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Math 20C Homework 2 Partial Solutions
Problem 1 (12.4.14). Calculate (j− k)× (j + k).
Solution. The basic properties of the cross product are found in Theorem 2 of Section 12.4.From these properties, we deduce
(j− k)× (j + k) = j× (j + k)− k× (j + k)
= j× j︸︷︷︸=0
+j× k− k× j + k× k︸ ︷︷ ︸=0
= j× k− k× j
= j× k + j× k.
Next we recall the following adaptation of Figure 7 (also Figure 14) from Section 12.4:
i
jk
=⇒
i× j = k
j× k = i
k× i = j
Therefore we know that j× k = i, so from the calculation above we conclude that
(j− k)× (j + k) = i + i = 2i = 〈2, 0, 0〉.
Problem 2 (12.4.24). Find v ×w, where v and w are vectors of length 3 in the xz-plane,oriented as in the following figure, and θ = π
6.
y
x
z
v
w
θ
1
Solution. From Theorem 1 of Section 12.4, we know that
‖v ×w‖ = ‖v‖‖w‖ sin θ = 3 · 3 · sin π6
=9
2.
Moreover, by the right-hand rule (cf. Theorem 1 of Section 12.4), we know that v×w pointsin the negative y direction:
y
x
z
v ×w
v
w
θ
Thus v ×w = 〈0,−a, 0〉 for some positive real number a. From knowing that the length ofv ×w is 9/2, we conclude that a = 9/2, so that
v ×w =
⟨0,−9
2, 0
⟩.
Problem 3 (12.4.42). Find the area of the parallelogram determined by the vectors 〈a, 0, 0〉and 〈0, b, c〉.
Solution. Let u = 〈a, 0, 0〉 and v = 〈0, b, c〉. By definition, the area of the parallelogramdetermined by u and v is
‖u× v‖.By Theorem 1(ii) of Section 12.4, we have
‖u× v‖ = ‖u‖‖v‖ sin θ, (3.1)
where θ is the angle between u and v (and 0 ≤ θ ≤ π). Next, note that
u · v = a · 0 + 0 · b+ 0 · c = 0,
so u and v are perpendicular by the properties of the dot product (see the bottom box onpage 680 in Section 12.3). That is, θ = π
2. Now from (3.1) we see that
‖u× v‖ = ‖u‖‖v‖ sin θ = a√b2 + c2 sin
π
2= a√b2 + c2,
and this is the area of the parallelogram determined by u and v. Note that we could havejust computed the cross product u×v directly and then taken its magnitude, but that wouldhave been less interesting.
2
Problem 4 (12.5.22). Find the equation of the plane passing through (4, 1, 9) and parallelto x+ y + z = 3.
Solution. Let’s call the plane we want to find P . The plane given by the equation
x+ y + z = 3
has normal vector n = 〈1, 1, 1〉 (obtained from just reading off the coefficients of x, y, andz in the equation above). Since P is parallel to this plane, n is also a normal vector forP . Thus, we know a normal vector of P (n) and a point on P (the point (4, 1, 9)). UsingTheorem 1 of Section 12.5, we can write the equation of P in the following way:
n · 〈x, y, z〉 = n · 〈4, 1, 9〉.
This is the vector form of the equation of P . To get the scalar form, we simply expand theequation above to get
x+ y + z = 14.
Problem 5 (12.5.24). Find the equation of the plane passing through (3, 5,−9) and parallelto the xz-plane.
Solution. Simply repeat the steps of the previous problem, noting that the xz-plane is givenby the equation y = 0.
Problem 6 (12.5.26). Find the equation of the plane containing the lines r1(t) = 〈t, 2t, 2t〉and r2(t) = 〈3t, t, 8t〉.
Solution. To find the equation of this plane, it suffices to find a normal vector to the planeand a point on the plane. For a normal vector n, we can take the cross product of r1(1) andr2(1) since these vectors lie on the plane (this is not the only option–any nonzero choice oft would work) to get
n = r1(1)× r2(1)
= 〈1, 2, 3〉 × 〈3, 1, 8〉
=
∣∣∣∣2 31 8
∣∣∣∣ i− ∣∣∣∣1 33 8
∣∣∣∣ j +
∣∣∣∣1 23 1
∣∣∣∣k= 〈13, 1,−5〉.
Next, a point on the plane can be found by plugging any value of t into either r1 or r2. Inparticular, we can take t = 0 in r1 to see that (0, 0, 0) is on the plane. Thus by Theorem 1of Section 12.5, an equation of the plane we want in vector form is
〈13, 1,−5〉 · 〈x, y, z〉 = 〈13, 1,−5〉 · 〈0, 0, 0〉.
Expanding this into the scalar form, we get
13x+ y − 5z = 0.
3
Problem 7 (12.5.21). Parametrize the intersection of the surfaces
y2 − z2 = x− 2, y2 + z2 = 9
using t = y as the parameter.
Solution. Right away, set y = t. From the second equation in the problem, we get
z2 = 9− t2 =⇒ z = ±√
9− t2.
We see now that we must have −3 ≤ t ≤ 3. Next, we substitute the formula for z in termsof t into the first equation to get
x = y2 − z2 + 2 = t2 − (9− t2) + 2 = 2t2 − 7.
Thus we have parametrized the intersection of the two surfaces by the following two functions:
r1(t) = 〈2t2 − 7, t,√
9− t2〉, r2(t) = 〈2t2 − 7, t,−√
9− t2〉, −3 ≤ t ≤ 3.
Problem 8 (12.5.22). Find a parametrization of the curve in the previous problem usingtrigonometric functions.
Solution. From the equationy2 + z2 = 9 = 32,
it makes sense to set
y = 3 cos t, z = 3 sin t, 0 ≤ t < 2π.
Now all we have to do is express x in terms of t. From the equation
y2 − z2 = x− 2,
we see thatx = y2 − z2 + 2 = 9 cos2 t− 9 sin2 t+ 2.
There might be a trig identity that simplifies this, but it’s not important. Now we finish theproblem by giving a trigonometric parametrization of the intersection of our two surfaces:
r(t) = 〈9 cos2 t− 9 sin2 t+ 2, 3 cos t, 3 sin t〉, 0 ≤ t < 2π.
Problem 9 (12.5.34). Find a parametrization of the horizontal circle of radius 1 with center(2,−1, 4).
Solution. The circle being horizontal means that the z-coordinate of our parametrization isconstant. A horizontal circle centered at the origin with radius 1 has the parametrization
r1(t) = 〈cos t, sin t, 0〉, 0 ≤ t < 2π.
The circle we want is this circle translated so that the center become the point (2,−1, 4).To do this, we simply add the vector 〈2,−1, 4〉 to the function r1 to get the parametrizationof the circle we want:
r(t) = 〈cos t+ 2, sin t− 1, 4〉, 0 ≤ t < 2π.
4
Problem 10 (11.1.28). Find a parametric equation for the curve(x5
)2+( y
12
)2= 1.
Solution. Any time that we see something of the form
A2 +B2 = C2,
our first instinct should be to set
A = C cos t, B = C sin t, 0 ≤ t < 2π.
For our problem we have A = x5, B = y
12, and C = 1, so we set
x
5= cos t,
y
12= sin t, 0 ≤ t < 2π.
Simplifying, we getx = 5 cos t, y = 12 sin t, 0 ≤ t < 2π.
These are parametric equations for our curve.
5
Math 20C HW3 Partial Solutions
Zev Finnley
April 17, 2014
13.2.28: Let ~v(s) =< s2, 2s, 9s−2 >. Evaluate dds~v(g(s)) at s = 4, assuming
g(4) = 3 and g′(4) = −9solution:~v′(s) =< 2s, 2,−18s−3 >dds~v(g(s)) = v′(g(s))g′(s) (By the Chain Rule)So at s = 4 this becomes:dds~v(g(4)) = v′(g(4))g′(4) = v′(3)(−9)
=< 2(3), 2,−18(3)−3 > (−9) = −9 < 6, 2,−18/27 >=< −54,−18, 6 >
13.2.56: dr/dt =< 2t−1/2, 6, 8t >, r(1) =< 4, 9, 2 > Find the location and thevelicoty of the particle at t=4.solution:velocity at t=4 is the derivative of the position vector evaluated at t=4:
velocity(4) = r′(4) =< 2(4)−1/2, 6, 8(4) >=< 1, 6, 32 >
To find the position vector we need to take the integral of r′(t):Integrating element wise gives: (Note: c is the constant of integration)
r(t) =< 4t1/2, 6t, 4t2 > +c
Evaluating for our initial condition:
r(1) =< 4(1)1/2, 6(1), 4(1)2 > +c =< 4, 9, 2 >
c =< 0, 3,−2 >
Plugging in our constant of integration and evaluating at t = 4 yields:
r(4) =< 4(4)1/2, 6(4), 4(4)2 > + < 0, 3,−2 >=< 8, 27, 62 >
11.2.6: Evaluate the length of (t3 + 1, t2 − 3), 0 ≤ t ≤ 1solution:
s =
∫ b
a
√(x′)2 + (y′)2dt
s =
∫ 1
0
√(3t2)2 + (2t′)2dt
1
s =
∫ 1
0
√(9t4 + 2t2dt
s =
∫ 1
0
t√
(9t2 + 2dt
Using u substitution with u = 9t2 + 2 and dt = du/18 and bounds 4 ≤ u ≤ 13
s =1
18
∫ 13
4
√udu
s =1
18(2
3u3/2)|134
s =1
27(133/2 − 43/2) ≈ 1.44
11.2.19: Find the minimum speed of c(t) = (t3 − 4t, t2 + 1) for t ≥ 0.solution:First we need to find the speed of the particle:
ds
dt=
√(3t2 − 4)2 + (2t)2 =
√9t4 − 20t2 + 16
Since the square root is a strictly increasing function we only need to checkthe critical points of the function f(t) = 9t4 − 20t2 + 16.
f ′(t) = 36t3 − 40t = 4t(9t2 − 10)
Which has critical points at t = 0 and t =√
10/9. Evaluating at these points:
ds
dt(0) =
√16 = 4
ds
dt(√
10/9) =√
44/9 ≈ 2.21
Therefore the minimum speed is√
44/9
13.3.23: Find the path that traces a curve in the y = 10 plane with radius 4,and center (2, 10,−3) with constant speed equal to 8.solution:First we need to find start with a basic parametrized curve.
r(t) =< 2, 10, 3 > +4 < cos t, 0, sin t >
So we have our center point (2, 10,−3) and our circle of radius 4. Note thatthe y component of our parametrized curve is constant for all t. (i.e. in theplane y = 10)Now we need to parametrize the curve by substituting t = g(s) so thatr(s) = r(g(s)) has the constant speed, ‖r′(s)‖ = 8. So we need to find r′(s)using the chain rule.
r′(s) =d
dsr(g(s)) = r′(g(s))g′(s) (1)
2
Since r′(t) =< −4 sin t, 0, 4 cos t > we can sub in g(s) for t and get.
r′(g(s)) =< −4 sin g(s), 0, 4 cos g(s) >
Plugging into equation (1) gives:
r′(s) =< −4 sin g(s), 0, 4 cos g(s) > g′(s) = 4g′(s) < − sin g(s), 0, cos g(s) >
‖r′(s)‖ = 4|g′(s)|√
(cos g(s))2 + (sin g(s))2 = 4|g′(s)|
So |g′(s)| must equal 2. Integration give g(s) = 2s. Putting it all together:
r(s) = r(g(s)) =< 2 + 4 cos (2s), 10,−3 + 4 sin (2s) >
3
Math 20C Homework 10 Solutions
Note. These solutions are not guaranteed to be free from typos and errors. If you seesomething wrong, report it on Piazza!
Problem 1 (15.4.1). Sketch the region D indicated and integrate f(x, y) over D using polarcoordinates.
f(x, y) =√x2 + y2; x2 + y2 ≤ 2
Solution. The region D indicated is the disk of radius√
2 centered at the origin:
x
y
D
√2
In polar coordinates D is given by
0 ≤ r ≤√
2, 0 ≤ θ ≤ 2π.
Moreover, the integrand f(x, y) =√x2 + y2 becomes
f(r, θ) =
√(r cos θ)2 + (r sin θ)2 =
√r2(cos2 θ + sin2 θ
)=√r2 = r
after changing to polar coordinates. Therefore our integral is∫∫Df(x, y) dA =
∫ 2π
0
∫ √20
f(r, θ) r dr dθ =
∫ 2π
0
∫ √20
r2 dr dθ
=
(∫ 2π
0
1 dθ
)(∫ √20
r2 dr
)= 2π
(r3
3
∣∣∣∣√2
0
)
=4√
2π
3.
1
Problem 2 (15.4.3). Sketch the region D indicated and integrate f(x, y) over D using polarcoordinates.
f(x, y) = xy; x ≥ 0, y ≥ 0, x2 + y2 ≤ 4
Solution. The region D indicated is the portion of the disk of radius 2 centered at the origin(x2 + y2 ≤ 4) restricted to the first quadrant (x ≥ 0 and y ≥ 0):
x
y
D
2
In polar coordinates D is given by
0 ≤ r ≤ 2, 0 ≤ θ ≤ π
2.
Moreover, the integrand f(x, y) = xy becomes
f(r, θ) = (r cos θ)(r sin θ) = r2 cos θ sin θ =r2 sin(2θ)
2
after changing to polar coordinates and using the trig identity
sin(2θ) = 2 cos θ sin θ.
Therefore our integral is∫∫Df(x, y) dA =
∫ π/2
0
∫ 2
0
f(r, θ) r dr dθ =
∫ π/2
0
∫ 2
0
r3 sin(2θ)
2dr dθ
=1
2
(∫ π/2
0
sin(2θ) dθ
)(∫ 2
0
r3 dr
)
=1
2
(−cos(2θ)
2
∣∣∣∣π/20
)(r4
4
∣∣∣∣20
)
=1
2
(−cosπ
2+
cos 0
2
)24
4
= 2.
2
Problem 3 (15.4.7). Sketch the region of integration and evaluate by changing to polarcoordinates. ∫ 2
−2
∫ √4−x20
(x2 + y2
)dy dx
Solution. From the outer integral, we can tell that the x values of the region of integrationrange from −2 to 2. From the inner integral, we can tell that for a given value of x (with−2 ≤ x ≤ 2), the y-values range between 0 and
√4− x2. As x ranges from −2 to 2,
y =√
4− x2 traces out the upper half of the circle of radius of 2 centered at the origin.Therefore the region of integration looks like the following:
x
y
2
Thus we see that the region is described in polar coordinates as
0 ≤ r ≤ 2, 0 ≤ θ ≤ π,
and so our integral is
∫ 2
−2
∫ √4−x20
r2︷ ︸︸ ︷(x2 + y2
)dy dx =
∫ π
0
∫ 2
0
r2 r dr dθ
=
(∫ π
0
1 dθ
)(∫ 2
0
r3 dr
)= π
r4
4
∣∣∣∣20
= 4π.
Problem 4 (15.4.9). Sketch the region of integration and evaluate by changing to polarcoordinates. ∫ 1/2
0
∫ √1−x2√3x
x dy dx
Solution. From the outer integral, we can tell that the x values of the region of integrationrange from 0 to 1/2. From the inner integral, we can tell that for a given value of x (with0 ≤ x ≤ 1/2), the y-values range between y =
√3x (the line with slope
√3 and y-intercept
0) and y =√
1− x2 (the upper half of the circle of radius 1 centered at the origin). Thereforethe region of integration looks like the following:
3
x
y
1
1/2
y =√
3x
=⇒
x
y
1
1/2
√3/2
=⇒
x
y
1
1/2
√3/2
π/3
Thus we see that the region is described in polar coordinates as
0 ≤ r ≤ 1, π/3 ≤ θ ≤ π/2,
and so our integral is∫ 1/2
0
∫ √1−x2√3x
x dy dx =
∫ π/2
π/3
∫ 1
0
r cos θ r dr dθ
=
(∫ π/2
π/3
cos θ dθ
)(∫ 1
0
r2 dr
)=
(sin θ
∣∣∣π/2π/3
)(r3
3
∣∣∣∣10
)
=
(1−√
3
2
)1
3.
Problem 5 (15.4.17). Calculate the integral over the given region by changing to polarcoordinates.
f(x, y) = |xy|; x2 + y2 ≤ 1.
Solution. The region D over which we’re integrating is the unit disk centered at the origin.In polar coordinates, D is given by
0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π.
Moreover, our integrand f(x, y) = |xy| is
f(r, θ) = |(r cos θ) (r sin θ)| = r2 |cos θ sin θ| = r2
2|sin(2θ)|
after changing to polar coordinates and using the trig identity
sin(2θ) = 2 cos θ sin θ.
We can determine the sign of sin 2θ = 2 cos θ sin θ, and hence the value of |sin(2θ)|, in eachof the four quadrants in the plane:
4
x
y
0 ≤ θ ≤ π/2
cos θ ≥ 0
sin θ ≥ 0
cos θ sin θ ≥ 0
π/2 ≤ θ ≤ π
cos θ ≤ 0
sin θ ≥ 0
cos θ sin θ ≤ 0
3π/2 ≤ θ ≤ 2π
cos θ ≥ 0
sin θ ≤ 0
cos θ sin θ ≤ 0
π ≤ θ ≤ 3π/2
cos θ ≤ 0
sin θ ≤ 0
cos θ sin θ ≥ 0
=⇒ |sin(2θ)| =
sin(2θ), if 0 ≤ θ ≤ π/2
− sin(2θ), if π/2 ≤ θ ≤ π
sin(2θ), if π ≤ θ ≤ 3π/2
− sin(2θ), if 3π/2 ≤ θ ≤ 2π
Finally, our integral is∫∫Df(x, y) dA =
∫ 2π
0
∫ 1
0
f(r, θ) r dr dθ =
∫ 2π
0
∫ 1
0
r3
2|sin(2θ)| dr dθ
=1
2
(∫ 1
0
r3 dr
)(∫ 2π
0
|sin(2θ)| dθ)
=1
2
(r4
4
∣∣∣∣10
)(∫ π/2
0
sin(2θ) dθ −∫ π
π/2
sin(2θ) dθ +
∫ 3π/2
π
sin(2θ) dθ −∫ 2π
3π/2
sin(2θ) dθ
)
=1
8
−cos(2θ)
2
∣∣∣∣∣π/2
0
+cos(2θ)
2
∣∣∣∣∣π
π/2
− cos(2θ)
2
∣∣∣∣∣3π/2
π
+cos(2θ)
2
∣∣∣∣∣2π
3π/2
=
1
16(− cosπ + cos 0 + cos(2π)− cosπ − cos(3π) + cos(2π) + cos(4π)− cos(3π))
=1
16(1 + 1 + 1 + 1 + 1 + 1 + 1 + 1) =
1
2.
Problem 6 (15.4.21). Find the volume of the wedge-shaped region contained in the cylinderx2 + y2 = 9, bounded above by the plane z = x and below by the xy-plane.
Solution. See your text for the picture of the region D. From the picture (and the descriptionof the region) you can conclude that D is given in cylindrical coordinates as
0 ≤ r ≤ 3, −π/2 ≤ θ ≤ π/2, 0 ≤ z ≤ x = r cos θ.
Therefore, the volume of D is∫∫∫D
1 dV =
∫ 3
0
∫ π/2
−π/2
∫ r cos θ
0
r dz dθ dr =
∫ 3
0
r
(∫ π/2
−π/2
(z∣∣∣r cos θ0
)dθ
)dr
=
∫ 3
0
r
(∫ π/2
−π/2r cos θ dθ
)dr =
(∫ 3
0
r2 dr
)(∫ π/2
−π/2cos θ dθ
)
=
(r3
3
∣∣∣∣30
)(sin θ
∣∣∣π/2−π/2
)= 18.
5
Problem 7 (15.4.23). Evaluate∫∫D
√x2 + y2 dA where D is the region
2x
y
Solution. First, we consider the inner circle. The inner circle Dinn is centered at (1, 0) andhas radius 1. Therefore, in rectangular coordinates it is given by the equation
(x− 1)2 + y2 = 1.
Changing to polar coordinates, we have
(r cos θ − 1)2 + r2 sin2 θ = 1
=⇒ r2 cos2 θ − 2r cos θ + 1 + r2 sin2 θ = 1
=⇒ r2 − 2r cos θ = 0
=⇒ r = 2 cos θ,
and π ranges between −π/2 and π/2 (since these are the values of θ that make r = 0). Thusthe integral of the integrand
√x2 + y2 = r over the inner circle Dinn is given by∫∫
Dinn
√x2 + y2 dA =
∫ π/2
−π/2
∫ 2 cos θ
0
r r dr dθ =
∫ π/2
−π/2
(r3
3
∣∣∣∣2 cos θ0
)dθ
=8
3
∫ π/2
−π/2cos3 θ dθ =
8
3
∫ π/2
−π/2(1− sin2 θ) cos θ dθ.
Now let u = sin θ, so that du = cos θ dθ. Also, when θ = −π/2, we have u = sin(−π/2) = −1,and when θ = π/2, we have u = sin(π/2) = 1. Therefore our integral over the inner circleDinn becomes∫∫
Dinn
√x2 + y2 dA =
8
3
∫ 1
−1(1− u2) du =
8
3
(u− u3
3
)∣∣∣∣1−1
=8
3
(2− 2
3
)=
32
9.
This finishes our consideration of the inner circle.Next we consider the outer circle Dout. It is given in polar coordinates as
0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π.
6
Therefore the integral of the integrand√x2 + y2 = r over Dout is∫∫
Dout
√x2 + y2 dA =
∫ 2π
0
∫ 2
0
r r dr dθ =
(∫ 2π
0
1 dθ
)(∫ 2
0
r2 dr
)=
16π
3.
This finishes our consideration of the outer circle.We can now compute
∫∫D
√x2 + y2 dA and finish the problem. Since D is obtained by
removing Dinn from Dout, it follows that∫∫D
√x2 + y2 dA =
∫∫Dout
√x2 + y2 dA−
∫∫Dinn
√x2 + y2 dA =
16π
3− 32
9.
Problem 8 (15.4.27). Use cylindrical coordinates to calculate∫∫∫
W f(x, y, z) dV for thegiven function and region.
f(x, y, z) = x2 + y2; x2 + y2 ≤ 9, 0 ≤ z ≤ 5
Solution. In three dimensions, the inequality x2 + y2 ≤ 9 is the filled-in cylinder of radius3 centered along the z-axis, and in cylindrical coordinates it is given by the inequalities0 ≤ r ≤ 3 and 0 ≤ θ ≤ 2π. Moreover, the inequality 0 ≤ z ≤ 5 gives us the portion of thiscylinder that we need to consider. The integrand f(x, y, z) = x2 + y2 is given in cylindricalcoordinates as f(r, θ, z) = r2, and so our integral is∫∫∫
Wf(x, y, z) dV =
∫ 5
0
∫ 2π
0
∫ 3
0
r2 r dr dθ dz
=
(∫ 5
0
1 dz
)(∫ 2π
0
1 dθ
)(∫ 3
0
r3 dr
)= 5 · 2π · 34
4=
405π
2.
Problem 9 (15.4.29). Use cylindrical coordinates to calculate∫∫∫
W f(x, y, z) dV for thegiven function and region.
f(x, y, z) = y; x2 + y2 ≤ 1, x ≥ 0, y ≥ 0, 0 ≤ z ≤ 2
Solution. The region W indicated here is the cylinder of radius 1 centered along the z-axis,restricted to the first octant, and with maximum height 2. In polar coordinates, it is givenby
0 ≤ r ≤ 1, 0 ≤ θ ≤ π/2, 0 ≤ z ≤ 2.
The integrand f(x, y, z) = y is given in cylindrical coordinates as f(r, θ, z) = r sin θ, and soour integral is∫∫∫
Wf(x, y, z) dV =
∫ 2
0
∫ π/2
0
∫ 1
0
r sin θ r dr dθ dz
=
(∫ 2
0
1 dz
)(∫ π/2
0
sin θ dθ
)(∫ 1
0
r2 dr
)=
2
3.
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Problem 10 (15.4.31). Use cylindrical coordinates to calculate∫∫∫
W f(x, y, z) dV for thegiven function and region.
f(x, y, z) = z; x2 + y2 ≤ z ≤ 9.
Solution. From the description of the region, we can deduce two inequalities:
x2 + y2 ≤ 9, x2 + y2 ≤ z ≤ 9.
The first inequality is the cylinder of radius 3 centered along the z-axis; in polar coordinatesit is 0 ≤ r ≤ 3 and 0 ≤ θ ≤ 2π. The second inequality means that the values of z that wewant to integrate over are above the paraboloid z = x2 + y2 and below the plane z = 9; inpolar coordinates this means that r2 ≤ z ≤ 9. Therefore we can already integrate:∫∫∫
Wf(x, y, z) dV =
∫ 2π
0
∫ 3
0
∫ 9
r2z r dz dr dθ =
(∫ 2π
0
1 dθ
)(∫ 3
0
r
(z2
2
∣∣∣∣9r2
)dr
)
= 2π
∫ 3
0
r
(81
2− r4
2
)dr = π
∫ 3
0
(81r − r5
)dr = 243π.
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