Dawkins Surface Integrals PDF

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CALCULU Surface Integ

Paul Dawkins

Calculus III

Table of Conten Preface............................................................................ Surface Integrals ........................................................... Introduction ........................................................................ Parametric Surfaces ............................................................. Surface Integrals ................................................................. Surface Integrals of Vector Fields ....................................... Stokes’ Theorem ................................................................. Divergence Theorem ...........................................................

Calculus III

Preface Here are my online notes for my Calculus III course tha Despite the fact that these are my “class notes”, they sho learn Calculus III or needing a refresher in some of the t These notes do assume that the reader has a good worki including limits, derivatives and integration. It also assu knowledge of several Calculus II topics including some equations, vectors, and knowledge of three dimensional Here are a couple of warnings to my students who may a day that you missed. 1. Because I wanted to make this a fairly complete calculus I have included some material that I do and because this changes from semester to seme find one of your fellow class mates to see if ther covered in class. 2. In general I try to work problems in class that ar with Calculus III many of the problems are diffi moment and so in this class my class work will worked problems go. With that being said I wil

Calculus III

Surface Integrals Introduction In the previous chapter we looked at evaluating integral points came from a curve in two- or three-dimensional s and integrate functions and vector fields where the poin dimensional space. These integrals are called surface in Here is a list of the topics covered in this chapter. Parametric Surfaces – In this section we will take a loo with parametric equations. We will also take a look at a Surface Integrals – Here we will introduce the topic of with surface integrals of functions in this section. Surface Integrals of Vector Fields – We will look at su section. Stokes’ Theorem – We will look at Stokes’ Theorem in Divergence Theorem – Here we will take a look at the

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Parametric Surfaces Before we get into surface integrals we first need to talk When we parameterized a curve we took values of t from in to

r r r r (t ) = x (t ) i + y (t ) j

and the resulting set of vectors will be the position vecto With surfaces we’ll do something similar. We will take dimensional space D and plug them into

r r r ( u, v ) = x ( u, v ) i + y ( u, v )

and the resulting set of vectors will be the position vecto are trying to parameterize. This is often called the para surface S. We will sometimes need to write the parametric equat nothing more than the components of the parametric rep

x = x (u , v )

y = y ( u, v )

Example 1 Determine the surface given by the parame r r r r (u , v ) = u i + u cos v j Solution Let’s first write down the parametric equations.

x=u

y = ucos v

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Solution (a) The elliptic paraboloid x = 5y 2 + 2z 2 - 10 . This one is probably the easiest one of the four to see ho x = f ( y, z ) we can quickly write down a set of parame

x = 5 y2 + 2 z2 - 10

y=

The last two equations are just there to acknowledge tha want them to be. The parametric representation is then,

r r ( y , z ) = ( 5 y 2 + 2z 2 - 10)

(b) The elliptic paraboloid x = 5 y2 + 2 z2 - 10 that is This is really a restriction on the previous parametric rep representation stays the same.

r r ( y , z ) = ( 5 y 2 + 2z 2 - 10)

However, since we only want the surface that lies in fro that x ³ 0 . This is equivalent to requiring,

5 y 2 + 2 z 2 - 10 ³ 0 2 2 2 (c) The sphere x + y + z = 30 .

or

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(d) The cylinder y2 + z2 = 25 . As with the last one this can be tricky until you see how to use cylindrical coordinates since they can be easily us cylinder. In cylindrical coordinates the equation of a cylinder of r

r =a

and so the equation of the cylinder in this problem is r = Next, we have the following conversion formulas.

x =x

y = r sin q

Notice that they are slightly different from those that we them up here since the cylinder was centered upon the x Finally, we know what r is so we can easily write down cylinder.

r r r r ( x ,q ) = x i + 5sinq j +

We will also need the restriction 0 £ q £ 2p to make su the cylinder. Since we haven’t put any restrictions on th any restriction on x.

In the first part of this example we used the fact that the quickly write down a parametric representation This ca

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r r r Now, provided ru ´ rv ¹ 0 it can be shown that the vecto S. This means that it can be used for the normal vector equation of a tangent plane. This is an important idea th the next couple of sections. Let’s take a look at an example.

Example 3 Find the equation of the tangent plane to th r r r 2 r ( u, v) = u i + 2 v j + ( at the point ( 2, 2,3) . Solution r r Let’s first compute ru ´ rv . Here are the two individual v

r r r ru (u , v ) = i + 2u k

r r

Now the cross product (which will give us the normal v

r i

r j

r k

r r r n = ru ´ rv = 1 0 2u = 0 4v 1 Now, this is all fine, but in order to use it we will need t give us the point in question. We can easily do this by s parametric representation equal to the coordinates of the

2= u

Þ

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r r r ( u, v) = x ( u, v) i + y ( u, v ) and as we will see it again comes down to needing the v So, provided S is traced out exactly once as (u , v ) range S is given by,

A = òò

r r ru ´rv dA

D

Let’s take a look at an example.

Example 4 Find the surface area of the portion of the s cylinder x 2 + y 2 = 12 and above the xy-plane. Solution Okay we’ve got a couple of things to do here. First we We parameterized a sphere earlier in this section so ther is the parameterization.

r r r ( q , j ) = 4 sin j cos q i + 4 sin j

Next we need to determine D. Since we are not restricti rotating with the sphere we can take the following range

0 £ q £ 2p Now, we need to determine a range for j . This will tak First, let’s start with the equation of the sphere.

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r

r

Finally, we need to determine rq ´ rj . Here are the two

r r rq ( q ,j ) = -4 sin j sin q i + 4 sin j r r rj (q ,j ) = 4 cos j cosq i + 4 cosj

Now let’s take the cross product.

r j

r k

4 sinj cosq

0

r i

r r rq ´ rj = -4 sinj sinq

4 cosj sinq - 4 sinj r r 2 2 = -16sin j cosq i - 16sin j cosj sin q k - 1 r r 2 2 = -16sin j cosq i - 16sin j sinq j - 16 sinj r r 2 2 = -16sin j cosq i - 16sin j sinq j - 16 sinj 4 cosj cosq

We now need the magnitude of this,

r r rq ´ rj = 256 sin 4 j cos 2 q + 256 sin 4 j = 256 sin 4 j (cos 2 q + sin 2 q ) + = 256 sin 2 j (sin 2 j + cos 2 j ) = 16 sin2 j = 16 sin j = 16 sin j

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Surface Integrals It is now time to think about integrating functions over s space. Let’s start off with a sketch of the surface S since once we get into it. Here is a sketch of some surface S.

The region S will lie above (in this case) some region D rectangle here, but it doesn’t have to be of course. Also a surface S that was in front of some region D in the yz-p locked into the xy-plane that you can’t do problems that

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Note as well that there are similar formulas for surfaces plane) and x = g ( y, z ) (with D in the yz-plane). We w examples and we’ll leave the other to you to write down The second method for evaluating a surface integral is f parameterization,

r r r ( u, v) = x ( u, v) i + y ( u, v )

In these cases the surface integral is,

r = f x y z dS f r , , ( ) ( òò òò (u, S

D

where D is the range of the parameters that trace out the Before we work some examples let’s notice that since w z = g ( x, y ) as,

r r r r ( x , y ) = xi + yj + g

we can always use this form for these kinds of surfaces 2

r r æ ¶g ö æ ¶ rx ´ ry = ç ÷ + ç è ¶x ø è ¶ for these kinds of surfaces. You might want to verify th cross products. Let’s work some examples.

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Here is a sketch of the region D.

Calculus III 2 xy d S y y 6 3 6 = -z ( òò òò S

D

1

1- y

= 6 3ó ò õ0 0

y- y

1

ó æ = 6 3 ô ç yz - zy 2 õ0 è 1

1 y - y2 + = 6 3ó ô õ0 2 1 æ1 = 6 3 ç y 2 - y3 3 è4 Example 2 Evaluate

òò z dS

where S is the upper half

S

Solution We gave the parameterization of a sphere in the previou for this sphere.

r r r ( q , j ) = 2 sin j cos q i + 2 sin j

Since we are working on the upper half of the sphere he

0 £ q £ 2p r

r

0

Next, we need to determine rq ´ rj . Here are the two ind

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r r rq ´ rj = 16 sin 4 j cos 2 q + 16 sin 4 j = 16 sin 4 j ( cos 2 q + sin 2 q ) + = 16 sin 2 j ( sin 2 j + cos 2 j ) = 4 sin2 j = 4 sin j = 4sin j We can drop the absolute value bars in the sine because are working with. The surface integral is then,

òò z dS = òò 2 cos j ( 4 s S

D

Don’t forget that we need to plug in for x, y and/or z in t needed to plug in z. Here is the evaluation for the doub 2p

òò z dS = ò ò S

0

=ò =ò

2p 0 2p

0

= 8p

p 2

0

4sin ( 2

(- 2 cos ( 2j 4 dq

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r r rz ´ rq =

r i 0 - 3 sinq

3

r = - 3 cosq i The magnitude of this vector is,

r r rz ´ rq = 3cos2 q + 3 si The surface integral is then,

òò y dS = òò S

3 sin q

(

D

= 3ò = 3ò

2p 0 2p 0

ò

6 0

sin q

6sinq

= ( -18cos q ) =0 Example 4 Evaluate

òò y + z dS

where S is the surfac

S 2 2 whose bottom is the disk x + y £ 3 in the xy-plane an

Solution

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Actually we need to be careful here. There is more to th We’re going to let S1 be the portion of the cylinder that other words, the top of the cylinder will be at an angle. lies inside (i.e. the cap on the cylinder) S2 . Finally, the is the disk of radius

3 in the xy-plane and is denoted b

In order to do this integral we’ll need to note that just lik

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The upper limit for the z’s is the plane so we can just plu cylinder we know what y is from the parameterization s Here is the integral for the cylinder.

òò y + z dS = òò ( S1

3 sin q + z

)( 3 ) dA

D

= 3ò

2p

0

ò

4 - 3 sinq

3 sin q + z

0

2p

= 3 ò0

(

3 sinq 4 - 3 sin

3 = 3 ò 8 - sin 2 q dq 0 2 2p 3 = 3 ò 8 - (1 - cos ( 2q ) ) 0 4 2p

3 æ 29 ö = 3 ç q + sin ( 2q ) ÷ 8 è 4 ø

2p

0

29 3 p = 2 S2 : Plane on Top of the Cylinder In this case we don’t need to do any parameterization si gave at the start of this section. Remember that the plan f

thi

f

D i th di k f

di

3

t

d t th

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òò y + z dS = 4 2òò d S2

D

(

=4 2 p = 12 2 p S3 : Bottom of the Cylinder Again, this is set up to use the initial formula we gave in equation for the bottom is given by g ( x, y ) = 0 and D i origin. Also, don’t forget to plug in for z. Here is the work for this integral.

òò y + z dS = òò ( y + 0) S3

( 0)

2

D

= òò y dA D

2p

=ó õ0

ò

0

3

r 2 sin q d

2p

ó æ1 3 ö = ô ç r sin q ÷ ø õ0 è 3 =ò

2p

0

3 sin q d q

Calculus III

Surface Integrals of Vector Fields Just as we did with line integrals we now need to move Recall that in line integrals the orientation of the curve w the answer. The same thing will hold true with surface doing surface integrals of vector fields we first need to i Let’s start off with a surface that has two sides (while th Mobius Strip is a surface that only has one side!) that ha possibly along the boundary). Making this assumption r r r normal vectors, n1 and n2 = - n1 . This means that every vectors. The set that we choose will give the surface an There is one convention that we will make in regards to we need to define a closed surface. A surface S is close region E. A good example of a closed surface is the sur surface S has a positive orientation if we choose the set from the region E while the negative orientation will be in towards the region E. Note that this convention is only used for closed surface In order to work with surface integrals of vector fields w formula for the unit normal vector corresponding to the with. We have two ways of doing this depending on ho First, let’s suppose that the function is given by z = g ( x function

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First, notice that the component of the normal vector in normal vector) is always positive and so this normal vec not point directly up, but it will have an upwards compo This will be important when we are working with a clos orientation. If we know that we can then look at the nor “positive” orientation should point upwards or downwa orientation must point out of the region and this may me turns out that we need the downward orientation we can vector and we’ll get the one that we need. Again, remem when choosing the unit normal vector. Before we move onto the second method of giving the s did this for surfaces in the form z = g ( x , y ) . We could surfaces in the form y = g ( x , z ) (so f ( x, y, z) = y - g

x = g ( y , z ) (so f ( x, y, z ) = x - g ( y, z ) ). Now, we need to discuss how to find the unit normal ve as,

r r r ( u, v) = x ( u, v) i + y( u, v ) r r In this case recall that the vector ru ´ rv will be normal to But if the vector is normal to the tangent plane at a poin at that point. So, this is a normal vector. In order to gua will also need to divide it by its magnitude. So in the case of parametric surfaces one of the unit nor

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where the right hand integral is a standard surface integr r F across S. Before we work any examples let’s notice that we can su get a somewhat easier formula to use. We will need to b formulas however as each will assume a certain orientat normal vector to match the given orientation. Let’s first start by assuming that the surface is given by

r r r assume that the vector field is given by F = P i + Q j + after is the “upwards” orientation. Under all of these as S is,

r r r r òò F gdS = òò F gn dS S

S

r óó æ r r r ç - gx i - gy ôô P i Q j R k gç = + + ôô 2 ôô ç (g x ) + (g õõ è D r r r r r = òò P i + Q j + R k g - g x i - g y j

(

(

)

)(

D

= òò -Pg x - Qg y + R dA D

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Let’s now take a quick look at the formula for the surfac r parametrically by r ( u, v) . In this case the surface integ

r r r r òò F gdS = òò F gn dS S

S

r r óó r æ ru ´ rv = ôô F g ç r r ç õõ è ru ´ rv D r r r = òò F g(ru ´rv ) dA D

r

r

Again note that we may have to change the sign on ru ´ r and so there is once again really two formulas here. Als in this case and so we won’t need to worry that in these

Note as well that there are even times when we will used directly. We will see an example of this below. Let’s now work a couple of examples.

r r r r r Example 1 Evaluate òò F gdS where F = y j - z k an S

paraboloid y = x 2 + z 2 , 0 £ y £ 1 and the disk x + z orientation. 2

2

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As noted in the sketch we will denote the paraboloid by order for unit normal vectors on the paraboloid to point point generally in the negative y direction. On the other will need to point in the positive y direction in order to p Since S is composed of the two surfaces we’ll need to d add the results to get the overall surface integral. Let’s have the surface in the form y = g (x , z ) so we will nee one given initially wasn’t for this kind of function. This define,

f ( x , y ,z ) = y - g ( x ,z ) =

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r r òò F gdS = S1

=

òò (

r r y j- zk

D

æ 2x , gç ç Ñ è

)

2 y z dA 2 òò D

=

2 2 2 + x z z d 2 ( ) òò D

= - òò x 2 + 3z 2 dA D

Don’t forget that we need to plug in the equation of the the integral. In this case D is the disk of radius 1 in the polar coordinates to complete this integral. Here are po

x = r cosq 0 £ q £ 2p

z 0

Note that we kept the x conversion formula the same as z be the formula that used the sine. We could have done at least working with one of them as we are used to wor Here is the evaluation of this integral.

r r 2 2 g = + F d S x z dA 3 òò òò S1

D

2p

= -ó õ0 ó

2p

ò (r 1

0

ò( 1

2

cos2 q + 3r 2 sin2

)

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also be orthogonal to the plane y = 1 then it must point but we already have a unit vector that does this. Namely

r r n= j

the standard unit basis vector. It also points in the corre have the vector field and the normal vector we can plug integral to get,

r r òò F gdS = S2

òò (

r r r y j - z k g( j

)

S2

At this point we need to plug in for y (since S 2 is a portio is) and we’ll also need the square root this time when w double integral. In this case since we are using the defin of the square root that we saw with the first portion. To acknowledge that

y = 1 = g ( x, z )

and so the square root is,

(gx ) +1 + (gz 2

The surface integral is then,

r r òòF gdS = S2

òò y dS S2

= òò1 0 + 1 + 0 D

At this point we can acknowledge that D is a disk of rad more than the double integral that will give the area of t

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We will call S1 the hemisphere and S 2 will be the botto on the sketch). Now, in order for the unit normal vector enclosed region they will all need to have a positive z co must be normal to the surface and if there is a positive z will have to be pointing away from the enclosed region. On the other ...