practical guide 12 surface integrals

8/12/2019 Practical Guide 12 Surface Integrals

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0.1 Practical Guide - Surface Integrals

Surface integral,means to integrate over a surface.We begin with the study of surfaces.The easiest way is to give as many familiar examples as possible

1) a plane surfaceany plane is completely dened by- 3 points (belonging to that plane),

( when the 3 points represent the intersections of the plane with Ox; Oy;Oz , then we have an easy drawing)

x

y

z

A (a,0,0)

A (0,b,0)

C (0,0,c)

- the "normal" to the plane and one point (belonging to the plane) , let the normal unit vector be !n = (a;b;c)and (x0; y0; z0) a point that belongs to the plane.

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O

M (x,y,z)

A(xo,yo,zo)

y

x

z n

the normal vector!n is orthogonal on every vector in the plane, we use the scalar product!n? !AM , < !n ;!AM >= !n !AM= 0 , (a;b;c) (x x0; y y0; z z0) = 0

ax+by+cz+ (ax0 by0 cz0)| {z }d

= 0 , ax+by+cz+d= 0

- 2 vectors in the plane and one point (belonging to the plane), let!u ;!v be 2 (linearly independent) vectorsin the plane and (x0; y0; z0) a point that belongs to the plane. We use the vector product (cross product) to get avector normal to the plane, since !u !v? !u , !u !v? !vconsequently apply the previous computation with!n = !u !v

2) a cylinder , we consider a simple case, the cylinder is orthogonal on the xOy plane, has radius = r and itsaxis is Oz

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y

x

z

the projection of the cylinder onto the xOy plane is clearly- a circle ( for a circular cylinder), which equation is x2 +y2 =r2

- an ellipse ( for an elliptical cylinder) which equation is x2

a2 + y2

b2 = 1

x x

yy

r a

b

this explains why in many text books you will nd the "equation" of such a cylinder as

x2 +y2 =r2 or x2

a2 +

y2

b2 = 1

which is a bit confusing, since these are equations for a circle or an ellipse.Actually the "full" equation of such a cylinder is

x2 +y2 =r2 and z2 R or x2

a2 +

y2

b2 = 1 and z2 R

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4) a cone, we consider a simple case, the axis of the cone is Oz and it is circular or elliptical, the equation is

z2 =a2(x2 +y2) or z2 = x2

a2+

y2

b2

x

z

3) a sphere or an ellipsoidcentered at the origin (0; 0; 0)

x x

zz

a

b

cr

r

r

the equations are

x2 +y2 +z2 =r2 or x2

a2 +

y2

b2 +

z2

c2 = 1

4) a paraboloid ( circular, elliptical, ...) in a simple case, with axis Oz

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x

z

the equations are

z= a2(x2 +y2) or z= x2

a2 +

y2

b2

Do we have to memorise all these equations ? No ! We have two easy problems.i) what kind of surface does an equation represent ?ii) what is the equation for a surface dened by its geometric properties ?Both problems are easily solved by cutting (intersecting) the surface with horizontal planes and with the "ver-

tical" plane yOz

Examples.

1. What kind of surface represents the equation z = 4(x2 +y2) ?Cut the surface with horizontal planes, that is z =constant=c2 , since in this casez 0we get 4(x2 +y2) =c2 , x2 +y2 = c22 which are clearly circles, "located" at z =c2, the greater is z ,

the greater the radius of these (horizontal) circles

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x

z

now cut the surface with the yOz plane, whose equation is x = 0 , put x = 0 in z = 4(x2 +y2) , we getz = 4y2 which is clearly a parabola in the y Oz plane

z z = 4y

nally the surface is "generated" by all these (horizontal) circles along a parabola, and we get a circularparaboloid

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x

z

2. what kind of surface represents the equation z2 = 4(x2 +y2) ?cut the surface with horizontal planes, that is z=constant=c ,

we get 4(x2 +y2) = c2 , x2 +y2 = c22 which are clearly (horizontal) circles

x

z

now cut the surface with the y Oz plane, whose equation is x= 0, put x = 0 in z2 = 4(x2 +y2) , we get

z2 = 4y2 , z2 4y2 = 0 , (z 2y)(z+ 2y) = 0 , z 2y= 0 or z+ 2y= 0

which are clearly the equations of two straight lines in the yOz plane

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z = 2yz = - 2y

y

z

nally the surface is "generated" by all these (horizontal) circles along (between) two straght lines, and we geta cone

y

x

z

Now for the converse problem, how to determine the equation of a surface dened by its geometric propertieswe proceed somehow "backwards" as before.

Examples.

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3. a sphere centered at(a;b;c) with radius r . All the points (x;y;z) on the sphere have a constant distance=r to the center(a;b;c), which leads to

distance =p

(x a)2 + (y b)2 + (z c)2 =r , (x a)2 + (y b)2 + (z c)2 =r2

4. A circular cone with axisOz . We need just the radius of one horizontal circle, say r = 2and the correspondingz = 3, we get the "cut" (intersection) with the y Oz plane, as two straight lines z= 32y , z = 32y ,

r = 2

z = 3

y

zz = (3/2)yz = - (3/2)y

which means that for x = 0 the equation of the cone should be

(z 32

y)( z +3

2y) = 0 , z2 9

4y2 = 0 , 4z2 = 9y2

since by cutting with horizontal planes ( z=constant ) we get circles, the "full" equation of the cone should be

9(x2 +y2) = 4z2

Now we get to "parametrized" surfaces. In simple words, it means we dene a surface using two coordinates(variables), that is we "map" the surface onto a plane surface (the "space" of the coordinates)

To get a better idea, think of the surface of the Earth and the "geographic" coordinates : longitude and latitude.They "map" the (curved) surface of the Earth onto a "plane" , (actually the map from a geographic atlas) denedby the two coordinates.

The "standard" denition is the following.

Denition. A function (correspondence, map) h : D R2 ! S R3 as h(u; v) = (x(u; v); y(u; v); z(u; v))or

x= x(u; v) y= y(u; v) z = z(u; v)

is called a parametrization of the surface S , or Sis a parametrized surface .We assume the function h to be injective (one to one).Also to be of class C1, that is, it has partial derivatives which are continuous.

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Also "nonsingular", that is, the Jacobi matrix has rank 2 at every point(u; v)

rank

0@ @x@u @x@v@y

@u@y@v

@z@u

@z@v

1A = 2

It looks a bit complicated, but actually simple enough.

Comments.We clearly need "one to one", every point on the surface (x;y;z)2 S should have its own coordinates, they

should not "overlap".Remember a parametrized path

y

z

x

(t)

(t)

: [a; b] ! R3 , (t) = (x(t); y(t); z(t)) is the "position" vector, and by taking the derivatives0(t) = (x0(t); y0(t); z0(t)) is the "velocity" vector, which is tangent to the path.

Now for a parametrized surface, we have a "similar" situation.Let!r = (x;y;z)be the "position" vector.

!r = (x;y;z) = x!i +y!j +z!k = (x(u; v) , y (u; v) , z (u; v))

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x

y

z

ru

rvru x rv

S

The plane which is tangent to the surface is dened by two vectors, the "two" partial derivatives, with respectto u and v

!ru = ( @x@u

;@ y

@u;@ z

@u) , !rv = ( @x

@v;@y

@v;@ z

@v)

To actually dene a plane these two vectors should be "linearly independent", which means their matrix hasrank 2

rank

0@

@x@u

@x@v

@y@u

@y@v

@z

@u

@z

@v

1A

= 2

And this is how we get the denition as stated before. Clearly the vector product!ru !rv is orthogonal to thetangent plane, thus a normal direction.

We should also add that D R2 is a "domain" , that is every point (; ) 2 D is "surounded" by an open disk

(; ) 2 f(u )2 + (v )2 < rg D

Each surface has innitely many parametrizations.In the following we basically consider just two kinds of parametrizations:

- projecting on the planesxOy , orxOz , or yOx , which works for almost all surfaces in "school" problems- using spherical coordinates for a sphere or an ellipsoid

To better understand, consider the surfaces we already studied in the begining.

Denition. Now we clearly state: the domainsSand D are bounded . The functionf :S!

R , is a continuousfunction on a neiborhood ofS

The "standard" notation for surface integral isZZS

f(x;y;z)ds or

ZZ

f(x;y;z)d or

Z

f(x;y;z)d

for the constant function 1 the surface integral represents the area of the surface

f(x;y;z) = 1 for all (x;y;z) 2 S , the "area" area(S) =ZZS

1ds

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and nally the surface integral is reduced to a double integralZZS

f(x;y;z)ds=

ZZD

f(x(u; v); y(u; v); z(u; v)) k!ru !rvk dudv

The value of the integral does not depend on the parametrization.

Comments.

To compute a surface integral we basically need to- nd a suitable parametrization,- determine the domainD ,- computek!ru !rvk

Case I. The surface is "projectable" onto one of the planesxOy , xOz or yOz , that is the projection is a"one to one" function.

Projecting the surface onto thexOy plane. (Whenever this is possible)

Let D R2 be the projection ofSon the plane xOy . An easy way to get the projection is to set z = 0 in theequation dening the surface (whenever this equation is given)

Then S= f(x;y;z) 2 R3 , (x; y) 2 D , z= z(x; y)gThe surface is parametrized as D 3 (x; y) ! (x;y;z(x; y)) 2 S(we actually identify R2 with the xOy plane by (x; y) $ (x;y; 0) )

D

S

y

z

x(x,y,0)

z = z(x,y)

(x,y,z(x,y))

In this case !r = (x;y;z) = (x;y;z(x; y)) and we take derivatives with respect to x and y

!rx = @@x

(x;y;z(x; y)) = (1; 0;@ z

@x) , !ry = @

@y(x;y;z(x; y)) = (0; 1;

@z

@y)

and the vector product is

!rx !ry =

!i !

j !

k

1 0 @z@x0 1 @z@y

= (@z

@x)!i ( @z

@y)!j + (1)

!k =

@z

@x;@z

@y; 1

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and the norm is

k!rx !ryk =s

@z@x

2+

@z

@y

2+ 12 =

s@z

@x

2+

@z

@y

2+ 1

and nally the surface integral computes as

ZZSf(x;y;z)ds= ZZD

f(x;y;z(x; y)) k!rx !ryk dxdy = ZZDf(x;y;z(x; y))

s1 +

@z

@x2

+@z

@y 2

dxdy

we denitely may use the nal formula directly any time we need.

Remark.A simple question rises naturally. If the surface S is a plane surface Sin the xOy plane, is there any dierence

between the surface integral and the double integral ?

x

y

z

S

The answer is No ! No dierence at all.In this case the projection of S onto the plane xOy is S itself, the surface is parametrized as (x; y)!

(x;y; 0) 2 S , therefore z = z(x; y) = 0 is a constant function and the derivatives are null@z

@x= 0 ,

@z

@y = 0

consequently the surface intergal computes asZZS

f(x;y;z)ds=

ZZS

f(x;y;z)dxdy

and the surface intergal turns out to be exactly a double integral.

The same happens if S is a plane surface parallel to the xOy plane, dened by z = c =constant. Again thepartial derivaties are null

@z

@x= 0 ,

@z

@y = 0

The cases when the surface is projectable onto the xOz or yOz planes are similar.

Examples.1) Compute the area of the hemisphere S= f(x;y;z) 2 R3 , x2 +y2 +z2 =R2 , z 0gThis surface is projectable onto the xOy plane and the projection is the disc D = fx2 +y2 R2g ( for z = 0 )You should notice that this hemisphere is not projectable onto the planesxOz or

From the equation of the surface x2 +y2 +z2 =R2 we get z = z(x; y) =p

R2 x2 y2 , (x; y) 2 D

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x

y

z

R

R

z = sqrt(R-x-y)

(x,y,0)

"sqrt" stands for "square root" , z =p

R2 x2 y2 , because some graphic facilities are not installed yet.Clearly we have

z = z(x; y) =p

R2 x2 y2 , @z@x

= xpR2 x2 y2 ,

@z

@y =

ypR2 x2 y2

The hemisphere projects onto the xOy plane into the disk D = f(x;y; 0) 2 R3 ; x2 +y2 R2g

D

x

y

R

R

t

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and the area is

area(S) =

ZZS

1ds=

ZZD

s1 +

@z

@x

2+

@z

@y

2dxdy=

ZZD

s1 +

x2

R2 x2 y2 + y2

R2 x2 y2 dxdy =

= ZZD s R2

R2

x2

y2

dxdy = R ZZD1

pR2 x2 y2dxdy=

use now polar coordinates to compute the double integral x= r cos t , y = r sin tin order to cover the disc D we need t 2 [0; 2] and r 2 [0; R] , so nally we get

=R

RZ0

0@2Z

0

1pR2 r2 rdt

1A dr = 2R RZ

0

1pR2 r2 rdr = 2R

p

R2 r2r=Rr=0

= 2R

pR2 = 2R2

which is indeed the area of an emisphere with radius = R .

2) Compute the area of the surface S= f(x;y;z) 2 R3 ; z = x2 +y2 4g , (a piece from a paraboloid)This a paraboloid. By projecting it on the xOy plane we get a disk D =f(x;y; 0)2 R3 ; x2 +y2 4g ,

centered at(0; 0; )with radius2 ( for z= 0)

The partial derivatives are

z = z(x; y) = x2 +y2 , @z

@x= 2x ,

@z

@y = 2y

The area ofS is

area(S) =

ZZS

1ds=

ZZD

s1 +

@z

@x

2+

@z

@y

2dxdy =

ZZD

q1 + (2x)2 + (2y)2dxdy =

= 2

ZZD

p1 +x2 +y2dxdy=

next we use polar coordinates to compute the double integral x= r cos t , y = r sin t , and we need t2[0; 2]and r 2 [0; R] to cover the disk, so nally we get

= 2

2Z0

0@2Z

0

p1 +r2rdt

1A dr= 4 2Z

0

p1 +r2rdr = 4

p1 +r2

r=2r=0

= 4(p

5 1)

3) Compute the area of the surface S which is the portion from the hemispherefx2 +y2 +z2 = 4 , z 0g"cut" by the cylinderf(x;y;z) 2 R3 , x2 +y2 = 2x , z2 Rg.

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x

z

y

S

D

2

This surface S is projectable on the planexOy and the projection is the disc D = f(x; y) 2 R2 , x2 +y2 2xg centered at (1; 0)and radius = 1, since

x2 +y2 2x , x2 2x+ 1 +y2 1 , (x 1)2 +y2 1

and we may dene the surface Sas followsS= f(x;y;z) 2 R3 , (x; y) 2 D , z = z(x; y) =

p4 x2 y2g.

Consequently the projection of the surface Sis a "smaller" disk, centered at (1; 0), tangent to Oy and to thecirclex2 +y2 = 4

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y

x

D

2

Consequently we get@z

@x=

xp4 x2 y2

, @z

@y =

yp4 x2 y2

area(S) =

ZZS

1ds=

ZZD

s1 +

@z

@x

2+

@z

@y

2dxdy=

ZZD

vuut1 +

xp4 x2 y2

!2+

yp4 x2 y2

!2=

= ZZD

s1 + x2 +y2

4 x2

y2

dxdy= ZZD

2

p4 x2 y2 dxdy=then use polar coordinatesx r cos t , r= r sin t to compute the double integralthe equation of the disc D in polar coordinates is r2 2r cos t , r 2cos t , so cos t 0 ) t2 [2 ; 2 ]

and 0 r 2cos t

area(S) = ::: =

2Z

2

0@2 cos tZ

0

2p4 r2 rdr

1A dt=

2Z

2

(2)

p4 r2

r=2cos tr=0

dt=

=

2Z

2

2

2p

4 4cos2 t

dt= 4

2Z

2

1

psin2 t

dt= 4

2Z

2

(1 jsin tj) dt=

= 4 40Z

2

sin tdt 4

2Z0

sin tdt= 4 4 cos tj02

4 ( cos t)j20 = 4 4 4 = 4 8

Case II. Using spherical coordinatesThe parametrization of a piece of a sphere centered at (0; 0; 0) with radius R is

x= x(; ') = R cos ' sin , y= y(; ') = R sin ' sin , z = z(; ') = R cos

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x

y

z

R

R

the partial derivatives are

!r =

@x

@;@y

@;@ z

@

= (R cos ' cos ; R sin ' cos ;R sin )

!r'= @x

@';@y

@';@z

@' = (R sin ' sin ; R cos ' sin ; 0)

!r !r' =

!i

!j

!k

R cos ' cos R sin ' cos R sin R sin ' sin R cos ' sin 0

==

R sin ' cos R sin R cos ' sin 0!i

R cos ' cos R sin R sin ' sin 0!j +

R cos ' cos R sin ' cos R sin ' sin R cos ' sin !k =

=R2 sin

0@ cos ' sin ; sin ' sin ; cos2 ' cos + sin2 ' cos | {z }

cos

1A =R2 sin ( cos ' sin ; sin ' sin ; cos )

k!r !r'k =R2 sin p( cos ' sin )2 + ( sin ' sin )2 + (cos )2 =R2 sin r(cos

2 '+ sin2 '| {z })sin2 + cos2 = R2 sin

and the surface integral becamesZZS

f(x;y;z)ds=

ZZD

f(R cos ' sin ; R sin ' sin ; R cos )R2 sin dd'

All we need to remember, are the spherical coordinates and the fact that k!r !r'k =R2 sin Examples.4) Compute the area of the sphere S= f(x;y;z) 2 R3 , x2 +y2 +z2 =R2 g , using spherical coordinates.To cover the whole (surface) sphere we clearly need , 2 [0; ] , ' 2 [0; 2]

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area(S) =

ZZS

1ds=

ZZ[0;][0;2]

1 k!r !r'k dd'=

=

Z0

0

@

2Z0

R2 sin d'

1

Ad= 2R2

Z0

sin d = 2R2 ( cos j0 ) = 2R2( cos + cos 0) = 4R2

5) Compute the area of the surface S from the spheref(x;y;z)2 R3 , x2 +y2 +z2 = 8g that is located"inside" the cone z2 =x2 +y2 .

x

y

z

R

R

We use spherical coordinates.Clearly the angle ' goes all around the Oz axis, that is ' 2 [0; 2] . The radius is R = 2p2We need to nd out the extent of the angle . Find the intersection between the sphere and the cone

x2 +y2 +z2 = 8z2 =x2 +y2

) , 2z2 = 8 ) z = 2

Therefore we have O A= 2p

2 , AB = 2 , sin = 22p2

= 1p2 ) = 4

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y

z

O C

AB

Finally the area of the surface "S" is

area(S) =

ZZS

1ds=

ZZ[0;

4][0;2]

1 k!r !r'k dd'=ZZ

[0;4][0;2]

R2 sin dd'=

=

=4Z0

0

@2Z0

R2 sin d'

1

Ad= 2R2

=4Z0

sin d = 2R2

cos j==4=0

= 2R2(1 1p

2)

practical guide 12 surface integrals

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