sin t)2 + 4t2 =√

1 + 5t2,

a(t) = v′(t) = 〈2 cos t− t sin t,−2 sin t− t cos t, 2〉.

9. Find the position and velocity vectors of a particle if its acceleration and initial velocity and position are

a(t) = i + 2j + 2tk, v(0) = 0, r(0) = i + k.

We have

v(t) =

∫a(t) dt =

∫(i + 2j + 2tk) dt = ti + 2tj + t2k + C.

Setting t = 0 and equating with v(0) = 0, we get C = 0. Thus v(t) = ti + 2tj + t2k. Next

r(t) =

∫v(t) dt =

∫(ti + 2tj + t2k) dt =

1

2t2i + t2j +

1

3t3k + D.

Setting t = 0 and equating to r(0) = i+ k, we get D = i+ k, thus r(t) =

(1

2t2 + 1

)i + t2j +

(1

3t3 + 1

)k.

10. Let g(s, t) = f(u(s, t), v(s, t)) where f, u, v are differentiable,

u(2, 1) = 1, v(2, 1) = 3, us(2, 1) = 5, vs(2, 1) = 7,

f(2, 1) = 0, f(1, 3) = 4, f(5, 7) = 11,

fu(1, 3) = −2, fv(1, 3) = 2, fu(2, 1) = −1, fv(2, 1) = 11.

Determine gs(2, 1). Some of the information provided is useless and only included to see if you can selectwhat you need.

gs(s, t) = fu(u(s, t), v(s, t))us(s, t) + fv(u(s, t), v(s, t))vs(s, t);

setting s = 2, t = 1 we get

gs(2, 1) = fu(1, 3)us(2, 1) + fv(1, 3)vs(2, 1) = (−2) · 5 + 2 · 7 = 4.

11. Find an equation of the tangent plane to the given surfaces at the specified point.

(a) z = x2 + y3 at the point where (x, y) = (1, 2); that is at (1, 2, 9).

zx = 2x, zy = 3y2 so that zx(1, 2) = 2, zy(1, 2) = 4. The equation is

z = 9 + 2(x− 1) + 12(y − 2).

5

(b) xy2 + yz2 + zx2 = 3 at (1, 1, 1).

The surface is given in the form f(x, y, z) = 3 where f(x, y, z) = xy2 + yz2 + zx2. Now ∇f(1, 1, 1) =〈3, 3, 3〉, thus the equation is 3(x− 1) + 3(y− 1) + 3(z − 1) = 0. This equation reduces to x+ y+ z = 3.

(c) x+ zey + yez = 4 at the point where (x, y) = (1, 0).

No value of z is given, so it has to be figured out. If x = 1, y = 0 the equation becomes 1 + z = 4,thus z = 3. The point at which the plane is tangent is (1, 0, 3). If f(x, y, z) = x + zey + yez, then∇f(1, 0, 3) = 〈1, 3 + e3, 1〉. The equation is

x− 1 + (3 + e3)y + z − 3 = 0, or x+ (3 + e3)y + z = 4.

12. Let f(x, y) = x2 + 4xy − y2. (a) Find the unit vectors giving the direction of steepest ascent and steepestdescent at (2, 1). (b) Find a vector that points in a direction of no change at (2, 1).

(a) ∇f(x, y) = 〈2x+ 4y, 4x− 2y〉; ∇f(2, 1) = 〈8, 6〉. Now |∇f(2, 1)| =√

82 + 62 = 10. The unit vector in the

direction of steepest ascent is 〈45,

3

5〉. , of steepest descent, −〈4

5,

3

5〉.

For (b), both 〈35,−4

5〉 or 〈−3

5,

4

5〉 point in directions of no change; either one can be used to answer part (b).

13. A mountain having an elliptical base can be described by the equation z = 25− x2 − 4y2. A climber will tryto reach the top starting at the point of coordinates (3, 2, 0). The climber wants to be always going in thedirection of steepest ascent. The climber decides to follow the path that can be parameterized by

x = 3e−t, y = 2e−4t, z = 25− 9e−2t − 16e−8t, t ≥ 0.

(a) Show that this is a path that is indeed on the mountain.

25− x(t)2 − 4y(t)2 = 25− 9e−2t − 16e−8t = z(t),

so it is a mountain path.

(b) Show that if the climber follows it, at each point the climber will be moving in the direction of steepestascent.

Let P (x, y, z) be a point on the path, so for some value of t ≥ 0 we have

x = 3e−t, y = 2e−4t, z = 25− 9e−2t − 16e−8t.

Since the climber is moving along the path in question, the climber’s direction is given by the tangentto the curve, namely

T = 〈−3e−t,−8e−4t, 18e−2t + 128e−4t〉.The last component is not too important here; it is positive showing the motion is upward, but that’sall. The two first components; that is, u = 〈−3e−t,−8e−4t〉, give the direction of motion. The directionof steepest ascent is given by the gradient; ∇z = 〈−2x,−8y〉. At the point in question we have

〈−2x,−8y〉 = 〈−2(3e−t),−8(2e−4t)〉 = 〈−6e−t,−16e−4t〉 = 2u.

This shows that the vector governing the direction of motion is parallel to the direction of steepest ascent(and pointing in the same direction), so the path is indeed a path of steepest ascent.

14. Find the critical points and use the second derivative test to classify them of f(x, y) = x4 +y4−4x−32y+10.

To find the critical points, we set the first order partials to 0.

fx = 4x3 − 4 = 0,

fy = 4y3 − 32 = 0.

There is a single solution, (x, y) = (1, 2). This is the only critical point. Now

D(x, y) =

∣∣∣∣ fxx(x, y) fxy(x, y)fxy(x, y) fyy(x, y)

∣∣∣∣ =

∣∣∣∣ 12x3 00 12y3

∣∣∣∣

6

so that

D(1, 2) =

∣∣∣∣ 12 00 96

∣∣∣∣ = 1152 > 0.

Because D(1, 2) is positive, we have a relative minimum or maximum at (1, 2). Because fxx(1, 2) < 0, it is arelative minimum.

15. Find the critical points and use the second derivative test to classify them of f(x, y) = xye−x−y.

We set the first order partial derivatives to 0:

fx = ye−x−y − xye−x−y = (1− x)ye−x−y = 0,

fy = xe−x−y − xye−x−y = (1− y)xe−x−y = 0.

The first equation is satisfied if (and only if) x = 1 or y = 0. Going to the second equation with x = 1, theonly solution is y = 1. So (1, 1) is a critical point, and the only one with first component 1. If we put y = 0in the second equation, we see that we must have x = 0. So (0, 0) is a critical point; the only one with secondcomponent 0. The critical points are (0, 0) and (1, 1).The Hessian determinant is

D(x, y) =

∣∣∣∣ y(x− 2)e−x−y (1− x)(1− y)e−x−y

(1− x)(1− y)e−x−y x(y − 2)e−x−y

∣∣∣∣ .From this D(0, 0) = −1, so (0, 0) is a saddle point. Next, D(1, 1) = e−4 > 0. so we have a relativeextremum. Since fxx(1, 1) = −e−2 < 0, (1, 1) is a relative maximum.

16. Find ALL critical points off(x, y) = 3x2 − 2x2y − 4y2 + 2y3

and classify each one of them as local maximum, local minimum, or saddle point.

We compute the partial derivatives and set them to zero:

fx : 6x− 4xy = 0,

fy : −2x2 − 8y + 6y2 = 0.

The first equation has one solution x = 0. Using this in the second equation we get −8y + 6y2 = 0, thusy(6y − 8) = 0, which implies y = 0 or y = 4/3. Two critical points are (0, 0) and (0, 4/3). If x 6= 0 then the

first equation has the only solution y = 3/2. Using this in the second equation gives x = ±√

3/2. The criticalpoints are

(0, 0), (0, 4/3), (−√

3/2, 3/2), (√

3/2, 3/2).

For the classification we compute the Hessian determinant

D(x, y) =

∣∣∣∣ 6− 4y −4x−4x 12y − 8

∣∣∣∣ = 104y − 48y2 − 48− 16x2.

Now for the classification.

D(0, 0) = −48 < 0, (0, 0) is a saddle point.

D(0, 4/3) = 16/3 > 0, , fxx(0, 4/3) = 2/3 > 0, (0, 4/3) is a relative minimum.

D(±√

3/2, 3/2) = −12 < 0, (−√

15/2, 3/2) and (√

15/2, 3/2) are saddle points.

17. Show that the second derivative test is inconclusive when applied to f(x, y) = x2y− 3 at (0, 0). Describe thebehavior of the function at the critical point.

fx(x, y) = 2xy, fy(x, y) = x2, so fx(0, 0) = fy(0, 0) = 0. The point (0, 0) is indeed critical. We also havefxx(0, 0) = 2y

∣∣(0,0)

= 0, fxy(0, 0) = 2x∣∣(0,0)

= 0, fyy(0, 0) = 0, so of course the Hessian determinant is 0 and

the test is inconclusive.Now f(0, 0) = −3. To describe the behavior at (0, 0) we might notice that if x > 0, then xy2 > 0 so that aswe move away from (0, 0) along the positive x axis, the function grows. But for the same reason, moving awayfrom (0, 0) along the negative x-axis, the function decreases. This means we can’t have a relative maximumor minimum at (0, 0). It follows that there is a saddle point at (0, 0).

7

18. Show that the second derivative test is inconclusive when applied to f(x, y) = sin(x2y2) at (0, 0). Describethe behavior of the function at the critical point.

I will begin describing the behavior of the function at the point. It is clear the function has a relative minimumat (0, 0). In fact, if both x, y are different from 0 and small, then sin(x2y2) > 0. If one of x or y is 0, thensin(x2y2) = 0. In either case sin(x2y2) ≥ 0 = sin(02 · 02), so there is a relative minimum at (0, 0).One sees that fx, fy, fxx, fxy, fyy are all 0 at(0, 0), so it is a critical point (as it had to be, being a relativeminimum) and the Hessian determinant is 0.

For the next few exercises things to know are:

1. In a closed and bounded region, a continuous function will assume a maximum value and it will assumea minimum value.

2. These values have to be assumed either at a critical interior point or on the boundary. They cannot beassumed anywhere else.

Maybe I should add a third thing, if a problem asks for the maximum or minimum VALUE of a function,one might consider that a value is a number, not a pair of numbers or a point in the plane. Any answer thatis not a number is essentially wrong.

19. Textbook, §12.8, Exercise # 48, p. 949: Find the absolute maximum and the absolute minimum values ofthe function f(x, y) = x2 + y2 − 2x− 2y on the closed triangle R of vertices (0, 0), (2, 0) and (0, 2).

Notice that it is the region bounded by the lines x = 0, y = 0 and y = 2 − x. The boundary can be thusdescribed as the union of three segments that I will label by I, II, and III, and they are

I) x = 0, 0 ≤ y ≤ 2,

II) 0 ≤ x ≤ 2, y = 0,

III 0 ≤ x ≤ 2, y = 2− x.

We begin finding critical points, setting the first order partial derivatives to 0:

fx = 2x− 2 = 0,

fy = 2y − 2 = 0.

The only critical point is (1, 1); it is on the boundary of the region. It will probably pop up again whenstudying the function on the boundary.Boundary Behavior.On I. Let h(y) = f(0, y) = y2 − 2y. Points where something could happen are the endpoints of the intervalor at a critical point of h. Now h′(y) = 2y − 2 = 0 for y = 1. So the points we have to consider here are thepoints (0, 0), (0, 1), (0, 2).On II. Since the function is symmetric in x and y, the points to consider on this interval will be (0, 0), (1, 0)and (2, 0).On III. Let h(x) = f(x, 2− x) = 2x2 − 4x. Now h′(x) = 4x− 4 = 0 for x = 1 so a point to consider is (1, 1).We also have to consider the endpoints, but these are (0, 2) and (2, 0); already considered before.All in all we have the following points to consider:

(1, 1), (0, 0), (0, 1), (0, 2), (1, 0), (2, 0).

We have:

f(1, 1) = −2, f(0, 0) = 0, f(0, 1) = −1, f(0, 2) = 0, f(1, 0) = −1, f(2, 0) = 0.

From this we see that the minimum value is −2 (assumed at (1, 1)), the maximum value is 0 (assumed at(0, 0), (0, 2), and at (2, 0)).

20. Find the maximum and the minimum value of the following function in the indicated domain. If there is nomaximum or minimum, say so.

f(x, y) = |3− x2 − y2| in the closed disc {(x, y) : x2 + y2 ≤ 9}.

8

This is a continuous function in a closed and bounded domain. It MUST assume a maximum value and itMUST assume a minimum value.f(x, y) is negative, but it is 0 for x2 + y2 = 3. The minimum value is 0, assumed one the circle of radius√

3 centered at the origin. Moving away from that circle, the function grows. It has a local maximum at(0, 0), with a value of 3. However on the boundary of the domain the value assumed is 6. The answer is: Theminimum value is 0, the maximum value is 6.

21. Compute∫∫Rxy dA; R is the region in the first quadrant bounded by x = 0, y = x2 and y = 8− x2.

∫∫R

(x+ y) dA =

∫ 2

0

∫ 8−x2

x2

(x+ y) dy dx =152

3.

22. Write as an iterated integral in the order dx dy the integral of f over the region in quadrants 2 and 3 boundedby the semicircle of radius 3 centered at (0, 0).∫ 3

−3

∫ 0

−√

9−y2f(x, y) dx dy.

23. Evaluate∫∫Ry2 dA where R is the region bounded by y = 1, y = 1− x, y = x− 1.

∫∫R

y2 dA =

∫ 1

0

y2∫ 1+y

1−ydx dy =

1

2.

24. Find the volume of the solid in the first octant bounded by the coordinate planes and the surface z = 1−y−x2.

The solid can be described as being {(x, y, z) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1− x2, 0 ≤ z ≤ 1− y − x2}. Thus

V =

∫ 1

0

∫ 1−x2

0

(1− y − x2) dy dx =4

15.

25. Sketch the region of integration and evaluate the

∫ 2

0

∫ 4−x2

0

xe2y

4− ydy dx by reversing the order of integration.

The region can be described by R = {(x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 4− x2}. Here’s a sketch.

From the sketch, or directly, we see that the region can also be described as R = {0 ≤ y ≤ 4, 0 ≤ x ≤√

4− y}.Thus∫ 2

0

∫ 4−x2

0

xe2y

4− ydy dx =

∫ 4

0

∫ √4−y

0

xe2y

4− ydx dy =

1

2

∫ 4

0

(√

4− y)2e2y

4− ydy =

1

2

∫ 4

0

e2y dy =e8 − 1

4.

9

26. Find the volume of the solid bounded by the paraboloids z = 2x2 + y2 and z = 27− x2 − 2y2.

The paraboloids intersect for x2 + y2 = 9 so the solid can be described by D = {(x, y, z) : x2 + y2 ≤9, 2x2 + y2 ≤ z ≤ 27− x2 − 2y2}. The description in terms of polar coordinates is D = {0 ≤ r ≤ 3, 0 ≤ θ ≤2π, r2(1 + cos2 θ) ≤ z ≤ 27− r2(1 + sin2 θ)}. Thus

V =

∫ 2π

0

∫ 3

0

[(27− r2(1 + sin2 θ))− r2(1 + cos2 θ)

]r dr dθ =

∫ 2π

0

∫ 3

0

(27− 3r2)r dr dθ =243

2π.

27. Sketch the region inside both the cardioid r = 1− cos θ and the circle r = 1, and find its area.

Sketch:

The region is symmetric with respect to the x axis, so the area is twice the area above the x-axis; that is for0 ≤ θ ≤ π. For 0 ≤ θ ≤ π/2, r goes from the origin to the cardioid; after that to the circle. Thus

A = 2

(∫ π/2

0

∫ 1−cos θ

0

r dr dθ +

∫ π

π/2

∫ 1

0

r dr dθ

)= 2

(1

2

∫ π/2

0

(1− cos θ)2 dθ +π

4

)=

5π

4− 2.

28. Find the average distance of points within the cardioid r = 1 + cos θ and the origin.

The distance of a point from the origin is r, so the average will be the integral of r over the cardioid, dividedby the area of the cardioid. The area of the cardioid is

A =

∫ 2π

0

∫ 1+cos θ

0

r dr dθ =3π

2.

The average is

Av =1

A

∫ 2π

0

∫ 1+cos θ

0

r2 dr dθ =1

A· 5π

3=

10

9.

29. Compute

∫∫R

x− yx2 + y2 + 1

dA, R is the unit disc.

In polar coordinates ∫∫R

x− yx2 + y2 + 1

dA =

∫ 2π

0

∫ 1

0

r(cosθ − sin θ)

1 + r2r dθ dr.

Since∫ 2π

0cos θ dθ =

∫ 2π

0sin θ dθ = 0, the integral works out to 0.

30. The figure shows the region of integration for the integral∫ 1

0

∫ 1−x

0

∫ 1−x2

0

f(x, y, z) dz dy dx.

10

Rewrite the integral as an iterated integral in the five other orders. That is, replace the question marks withappropriate expressions so the following five integrals are equal to the one given above.∫ ?

?

∫ ?

?

∫ ?

?

f(x, y, z) dy dz dx,

∫ ?

?

∫ ?

?

∫ ?

?

f(x, y, z) dy dx dz,

∫ ?

?

∫ ?

?

∫ ?

?

f(x, y, z) dz, dx dy,

∫ ?

?

∫ ?

?

∫ ?

?

f(x, y, z) dx dy dz,

∫ ?

?

∫ ?

?

∫ ?

?

f(x, y, z) dx dz dy.

The region can be described by

D = {(x, y, z) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1− x, 0 ≤ z ≤ 1− x2}.

I usually prefer to work from the description. If there is a picture I’ll use it as a guide, but pictures are notessential. Here are the other five integrals, in the same order as in the question. Actually, the last two are abit tricky and will result in more than one integral. You will see. Because of this, see also the grading noteat the end.

dy dz dx) There is no relation between y and z in the description, so this is quite easy,∫ 1

0

∫ 1−x

0

∫ 1−x2

0

f(x, y, z) dz dy dx =

∫ 1

0

∫ 1−x2

0

∫ 1−x

0

f(x, y, z) dy dz dx.

dy dx dz) z can range from 0 to 1; since 0 ≤ z ≤ 1− x2 we get 0 ≤ x ≤√

1− z. For every value of (x, z), ystill can go from 0 to 1− x. Thus∫ 1

0

∫ 1−x

0

∫ 1−x2

0

f(x, y, z) dz dy dx =

∫ 1

0

∫ √1−z

0

∫ 1−x

0

f(x, y, z) dy dx dz.

dz dx dy) There isn’t much of a problem here.∫ 1

0

∫ 1−x

0

∫ 1−x2

0

f(x, y, z) dz dy dx =

∫ 1

0

∫ 1−y

0

∫ 1−x2

0

f(x, y, z) dz dx dy.

dx dy dz) This one is tricky. We have 0 ≤ z ≤ 1 and 0 ≤ y ≤ 1. But when it comes to x we have to haveboth 0 ≤ x ≤ 1− y and 0 ≤ x ≤

√1− z. For a given value of z, 0 ≤ z ≤ 1, where x ranges depends on

whether 1− y ≤√

1− z; i.e., y ≥ 1−√

1− z, in which case x goes from 0 to 1− y, or if 1− y >√

1− z,in which case x ranges from 0 to

√1− z. The answer in this case is∫ 1

0

∫ 1−x

0

∫ 1−x2

0

f(x, y, z) dz dy dx =

∫ 1

0

∫ 1−√1−z

0

∫ √1−z

0

f(x, y, z) dx dy dz

+

∫ 1

0

∫ 1

1−√1−z

∫ 1−y

0

f(x, y, z) dx dy dz.

11

dx dz dy) The analysis here is similar as in the previous case. The answer works out to∫ 1

0

∫ 1−x

0

∫ 1−x2

0

f(x, y, z) dz dy dx =

∫ 1

0

∫ 1−(1−y)2

0

∫ 1−y

0

f(x, y, z) dx dz dy

+

∫ 1

0

∫ 1

1−(1−y)2

∫ √1−z

0

f(x, y, z) dx dz dy.

31. Suppose f is continuous and a > 0. Show that∫ a

0

∫ y

0

∫ z

0

f(x) dx dz dy =1

2

∫ a

0

(a− x)2f(x) dx.

The region of integration is D = {(x, y, z) : 0 ≤ y ≤ a, 0 ≤ z ≤ y, 0 ≤ x ≤ z}. It can also be described byD = {(x, y, z) : 0 ≤ x ≤ a, x ≤ z ≤ a, z ≤ y ≤ a}. Thus∫ a

0

∫ y

0

∫ z

0

f(x) dx dz dy =

∫ a

0

∫ a

x

∫ a

z

f(x) dy dz dx =

∫ a

0

f(x)

∫ a

x

∫ a

z

dy dz dx

=

∫ a

0

f(x)

∫ a

x

(a− z) dz dx =1

2

∫ a

0

(a− x)2f(x) dx.

An alternative way of doing this exercise is describing the region in the form D = {(x, y, z) : 0 ≤ x ≤ a, x ≤z ≤ a, x ≤ y ≤ z}. In this case∫ a

0

∫ y

0

∫ z

0

f(x) dx dz dy =

∫ a

0

∫ a

x

∫ y

x

f(x) dz dy dx =

∫ a

0

f(x)

∫ a

x

∫ y

x

dz dy dx

=

∫ a

0

f(x)

∫ a

x

(x− y) dy dx =1

2

∫ a

0

(a− x)2f(x) dx.

32. Let R = {(x, y) : 0 ≤ x ≤√π, 0 ≤ y ≤

√π}. Evaluate∫∫

R

sin(max(x2, y2)

)dA.

By max(a, b), if a, b are numbers, one understands a if a ≥ b; otherwise it is b.

We divide the square R up into the triangle T1 where x < y, and the triangle T2 where x > y:

T1 = {(x, y) : 0 ≤ y ≤√π, 0 ≤ x ≤ y}, T2 = {(x, y) : 0 ≤ x ≤

√π, 0 ≤ y ≤ x}.

Then∫∫R

sin(max(x2, y2)

)dA =

∫∫T1

sin y2 dA+

∫∫T2

sinx2 dA =

∫ √π0

∫ y

0

sin y2 dx dy +

∫ √π0

∫ x

0

sinx2 dy dx

=

∫ √π0

y sin y2 dy +

∫ √π0

x sinx2 dy = − 1

2cos(y2)

∣∣∣∣√π

0

+

(− 1

2cos(x2)

∣∣∣∣√π

0

)= 2.

33. Use cylindrical coordinates to evaluate∫ 3

−3

∫ √9−x2

0

∫ 2

0

1

1 + x2 + y2dz dy dx.

The z integral is totally independent from the x, y integrals, and {(x, y) : −3 ≤ x ≤ 3, 0 ≤ y ≤√

9− x2} isthe half circle of radius 3 above the x-axis, centered at the origin.∫ 3

−3

∫ √9−x2

0

∫ 2

0

1

1 + x2 + y2dz dy dx = 2

∫ 3

−3

∫ √9−x2

0

1

1 + x2 + y2dy dx = 2

∫ 3

0

∫ π

0

1

1 + r2r dθ dr

= 2

(∫ 3

0

r

1 + r2dr

)(∫ π

0

dθ

)= π ln 10.

12

34.

Let D be the region in the first octant of R3

bounded by the paraboloid z = 4−(x2+y2). Com-

pute

∫∫∫D

x dV .

The picture on the right shows the paraboloid andthe plane z = 0.

Hint: Use cylindrical coordinates.

∫∫∫D

x dV =

∫ 2

0

∫ π/2

0

∫ 4−r2

0

r2 cos θ dz dθ dr =

∫ 2

0

∫ π/2

0

(4− r2)r2 cos θ dθ dr =

∫ 2

0

(4r2 − r4) dr =64

15.

35. Let D = {(x, y, z) : x ≥ 0, y ≥ 0, z ≥ 0, x2 + y2 + z2 ≤ 9}. Compute∫∫∫D

ez dV.

using spherical coordinates.

Working with spherical coordinates,∫∫∫D

ez dV =

∫ 3

0

∫ π/2

0

∫ π/2

0

eρ cosϕρ2 sinϕdθ dϕ dρ =π

2

∫ 3

0

∫ π/2

0

eρ cosϕρ2 sinϕdϕdρ

=π

2

∫ 3

0

ρ2

(− 1

ρeρ cosϕ

∣∣∣∣ϕ=π/2ϕ=0

)dρ =

π

2

∫ 3

0

ρ(eρ − 1) dρ = π

(e3 − 7

4

).

36. Find the volume of the region common to the three cylinders

x2 + y2 ≤ 1, y2 + z2 ≤ 1, z2 + x2 ≤ 1.

There is a picture of the region on page 1044 of the textbook.

This is not as hard as it seems, one merely has to be systematic. In the first place, we can simplify the problemby using some symmetry, finding out the volume of the intersection in the first octant x|ge0, y ≥ 0, z ≥ 0; thetotal volume is then 8 times the volume in the first octant.If (x, y, z) are in the intersection of the three cylinders then they verify simultaneously

x2 + y2 ≤ 1, y2 + z2 ≤ 1, z2 + x2 ≤ 1.

So, since we are in the first octant, beginning with x, we have that we can have any value of x in the interval[0, 1], so 0 ≤ x ≤ 1. Next we see that since x2 + y2 ≤ 1, we’ll have 0 ≤ y ≤

√1− x2. The problem is now

what happens with z. The answer is that since we must have both z ≤√

1− x2 and z ≤√

1− y2 we need

to break up the unit disc (more precisely the quarter unit disc) into the part where√

1− x2 ≤√

1− y2 and

where√

1− x2 ≤√

1− y2.A simple College Algebra (i.e., High School) level computation shows that√1− x2 ≤

√1− y2 if and only if y ≤ x.

We could continue with cartesian coordinates, but here is where switching t cylindrical coordinates makeslife easier. In the quarter disc below, the part in red is where y ≤ x, so there

√1− x2 ≤

√1− y2 and z

will range from 0 to√

1− x2 =√

1− r2 cos2 θ. In the green part, y ≥ x, so there z will range from 0 to√1− y2 =

√1− r2 sin2 θ.

13

In polar coordinates, the red region is described by {(r, θ), : 0 ≤ r ≤ 1, 0 ≤ the ≤ π/4} while the green partis {(r, θ), : 0 ≤ r ≤ 1, π/4 ≤ the ≤ π/2}. Putting this all together, remembering to multiply by 8, we seethat

V = 8

(∫ π/4

0

∫ 1

0

√1− r2 cos2 θ r dr dθ +

∫ π/2

π/4

∫ 1

0

√1− r2 sin2 θ r dr dθ

).

The integrals seem hard, but they are not really. In the first iterated integral, we can perform the integrationwith respect to r by the substitution s = 1− r2 cos2 θ, so rdr = −1/(2 cos2 θ) ds and∫ π/4

0

∫ 1

0

√1− r2 cos2 θ r dr dθ =

1

2

∫ π/4

0

1

cos2 θ

∫ 1

sin2 θ

√s ds =

1

3

∫ π/4

0

1

cos2 θs3/2

∣∣∣∣1sin2 θ

dθ

=1

3

∫ π/4

0

1− sin3 θ

cos2 θdθ =

1

3

∫ π/4

0

1− sin θ(1− cos2 θ)

cos2 θdθ

=1

3

∫ π/4

0

(1

cos2 θ− sin θ

cos2 θ+ sin θ

)dθ =

1

3(tan θ − sec θ − cos θ)

∣∣∣∣π/40

=1

3

(1− (

√2− 1)− (

1√2− 1)

)=

2−√

2

2.

Well, maybe it got a bit hairy at the end, but there is always Wolfram alpha to compute integrals (or Maple,or Mathematica, or tables). A similar computations shows that one also has∫ π/2

π/4

∫ 1

0

√1− r2 sin2 θ r dr dθ =

2−√

2

2,

so that finally one gets V = 8(2−√

2).

37. The result of this exercise will be used in Exercise 38. Find the coordinates of the center of mass (centroid)of the portion of the cardioid r = 1 + cos θ above the x-axis; that is, the region described by R = {(r, θ) :0 ≤ θ ≤ π, 0 ≤ r ≤ 1 + cos θ}.

14

We get

A =

∫ π

0

∫ 1+cos θ

0

r dr dθ =3π

4,

My =

∫ π

0

∫ 1+cos θ

0

r2 cos θ dr dθ =5π

8,

Mx =

∫ π

0

∫ 1+cos θ

0

r2 sin θ dr dθ =4

3.

thus

x =My

A=

5

6, y =

Mx

A=

16

9π.

38. We are going to work here with the solid obtained by revolving a cardioid. We flip the cardioid of Exercise37 so it looks like:

We then rotate it about the z-axis.

(a) Show, or at least convince yourself, that in spherical coordinates the solid so obtained is described by

D = {(ρ, ϕ, θ) : 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π, 0 ≤ ρ ≤ 1 + cosϕ}.

(b) Use spherical coordinates to find its volume.

(c) A theorem due to Pappus (who lived some 1,700 years ago) states that the volume of a solid obtained byrotating a plane figure of area A about an axis equals A×C, where C is the circumference of the circledescribed by the centroid of the figure. The solid of this problem is obtained by revolving about thez-axis the portion of the cardioid described in Exercise 37, except that what was the x-axis in Exercise37 is now the z-axis. If the coordinates of the center of mass you found in Exercise 37 are (a, b), thecenter of mass of the figure rotated about the z-axis are at (b, 0, a). Think about it. Use this informationto verify Pappus’ Theorem in this case.

15

(a) Convince yourself!

(b) By the description in spherical coordinates,

V =

∫ 2π

0

∫ π

0

∫ 1+cosϕ

0

ρ2 sinϕdρ dϕdθ =2π

3

∫ π

0

(1 + cosϕ)3 =8π

3.

(c) A bit of reflection (no pun intended) shows that the cardioid to be rotated comes from the one in theprevious exercise if the xaxis is replaced by the z axis and is placed in the plane y = 0:The coordinates of the center of mass of the cardioid in this position are ( 16

9π , 0,56 ); in particular, the

distance of the center of mass from axis of rotation is 169π . This means that the center of mass will

describe a circle of radius 16/9π. By Pappus theorem (recalling that the area was 3π/4),

V = 2π · 16

9π· 3π

4=

8π

3.

39. A solid is bounded above by the paraboloid z = 4−x2−y2 and below by the cone z =√x2 + y2. The density

of the solid satisfies ρ(x, y, z) = z. Find the coordinates of the center of mass.

Note: This exercise results into some really nasty integrals. While you will not have access (at least notlegally) to any computer algebra for the final, you might consider using something like Wolfram alpha here.

In cylindrical coordinates the two surfaces bounding the region are z = 4−r2 and z = r. They come togetherfor r = (

√17− 1)/2. The region can be described as

D = {(r, θ, z) : 0 ≤ r ≤√

17− 1

2, 0 ≤ θ ≤ 2π, r ≤ z ≤ 4− r2}.

By symmetry, x = y = 0. We have (integrals computed by Maple)

V =

∫ (√17−1)/2

0

∫ 2π

0

∫ 4−r2

r

z r dz dθ dr =135 + 17

√17

24π,

Mxy =

∫ (√17−1)/2

0

∫ 2π

0

∫ 4−r2

r

z2 r dz dθ dr =3123− 187

√17

120π

Finally, after some simplification

z =929− 153

√17

130≈ 2.2935756867730610452101868618151

(Here finding an approximate value was a good idea, to see if the answer is at least likely. It is.)

40. Let R be the region in the first quadrant bounded by the curves xy = 1, xy = 5, y = 3x− 2, y = 3x+ 2.

16

Show that changing variables by u = xy, v = y − 3x, changes the region into a rectangle {(u, v) : a ≤ u ≤b, c ≤ v ≤ d}. Find the correct values of c and d and use the change of variables to compute∫∫

R

(y + 3x) dA.

Compute also the integral without changing variables and verify that you get the same result.

Here is a fact that could ease computations:∂(x, y)

∂(u, v)=

(∂(u, v)

∂(x, y)

)−1.

It is easy to see that the indicated change maps the region R into the rectangle {1 ≤ u ≤ 5,−2 ≤ v ≤ 2}.The only place where things might get slightly hairy is when trying to invert the map, get x, y in terms ofu, v. From u = xy we have y = u/x; substituting into the expression for v we get 3x2 + vx − u = 0, fromwhich

x =−v ±

√v2 + 12u

6.

Clearly x has to be positive so x =

√v2 + 12u− v

6. Then

y =u

x=

6u√v2 + 12u

=3(√v2 + u− v)

2.

We now have to compute the Jacobian determinant and here is where things get really nasty, having tocompute

∂(x, y)

∂(u, v)=∂x

∂u

∂y

∂v− ∂x

∂v

∂y

∂u.

But perhaps the hint can help. We have

∂(u, v)

∂(x, y)=∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x= y − x(−3) = y + 3x.

Usually this would be of some help; to get∂(x, y)

∂(u, v)from this we have to use that

∂(x, y)

∂(u, v)=

1∂(u,v)∂(x,y)

=1

3x+ y,

and replace x, y by their expressions in u, v. We also have to replace any other x, y in the integral by theirexpressions in u, v. But in this case the simplification is absolute,∫∫

R

(y + 3x) dA =

∫ 5

1

∫ 2

−2(y(u, v) + 3y(u, v))

∣∣∣∣∂(x, y)

∂(u, v)

∣∣∣∣ dv du=

∫ 5

1

∫ 2

−2(y(u, v) + 3y(u, v))

1

y(u, v) + 3y(u, v)dv du =

∫ 5

1

∫ 2

−2dv du = 16 .

And solving for x, y was unnecessary! (But that’s because the integral was so set up.)If we find the integral without changing we get

∫∫R

(y + 3x) dA =

∫ 1

1/3

∫ 3x+2

1/x

(y + 3x) dy dx+

∫ 5/3

1

∫ 5/x

3x−2(y + 3x) dy dx = 16,

same as before, of course.

17

41. Evaluate

∫∫R

xy dA where R is the region in the plane bounded by the curves y = x2/25, y = (x−1)2/25+1,

y = x and y = x− 4 (see picture below) by changing variables by

x = u+ 5v,

y = u+ v2.

In this, as in all change of variables exercises, you must clearly indicate what the transformation does; intowhat it changes the region of integration, and give at least a good reason why the new region is what you sayit is.

Computing the integral by any method except the indicated one does not count!.

Let R′ = {(u, v) : 0 ≤ u ≤ 1, 0 ≤ v ≤ 1}. The boundary of R′ consists of the 4 segments

I1 = {(u, 0) : 0 ≤ u ≤ 1},I2 = {(1, v) : 0 ≤ v ≤ 1},I3 = {(u, 1) : 0 ≤ u ≤ 1},I4 = {(0, v) : 0 ≤ v ≤ 1}.

The change carries I1 to points satisfying x = u, y = u, 0 ≤ u ≤ 1. This is a parametrization of thesegment y = x from (0, 0) to (1, 1), the left boundary of R. The segment I2 gets mapped to points satisfyingx = 1 + 5v, y = 1 + v2, 0 ≤ v ≤ 1. This is a parametrization of the arc of parabola y = 1 + (x − 1)2/25,from (1, 1) to (6, 2), the upper boundary of R. The segment I3 maps to x = u + 5, y = u + 1, 0 ≤ u ≤ 1,a parametrization of the line segment from (5, 1) to (6, 2), the right boundary of R. Finally the segment I4maps to the curve parameterized by x = 5v, y = v2, 0 ≤ v ≤ 1, the arc of parabola of equation y = x2/25from (0, 0) to (5, 1), the lower boundary of R.When a smooth map maps the boundary of a simple region to the boundary of a simple region. it will mapthe interior either to the interior or to the exterior of the second region. It is easy to see that points insideR′ stay inside R; it suffice to check at a single point.Let us compute the Jacobian of the change:

∂(x, y)

∂(u, v)=

∣∣∣∣ 1 51 2v

∣∣∣∣ = 2v − 5

Incidentally, the fact that the Jacobian is negative (2v − 5 < 0 if 0 ≤ v ≤ 1) indicates that the mapping isorientation reversing. You can notice that if you go through the boundary of R′ with the interior to your left,you traverse the boundary of R with the interior to your right.The rest is easy, I think:∫∫

R

xy dA =

∫ 1

0

∫ 1

0

(u+ 5v)(u+ v2)|2v − 5| du dv =

∫ 1

0

∫ 1

0

(u+ 5v)(u+ v2)(5− 2v) du dv =43

4.

42. D is the region bounded by the planes

x+ 2y + 3z = 0, x+ 2y + 3z = 1, 4x− y = 0, 4x− y = 2, z = −3, z = 2.

Use a change of variables to compute ∫∫∫D

(5x+ y) dV.

18

An obvious choice is

u = x+ 2y + 3z,

v = 4x− y,w = z

This changes the region of integration to 0 ≤ u ≤ 1, 0 ≤ v ≤ 2,−3 ≤ w ≤ 2. We need to rewrite 5x + y in

terms of u, v, w and calculate the Jacobian determinant∂(x, y, z)

∂(u, v, w). For both it may seem necessary to express

x, y, z in terms of u and v. In general, this may be necessary, but there are shortcuts. For example we see

easily that u+ v = 5x+ y + 3z so 5x+ y = u+ v − 3w. To compute∂(x, y, z)

∂(u, v, w)we can use the following rule

∂(x, y, z)

∂(u, v, w)=∂(u, v, w)

∂(x, y, z),

but then replace in the last expression every appearance of x, y, z by its expression in terms of u, v, w. If weproceed this way we have

∂(u, v, w)

∂(x, y, z)=

∣∣∣∣∣∣1 2 34 −1 00 0 1

∣∣∣∣∣∣ = −9

Since no x, y, z appear, we now can say that∂(x, y, z)

∂(u, v, w)= −1/9. We thus have

∫∫∫D

(5x+ y) dV =

∫ 1

0

∫ 2

0

∫ 2

−3(u+ v − 3w)

∣∣∣∣−1

9

∣∣∣∣ du dv du =10

3.

A more systematic approach has one solving x, y, z in terms of u, v, w. This works out to

x =1

9u+

2

9v − 1

3w,

y =4

9u− 1

9v − 4

3w,

z = w.

Thus

∂(x, y, z)

∂(u, v, w)=

∣∣∣∣∣∣∣∣∣∣

19

29 − 1

3

49 − 1

9 − 43

0 0 1

∣∣∣∣∣∣∣∣∣∣= −1

9

(as before) and

5x+ y = 5

(1

9u+

2

9v − 1

3w

)+

4

9u− 1

9v − 4

3w = u+ v − 3w

and the rest is as before.

43. The picture shows a number of level curves of the function z = x2 +y or, as is equivalent, equipotential curvesof the potential V (x, y) = x2 + y

19

(a) Show that at every point (x, y) the equipotential curve through that point has a tangent perpendicularto the gradient.

(b) Illustrate this by selecting several points (at least 3, no more than 5) and drawing at those points thegradient and the tangent to the curve.

(a) The level curves have equations x2 + y = C. By implicit differentiation y′ = −2x. So at every point(x, y) the slope of the tangent line of the level curve through that point is −2x. We need a vector withthat slope; the easiest one is probably 〈1,−2x〉. At every point (x, y), the vector 〈1,−2x〉 is tangent tothe level curve through that point. On the other hand the gradient of V at (x, y) is ∇V (x, y) = 〈2x, 1〉.Since

〈2x, 1〉 · 〈1,−2x〉 = 2x− 2x = 0,

we see that at every point (x, y) the equipotential curve through that point has a tangent perpendicularto the gradient.

(b) In the picture below 5 points were marked; the tangent vectors are in red, the gradients in blue.

20

44. A particle of mass m is moving under the action of a conservative force field F = −∇V . Let r(t) =x(t)i+y(t)j+ z(t)k be the position of the particle at time t and v(t) = |r′(t)| its speed. Show that the energyE(t) = 1

2mv(t)2 + V (x(t), y(t), z(t)) remains constant.

Notice thatv(t)2 = r′(t) · r′(t)

so that,d

dt

(v(t)2

)= r′′(t) · r′(t) + r′(t) · r′′(t) = 2r′′(t) · ε′(t)

and, taking also into consideration Newton’s 2nd law: mr′′ = F, and the chain law,

dE

dt(t) =

1

2md

dt

(v(t)2

)+d

dtV (x(t), y(t), z(t))

= mr′′(t) · r′(t) +∂V

∂x(x(t), y(t), z(t))x′(t) +

∂V

∂y(x(t), y(t), z(t))y′(t) +

∂V

∂z(x(t), y(t), z(t))z′(t)

= F(x(t), y(t), z(t)) · r′(t) +∇V (x(t), y(t), z(t)) · r′(t)= (F(x(t), y(t), z(t)) +∇V (x(t), y(t), z(t))) · r′(t) = 0

since F(x(t), y(t), z(t)) + ∇V (x(t), y(t), z(t)) = −∇V (x(t), y(t), z(t)) + ∇V (x(t), y(t), z(t)) = 0. Since thederivative of E is identically 0, we conclude E is constant.

45. Escape Velocity. We compute the escape velocity, or rather the escape speed, from a planet of radius R,surface acceleration of gravity g. In this model, only the planet exists in the whole universe. This model is nottoo bad as long as we don’t get too far away from the planet. Infinity in physics may just be a large number.We assume a rocket is fired from the surface of the planet, in a direction perpendicular to the surface, withan initial speed v0. The question to be answered is: What should v0 be so that the rocket does not returnto the planet? We assume the rocket is so small that its gravitational pull on the planet is negligible. Thegravitational force field created by the planet is then, according to Newton’s law of universal gravitation:

F = −GmM|r|3

r,

where

m = mass of the rocket,

M = mass of the planet,

r = the radius vector with origin at the center of the planet,

G = the gravitational constant.

Here are your instructions:

(a) Because G,M can be hard to compute, while g,R are not, get rid of G,M by using the fact that on theplanet surface |F| = mg; using this show that GM = gR2.

(b) Show that F is conservative and find V such that F = −∇V . V is only determined up to a constant; aconvenient constant is one such that limx2+y2+z2→∞ V (x, y, z) = 0. Using this constant will result in anegative V .

(c) As seen in Exercise 44, the energy

E(t) = V (r(t)) +1

2m|r′(t)|2

is constant. As the speed of the rocket decreases, V increases accordingly (becoming less negative).Once the speed of the rocket is 0, the rocket will start returning to the planet. The idea now is to figureout what v0 should be so that this only happens at infinity, where V is equal to 0. Equating the energywith the limit for |r| → ∞ and the value when |r| = R should produce the escape speed.

21

(d) For the earth g ≈ 9.8m.s2 and R ≈ 6, 400, 000 m. Calculate the escape speed from earth.

Most of everything is immediate. Part of the exercise (seeing the gravitational field is conservative andcalculating the potential was done in class).

(a) On the surface of the planet we have r = R so that the intensity of the force is given both by |F| = GmM

R2

and by mg. Equating one gets MG = gR2

(b) Writing F = 〈F1, F2, F3〉, r = 〈x, y, z〉, we have

F1 = − mgR2x

(x2 + y2 + z2)3/2, F2 = − mgR2y

(x2 + y2 + z2)3/2, F3 = − mgR2z

(x2 + y2 + z2)3/2

and

∂F1

∂y=

3mgR2xy

(x2 + y2 + z2)5/2=∂F2

∂x,

∂F1

∂z=

3mgR2xz

(x2 + y2 + z2)5/2=∂F3

∂x,

∂F2

∂y=

3mgR2yz

(x2 + y2 + z2)5/2=∂F3

∂y.

The field is conservative. We want ∇V = −F so we begin with

V = −∫F1 dx = mgR2

∫x(x2 + y2 + z2)−3/2 dx = −mgR2(x2 + y2 + z2)−1/2 + h(y, z).

Proceeding as usual one gets as next step that ∂h∂y = 0, so h(z) = g(z). Finally one also gets g′(z) = 0.

So we can take V (x, y, z) = −mgR2(x2 + y2 + z2)−1/2; this is in fact the determination that satisfieslim|r|→∞ V = 0.

(c) For the rocket we will have

E(t) = V (x(t), y(t), z(t)) +1

2m|v(t)|2

is constant. At time t = 0 (launching time) we have |v(0)| = v0 while the potential energy is −mgR2/R =−mgR. The total energy, which will remain constant during the total life of the rocket, is

E = −mgR+1

2v20 . We want the kinetic energy given to the rocket to be sufficient to take it to infinity,

so it shouldn’t be exhausted before then. At infinity the potential energy is maximum (equal to 0, whichis maximum since it is otherwise negative) so the total energy, if the rocket’s velocity is zero at infinity(it just got there!) will also be zero, so E = 0. Constancy of energy now implies −mgR + 1

2mv20 = 0.

Thus v0 =√

2gR.

(d) Taking the given values of g,R one gets v0 ≈ 11, 200m/s, or approximately 11 km/s.

46. Evaluate the following line integrals over the given curves C

(a) (10 points)

∫C

xeyz ds where C is the curve parameterized by r(t) = 〈t, 2t,−4t〉, 0 ≤ t ≤ 1.∫C

xeyz ds =

∫ 1

0

te−8t2√

1 + 4 + 16 =√

21

∫ 1

0

te−8t2

dt =

√21

16(1− e−8).

(b) (10 points)

∫C

x ds, where C is given by r(t) =⟨t, 4t, t2

⟩, t = 0, 2. It may help to know that∫

x√a+ bx2 dx =

1

3b(a+ bx2)3/2 + C.∫

C

x ds =

∫ 2

0

t√

1 + 16 + 4t2 dt =

∫ 2

0

t√

17 + 4t2 dt =1

12(333/2 − 173/2).

22

(c)

∫C

(x− y + 2z) ds, where C is a circle of radius 3 in the plane x = 1, centered at (1, 0, 0).

∫C

(x− y + 2z) ds =

∫ 2π

0

(1− 3 cos t+ 6 sin t)√

02 + (−3 sin t)2 + (3 cos t)2 dt

= 3

∫ 2π

0

(1− 3 cos t+ 6 sin t) dt = 6π.

(d)

∫C

xy

zds where C is the line segment from (1, 4, 1) to (3, 6, 3), followed by the arc of circle centered at

(0, 0, 3) in the plane z = 3 from (3, 6, 3) to (6, 3, 3); clockwise.

Finding a parametrization for the arc of circle is slightly complicated. It is an arc of a circle of radius√62 + 32 =

√45. If we were going counterclockwise we would go from θ = arctan 1/2 to θ = arctan 2.

But the orientation plays no role in this type of line integral, so we may use the counterclockwiseparameterization.∫

C

xy

zds =

∫ 1

0

(1 + 2t)(4 + 2t)

1 + 2t

√22 + 22 + 22 dt+

∫ arctan 2

arctan 1/2

(√

45 cos t)√

45 sin t

3

√45 dt

= 8

∫ 1

0

(4 + 2t) dt+ 45√

5

∫ arctan 2

arctan 1/2

cos t sin t dt = 40 +1

245√

5 sin2 t

∣∣∣∣arctan 2

arctan 1/2

=125

2.

(e) Evaluate

∫C

z ds where C is the curve r(t) = 〈cos t− sin t, cos t+ sin t,−2 cos t〉, 0 ≤ t ≤ 2π.

∫C

z ds =

∫ 2π

0

(−2 cos t)√

(− sin t− cos t)2 + (− sin t+ cos t)2 + (2 sin t)2 dt

= −2

∫ 2π

0

cos(t)√

2 + 4 sin2 t dt = 0.

47. Find the work done by the force field F = 〈−y, x, z〉 in moving a particle along the helix

x = 2 cos t, y = 2 sin t, z =t

2π

for 0 ≤ t ≤ 2π.

W =

∫helix

〈−y, x, z〉 · dr =

∫ 2π

0

〈−2 cos t, 2 sin t,t

2π〉 · 〈−2 sin t, 2 cos t,

1

2π〉 dt

=

∫ 2π

0

(4 +

t

4π2

)dt = 8π +

1

2.

48. (a) For what values of a, b, c and d is the field F = 〈ax+ by, cx+ dy〉 conservative?

∂(ax+ by)

∂y= b,

∂(cx+ dy)

∂x= c

so the answer is b = c.

23

(b) For what values of a, b, and c is the field F = 〈ax2 − by2, cxy〉 conservative?

∂(ax2 − by2)

∂y= −2by,

∂(cxy)

∂x= cy

so the answer is c = −2b.

49. Consider the vector fieldF(x, y, z) = 〈yz + 1, xz + 2y, xy + 3z2〉.

(a) Show that F is conservative and determine a function f so that F = ∇f .

(b) Let C be the curve given in vector notation by

r(t) = t2i + t3j +2

1 + t2k, 0 ≤ t ≤ 1.

Evaluate ∫C

F · dr.

Realizing that these questions are related may save some time.

(a) We have

∂

∂y(yz + 1) = z

∂

∂x(xz + 2y),

∂

∂z(yz + 1) = y =

∂

∂x(xy + 3z2),

∂

∂z(xz + 2y) = x =

∂

∂y(xy + 3z2).

Since the functions are nicely differentiable everywhere, the fact that the cross derivatives are equal isnot only necessary, but also sufficient for the field to be conservative.We proceed to find f so ∇f = F. To satisfy ∂f

∂x = yz + 1 we integrate with respect to x.

f(x, y, z) =

∫(yz + 1) dx = xyz + x+ h(y, z).

To get ∂f∂y = xz + 2y we need

∂

∂y(xyz + x+ h(y, z)) = xz + 2y, hence xz +

∂h

∂y(y, z) = xz + 2y, thus

∂h

∂y(x, z) = 2y

so that

h(y, z) =

∫(2y) dy = y2 + g(z).

At this point, f(x, y, z) = xyz+ x+ y2 + g(z); to get the final condition, namely,∂ϕ

∂z(x, y, z) = xy+ 3z2

we need (and it suffices) to have

∂

∂z

(xyz + x+ y2 + g(z)

)= xy + 3z2,

integrating with respect to z we see that we can take g(z) = z3. Thus

f(x, y, z) = xyz + x+ y2 + z3.

(b) The path starts at (0, 0, 2) and ends at (1, 1, 1). Since the vector field is conservative,∫C

F · dr = f(1, 1, 1)− f(0, 0, 2) = 4− 8 = −4.

24

50. (a) Show that the vector field F = 〈2xey + z2ex, x2ey + 2yez, y2ez + 2zex + 1〉 is conservative and find thefunction f such that ∇f = F.

We have

∂

∂y(2xey + z2ex) = 2xey =

∂

∂x(x2ey + 2yez),

∂

∂z(2xey + z2ex) = 2zex =

∂

∂x(y2ez + 2zex + 1),

∂

∂z(x2ey + 2yez) = 2yez =

∂

∂y(y2ez + 2zex + 1).

The field is conservative. Now

f(x, y, z) =

∫(2xey + z2ex) dx = x2ey + z2ex + h(y, z).

The function h has to satisfy

∂

∂y

(x2ey + z2ex + h(y, z)

)= x2ey + 2yez; i.e. x2ey +

∂h

∂y(y, z) = x2e+2yez

so that∂h

∂y(y, z) = 2yez, hence h(y, z) = y2ez + g(z).

Returning with this to the expression for f we have so far f(x, y, z) = x2ey + z2ex + y2ez + g(z). Westill need to satisfy ∂f

∂z (x, y, z) = y2ez + 2zex + 1 or

∂

∂z(x2ey + z2ex + y2ez + g(z)) = y2ez + 2zex + 1; i.e.2zex + y2ez + g′(z) = y2ez + 2zex + 1.

This forces g′(z) = 1; we take g(z) = z and we get

f(x, y, z) = x2ey + z2ex + y2ez + z.

(b) An object is moved along the path parameterized by

x = t , t2, z = t3

from time t = 0 to time t = 2. Find the work done by the force field of part (a).

The particle is at (0, 0, 0) for t = 0 and at (2, 4, 8) for t = 2. Since the field is conservative, the workdone is

f(2, 4, 8)− f(0, 0, 0) = 4e4 + 64e2 + 16e8 + 8.

51. Evaluate

∮C

xy2 dx + xy dy in two ways where C is the positively (counterclockwise) oriented triangle of

vertices (0, 0), (0, 1), and (1, 1).:

(a) Directly, using the definition.

(b) Using Green’s Theorem.

(a) The triangle is the sum of three line segments that I will denote by I, II, and III, where the parame-terizations are

I. x(t) = t, y(t) = t, 0 ≤ t ≤ 1 (Segment from (0, 0) to (1, 1).)

II. x(t) = 1− t, y(t) = 1, 0 ≤ t ≤ 1 (Segment from (1, 1) to (0, 1).)

III. x(t) = 0, y(t) = 1− t, 0 ≤ t ≤ 1 (Segment from (0, 1) to (0, 0).)

25

We now have ∫I

xy2 dx+ xy dy =

∫ 1

0

(t3 + t2) dt =7

12,∫

II

xy2 dx+ xy dy = −∫ 1

0

(1− t) dt = −1

2,∫

III

xy2 dx+ xy dy = 0

Thus∮C

xy2 dx+ xy dy =

∫I

xy2 dx+ xy dy +

∫II

xy2 dx+ xy dy +

∫III

(y2 dx+ xy dy =7

12− 1

2=

1

12.

(b) By Green’s Theorem, since the triangle can be described by {(x, y) : 0 ≤ y ≤ 1, 0 ≤ x ≤ y}, we have∮C

xy2 dx+ xy dy =

∫ 1

0

∫ y

0

(∂xy

∂x− ∂xy2

∂y

)dy dx =

∫ 1

0

∫ y

0

(y − 2xy) dy dx =

∫ 1

0

(y2 − y3) dy =1

12.

52. Let R be the triangle in the (x, y)-plane of vertices (0, 0), (4, 2), and (8, 8). You are supposed to calculate∫∫R

x dA in four different ways, which are parts (a),(b),(c), and (d) of this exercise.

(a) Calculate

∫∫R

x dA if R is the triangle of vertices (0, 0), (4, 2), (8, 8) directly, as an iterated integral (or

a sum of iterated integrals). As a help, the right boundary of the triangle lies on the line y = 32x − 4,

the lower boundary on the line y = x/2.

The region can be described as the union of {(x, y) : 0 ≤ x ≤ 4,1

2x ≤ y ≤ x} and {(x, y) : 4 ≤ x ≤

8, 32x− 4 ≤ y ≤ x}. Thus∫∫R

x dA =

∫ 4

0

∫ x

x/2

x dy dx+

∫ 8

4

∫ x

32x−4

x dy dx =

∫ 4

0

1

2x2 dx+

∫ 8

4

x(4− 1

2x) dx = 32.

(b) Calculate

∫∫R

x dA if R is the triangle of vertices (0, 0), (4, 2), (8, 8) by Green’s Theorem, using the

fact that x =∂

∂x

(1

2x2)

.

By Green’s Theorem ∫∫R

∂

∂x

(1

2x2)dA =

∫C

1

2x2 dy

where C is the positively oriented boundary of the triangle. The boundary C is made up of the segmentsL1, followed by L2, followed by L3 where L1 joins (0, 0) to (4, 2), L2 joins (4, 2) to (8, 8) and L3 joins(8, 8) to (0, 0). the corresponding parameterizations are

L1 : x(t) = 4t, y(t) = 2t, 0 ≤ t ≤ 1,

L2 : x(t) = 4 + 4t, y(t) = 2 + 6t, 0 ≤ t ≤ 1,

L3 : x(t) = 8(1− t), y(t) = 8(1− t), 0 ≤ t ≤ 1.

We have ∫L1

1

2x2 dy =

∫ 1

0

1

2(4t)2(2) dt = 16

∫ 1

0

t2 dt =16

3.∫

L2

1

2x2 dy =

∫ 1

0

1

2(4 + 4t)2(6) dt = 48

∫ 1

0

(1 + t)2 dt = 112.∫L3

1

2x2 dy =

∫ 1

0

1

2[8(1− t)]2(−8) dt = −256

∫ 1

0

(1− t)2 dt = −256

3.

26

Thus ∫∫R

x dA =

∫L1

1

2x2 dy +

∫L2

1

2x2 dy +

∫L3

1

2x2 dy =

16

3+ 112− 256

3= 32.

(c) Calculate

∫∫R

x dA if R is the triangle of vertices (0, 0), (4, 2), (8, 8) by Green’s Theorem, using the

fact that x =∂(xy)

∂y.

By Green’s Theorem ∫∫R

∂(xy)

∂ydA = −

∫C

dx

where C is the positively oriented boundary of the triangle. We use the same parameterizations of theprevious point, getting∫

L1

xy dx =

∫ 1

0

8t2(4) dt = 32

∫ 1

0

t2 dt =32

3.∫

L2

xy dx =

∫ 1

0

(4 + 4t)(2 + 6t)(4) dt = 32

∫ 1

0

(3t2 + 4t+ 1) dt = 128.∫L3

xy dx =

∫ 1

0

[8(1− t)]2(−8) dt = −512

∫ 1

0

(1− t)2 dt = −512

3.

Thus ∫∫R

x dA = −∫L1

xy dy −∫L2

xy dy −∫L3

xy dy = −32

3− 128 +

512

3= 32.

(d) Calculate

∫∫R

x dA if R is the triangle of vertices (0, 0), (4, 2), (8, 8) by changing variables by

u = 3x− 2y

v = −x+ 2y

In this, as in all change of variables exercises, you must clearly indicate what the transformation does;into what it changes the region of integration, and give at least a good reason why the new region iswhat you say it is.

We see that the transformation keeps (0, 0) and (8, 8) fixed and carries (4, 2) into (8, 0). Being linear, itwill carry lines to lines, so the sides go to the sides of the triangle R′, and so does the interior. Invertingthe transformation, we get

x =1

2u+

1

2v

y =1

4u+

3

4v.

The Jacobian of the transformation is

∂(x, y)

∂(u, v)=

∣∣∣∣∣∣1/2 1/2

1/4 3/4

∣∣∣∣∣∣ =1

4.

Thus ∫R

x dA =

∫ 8

0

∫ u

0

(1

2(u+ v)

1

4dv du =

1

8

∫ 8

0

∫ u

0

(u+ v) dv du =3

16

∫ 8

0

u2 du = 32.

53. Use Green’s Theorem to evaluate the following line integrals.

27

(a)

∮C

(−3y + x3/2) dx+ (x− y2/3) dy where C is the boundary of the halfdisc {(x, y) : x2 + y2 ≤ 2, y ≥ 0}

with positive (counterclockwise) orientation.

Let R = {(x, y) : x2 + y2 ≤ 2, y ≥ 0}. By Green’s Theorem,∮C

(−3y + x3/2) dx+ (x− y2/3) dy =

∫∫R

(1− (−3)) dA = 4

∫∫R

dA4π.

(b)

∮C

(x3− y2) dx− (3x2y+xy) dy, where C is the square of vertices (±1,±1), with positive (counterclock-

wise) orientation.

By Green’s Theorem∮C

(x3 − y2) dx− (3x2y + xy) dy =

∫ 1

−1

∫ 1

−1(−6xy + y) dA = 0.

54. (10 points) Use Green’s Theorem to find the area of the region bounded by the counterclockwise circulararc from (1, 0) to (0,−1) of radius 1, centered at (0, 0), and the clockwise semicircle of radius 1/

√2 centered

at (1/2,−1/2). See picture.

Check your answer computing the area using elementary school geometry.

The large circular arc can be parameterized by x = cos t, y = sin t, 0 ≤ t ≤ 2π while the counterclockwise

version of the lower arc can be parameterized by x =1

2+

1√2

cos t,−1

2+

1√2

sin t, π/4 ≤ t ≤ 5π/4. Then,

using the version A =

∮x dy,

A =

∫ 3π/2

0

(cos t)(cos t) dt−∫ 5π/4

π/4

(1

2+

1√2

cos t)(1√2

cos t) dt

=

∫ 3π/2

0

cos2 t dt− 1

2√

2

∫ 5π/4

π/4

cos t dt− 1

2

∫ 5π/4

π/4

cos2 t dt =3π

2+

1

2− π

4=π + 1

2.

An elementary school approach could be as follows. The region can be obtained by first cutting of from the

unit circle a circular segment spanning an inner angle of 90◦:What remains is three quarters of the circle and a right triangle of legs of length 1. By very elementary

geometry, the area is3

4π+

1

2. To get the figure of the problem we have to cut out next a semicircle of radius

1/√

2, which has area1

2

(1√2

)2

π =π

4. Thus the area of the region is

3π

4+

1

2− π

4=π + 1

2.

55. Find the area of the shaded figure in two ways: (a) Using Green’s Theorem. (b) Not using Green’s Theorem,for example using double integrals and or elementary geometry

28

The boundary of the figure consists of 4 one quarter circle arcs as shown in the picture below

Red: Quarter circle of radius 2 centered at (0, 2)

Blue: Quarter circle of radius 1 centered at (3, 2)

Green: Quarter circle of radius 2 centered at (3, 1)

Yellow: Quarter circle of radius 1 centered at (0, 1)

By Green’s Theorem Denote the arcs by C1 (red), C2 (blue), C3 (green), and C4 yellow. It will probablybe easiest to parameterize them as arcs of a circle gone through counterclockwise, then subtract the integralif this direction is not the one we want. If this is unclear, I hope to clear it up in the execution.

C1. We parameterize by

x = 2 cos t, y = 2 + 2 sin t, −π2≤ t ≤ 0.

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Thus ∫C1

x dy =

∫ 0

−π/2(2 cos t)(2 cos t) dt = 4

∫ π/2

0

cos2 t dt = π.

C2. The easy thing is to parameterize by

x = 3 + cos t, y = 2 + sin t,π

2≤ t ≤ π.

The problem is that then we are going through the arc in the wrong direction. Rather than figure outhow to parameterize so as to get the right direction of traversal, we simply subtract the integral ratherthan add it. ∫

C2

x dy =

∫ π

π/2

(3 + cos t)(cos t) dt =

∫ π

π/2

(3 cos t+ cos2 t) dt = −3 +π

4.

C3. We parameterize by

x = 3 + 2 cos t, y = 1 + 2 sin t,π

2≤ t ≤ π.

Thus ∫C1

x dy =

∫ 0

−π/2(3 + 2 cos t)(2 cos t) dt =

∫ π/2

0

(6 cos t+ 4 cos2 t dt = −6 + π.

C4. We parameterize by

x = cos t, y = 1 + sin t, −π2≤ t ≤ 0,

but this is again the wrong direction. So∫C4

will have to be subtracted.∫C4

x dy =

∫ 0

−π/2(cos t)(cos t) dt =

∫ π/2

0

cos2 t dt =π

4.

We thus have

A =

∫C

x dy =

∫C1

x dy −∫C2

x dy +

∫C3

x dy −∫C4

x dy = π − (−3 +π

4)− 6 + π − π

4=

3π

2− 3.

Not by Green’s Theorem. We see that the region is what remains of the 3× 3 square after removal of thebrown an yellow regions.

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These regions are symmetric. The brown region is divided into 3 regions I labeled S,R,D. The area of theregion we are considering is equal to

A = 9− 2(S +R+D),

where I am using the same letters to denote the regions that make up the brown region as their areas. Now,R is a 1 × 2 rectangle, so R = 2. D is a quarter circle of radius 1, so D = π/4. The region S extends fromx = 0 to x = 2 and it is not too hard to figure out the equation of the curve bounding it at the top, namely.It is part of the circle x2 + (y − 2)2 = 4, so we see it is y = 2−

√4− x2. Thus

S =

∫ 2

0

∫ 2−√4−x2

0

dy dx =

∫ 2

0

(2−√

4− x2) dx = 4−∫ 2

0

√4− x2 dx.

We can look this last integral up in tables. Or we can substitute x = sin t. Or we can realize it is the area ofa quarter circle of radius 2, thus equal to π. Thus S = 4 − π. Finally, the area A we are looking for worksout to

A = 9− 2(4− π + 2 +π

4) =

3π

2− 3

as before.

56. The cycloid is a curve apparently first mentioned by Galileo. It became famous thanks to the Bernoullibrothers discovering that it was both the brachistochrone and the tautochrone (look it up!). Possibly thearches of the Ponte Vecchio in Florence Italy have a cycloidal shape.

Ponte Vecchio over the Arno

A parametrization of a single arc of the cycloid is given by

x = a(t− sin t), y = a(1− cos t), 0 ≤ t ≤ 2π,

where a > 0 is a constant. Use Green’s Theorem to find the area of the region bounded by the arc of a cycloidand the x axis.

(The first person to compute this area, as well as finding the length of an arc of the cycloid, was Gilles Personnede Roberval (1602-1675); they were also computed only a little bit later, independently, by EvangelistaTorricelli (1608-1647), more famous for having invented the barometer.)

We’ll use the formula A =1

2

∮(x dy− y dx) where C is the boundary of the region. Notice that this integral

is zero on the bottom part of the boundary; the segment from (0, 0) to (2π, 0) because dy = 0, x = 0 on thatsegment. However, one thing that we have to careful with is the orientation. I’ll do the calculationsand we’ll see why.

A =1

2

∫ 2π

0

(a(t− sin t)(a sin t)− a(1− cos t)a(1− cos t)) dt

=a2

2

∫ 2π

0

(t sin t− sin2 t− 1 + 2 cos t− cos2 t

)dt

= −3πa2.

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But this can’t be, area cannot be negative!!! Something must be wrong, and the wrong isn’t justrighted by merely changing the sign for no good reason. A bit of thinking shows where we went wrong. Theparametrization has us go through the arc of the cycloid starting at (0, 0) and ending at (2πa, 0). Doing this,the region bounded by this arc is to our right. That means we are going through the boundary in a negativeorientation. That means we get the negative of what we would get by going through in a positive orientation.

Now we can honestly change the sign and say A = 3πa2.

57. Find the area of the region bounded by the curves y = sinx, x = π + sin y, y = π and x = 0.

We’ll use the formula

A =

∮C

x dy,

where C is the boundary of the region. The boundary is made up of 4 parts I’ll denote by A,B,C,D, aspictured.

The very unskillfully drawn arrowheads indicate the positive orientation. The A section can be parameterizedby x = t, y = sin t, 0 ≤ t ≤ π, to give ∫

A

x dy =

∫ π

0

t sin t dt = −2.

Or, we could have just said we use x as a parameter and written it directly as

∫ π

0

x d(sinx) =

∫ π

0

x cosx dx.

For B we can use y as a parameter; the range is again 0→ π, so that∫B

x dy =

∫ π

0

(π + sin y) dy = π2 + 2.

And we are done because on C we have dy = 0, on D we have x = 0, so that the integrals over C and D are

zero. Adding up we see the area is π2.