CalcII_SurfaceArea

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Calculus II

2007 Paul Dawkins i http://tutorial.math.lamar.edu/terms.aspx

PrefaceHere are my online notes for my Calculus II course that I teach here at Lamar University.Despite the fact that these are my class notes, they should be accessible to anyone wanting to

learn Calculus II or needing a refresher in some of the topics from the class.

These notes do assume that the reader has a good working knowledge of Calculus I topics

including limits, derivatives and basic integration and integration by substitution.

Calculus II tends to be a very difficult course for many students. There are many reasons for this.

The first reason is that this course does require that you have a very good working knowledge of

Calculus I. The Calculus I portion of many of the problems tends to be skipped and left to thestudent to verify or fill in the details. If you dont have good Calculus I skills, and you are

constantly getting stuck on the Calculus I portion of the problem, you will find this course verydifficult to complete.

The second, and probably larger, reason many students have difficulty with Calculus II is that youwill be asked to truly think in this class. That is not meant to insult anyone; it is simply an

acknowledgment that you cant just memorize a bunch of formulas and expect to pass the courseas you can do in many math classes. There are formulas in this class that you will need to know,

but they tend to be fairly general. You will need to understand them, how they work, and moreimportantly whether they can be used or not. As an example, the first topic we will look at isIntegration by Parts. The integration by parts formula is very easy to remember. However, just

because youve got it memorized doesnt mean that you can use it. Youll need to be able to lookat an integral and realize that integration by parts can be used (which isnt always obvious) and

then decide which portions of the integral correspond to the parts in the formula (again, notalways obvious).

Finally, many of the problems in this course will have multiple solution techniques and so youllneed to be able to identify all the possible techniques and then decide which will be the easiest

technique to use.

So, with all that out of the way let me also get a couple of warnings out of the way to my studentswho may be here to get a copy of what happened on a day that you missed.

1. Because I wanted to make this a fairly complete set of notes for anyone wanting to learncalculus I have included some material that I do not usually have time to cover in class

and because this changes from semester to semester it is not noted here. You will need tofind one of your fellow class mates to see if there is something in these notes that wasntcovered in class.

2. In general I try to work problems in class that are different from my notes. However,with Calculus II many of the problems are difficult to make up on the spur of the momentand so in this class my class work will follow these notes fairly close as far as worked

problems go. With that being said I will, on occasion, work problems off the top of my

head when I can to provide more examples than just those in my notes. Also, I often
http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx


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Calculus II

2007 Paul Dawkins ii http://tutorial.math.lamar.edu/terms.aspx

dont have time in class to work all of the problems in the notes and so you will find thatsome sections contain problems that werent worked in class due to time restrictions.

3. Sometimes questions in class will lead down paths that are not covered here. I try toanticipate as many of the questions as possible in writing these up, but the reality is that Icant anticipate all the questions. Sometimes a very good question gets asked in class

that leads to insights that Ive not included here. You should always talk to someone whowas in class on the day you missed and compare these notes to their notes and see whatthe differences are.

4. This is somewhat related to the previous three items, but is important enough to merit itsown item. THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!!

Using these notes as a substitute for class is liable to get you in trouble. As already notednot everything in these notes is covered in class and often material or insights not in these

notes is covered in class.


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Calculus II

2007 Paul Dawkins 3 http://tutorial.math.lamar.edu/terms.aspx

SurfaceAreaIn this section we are going to look once again at solids of revolution. We first looked at them back inCalculus I when we found the volume of the solid of revolution. In this section we want to find thesurface area of this region.

So, for the purposes of the derivation of the formula, lets look at rotating the continuous function

( )f x= in the interval [ ],a b about thex-axis. Below is a sketch of a function and the solid ofrevolution we get by rotating the function about thex-axis.

We can derive a formula for the surface area much as we derived the formula for arc length. Well start

by dividing the integral into nequal subintervals of width D . On each subinterval we will approximatethe function with a straight line that agrees with the function at the endpoints of the each interval. Here is

a sketch of that for our representative function using 4n= .

Now, rotate the approximations about thex-axis and we get the following solid.


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The approximation on each interval gives a distinct portion of the solid and to make this clear eachportion is colored differently. Each of these portions are called frustumsand we know how to find the

surface area of frustums.

The surface area of a frustum is given by,

2A rlp= where,

( )1 2 1

2

1radius of right end

2

radius of left end

r r r r

r

= + =

=

and lis the length of the slant of the frustum.

For the frustum on the interval [ ]1,i ix- we have,

( )

( )

( )

1

2 1

1 1length of the line segment connecting and

i

i

i i i i

r f x

r f x

l P P P P

-

- -

=

=

=

and we know from the previous section that,

( ) [ ]2

* *

1 11 where is some point in ,i i i i i iP P f x x x x x- - = + D

Before writing down the formula for the surface area we are going to assume that xD is small andsince ( )f x is continuous we can then assume that,

( ) ( ) ( ) ( )* *1andi i i if x f x f x f x-

So, the surface area of the frustum on the interval [ ]1,i ix- is approximately,


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Calculus II


( ) ( )

( ) ( )

1

1

2* *

22

2 1

i i

i i i

i i

f x f xA P P

f x f x x

p

p

--

+ =

+ D

The surface area of the whole solid is then approximately,

( ) ( )2

* *

1

2 1n

i i

i

S f x f x xp=

+ D and we can get the exact surface area by taking the limit as ngoes to infinity.

( ) ( )

( ) ( )

2* *

1

2

lim 2 1

2 1

n

i in

i

b

a

S f x f x x

f x f x dx

p

p

=

= + D

= +

If we wanted to we could also derive a similar formula for rotating ( )h y= on [ ],c d about they-axis.This would give the following formula.

( ) ( )2

2 1d

c

S h y h y dyp = +

These are not the standard formulas however. Notice that the roots in both of these formulas are

nothing more than the two dss we used in the previous section. Also, we will replace ( )f x withyand

( )h y withx. Doing this gives the following two formulas for the surface area.

Surface Area Formulas

2 rotation about axis

2 rotation about axis

S y ds x

S x ds y

p

p

= -= -

where,

( )

( )

2

2

1 if ,

1 if ,

dyds dx y f x a x b

dx

dxds dy x h y c y d

dy

= + =

= + =

There are a couple of things to note about these formulas. First, notice that the variable in the integralitself is always the opposite variable from the one were rotating about. Second, we are allowed to use

either dsin either formula. This means that there are, in some way, four formulas here. We will choosethe dsbased upon which is the most convenient for a given function and problem.

Now lets work a couple of examples.

Example 1 Determine the surface area of the solid obtained by rotating 29 x= - , 2 2x-


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Calculus II


about thex-axis.

SolutionThe formula that well be using here is,

2S y dsp= since we are rotating about thex-axis and well use the first dsin this case because our function is in thecorrect form for that dsand we wont gain anything by solving it forx.

Lets first get the derivative and the root taken care of.

( ) ( )( )

12 2

12 2

19 2

29

dy xx x

dxx

-= - - = -

-

2 2

2 2 2

9 31 1

9 9 9

dy x

dx x x

+ = + = = - - -

Heres the integral for the surface area, 2

22

32

9S y dx

xp

-

=-

There is a problem however. The dxmeans that we shouldnt have anyys in the integral. So, beforeevaluating the integral well need to substitute in foryas well.

The surface area is then,2

2

22

2

2

32 9

9

6

24

S x dxx

dx

p

p

p

-

-

= --

=

=

Previously we made the comment that we could use either dsin the surface area formulas. Lets work anexample in which using either dswont create integrals that are too difficult to evaluate and so we cancheck both dss.

Example 2 Determine the surface area of the solid obtained by rotating 3 x= , 1 2y about they-axis. Use both dss to compute the surface area.

SolutionNote that weve been given the function set up for the first dsand limits that work for the second ds.

Solution 1This solution will use the first dslisted above. Well start with the derivative and root.

2

31

3

dyx

dx

-=


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Calculus II


4 42

3 3

4 4 2

3 3 3

1 9 1 9 11 1

9 9 3

dy x x

dxx x x

+ + + = + = =

Well also need to get new limits. That isnt too bad however. All we need to do is plug in the givenys

into our equation and solve to get that the range ofxs is 1 8x . The integral for the surface area isthen,

8 4

3

2

31

1 48

3 3

1

9 12

3

29 1

3

xS x dx

x

x dx

p

p

+=

= +

Note that this time we didnt need to substitute in for thexas we did in the previous example. In this case

we picked up a dxfrom the dsand so we dont need to do a substitution for thex. In fact if we hadsubstituted forxwe would have putys into the integral which would have caused problems.

Using the substitution4 1

3 39 1 12u x du x dx= + = the integral becomes,

145

10

1453

2

10

3 3

2 2

18

27

145 10 199.4827

S u du

u

p

p

p

=

=

= - =

Solution 2This time well use the second ds. So, well first need to solve the equation forx. Well also go ahead

and get the derivative and root while were at it.

3 23dx

y ydy

= =

2

41 1 9dx

ydy

+ = +

The surface area is then,2

4

12 1 9S x y dyp= +

We used the originalylimits this time because we picked up a dyfrom the ds. Also note that the presenceof the dymeans that this time, unlike the first solution, well need to substitute in for thex. Doing that


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CalcII_SurfaceArea