CalcI Volume Rings Solutions

8/2/2019 CalcI Volume Rings Solutions

1/27

Calculus I

2007 Paul Dawkins 1 http://tutorial.math.lamar.edu/terms.aspx

PrefaceHere are the solutions to the practice problems for my Calculus I notes. Some solutions will have

more or less detail than other solutions. The level of detail in each solution will depend up on

several issues. If the section is a review section, this mostly applies to problems in the first

chapter, there will probably not be as much detail to the solutions given that the problems really

should be review. As the difficulty level of the problems increases less detail will go into the

basics of the solution under the assumption that if youve reached the level of working the harder

problems then you will probably already understand the basics fairly well and wont need all the

explanation.

This document was written with presentation on the web in mind. On the web most solutions are

broken down into steps and many of the steps have hints. Each hint on the web is given as apopup however in this document they are listed prior to each step. Also, on the web each step can

be viewed individually by clicking on links while in this document they are all showing. Also,

there are liable to be some formatting parts in this document intended for help in generating the

web pages that havent been removed here. These issues may make the solutions a little difficult

to follow at times, but they should still be readable.

VolumesofSolidsofRevolution/MethodofRings

1. Use the method of disks/rings to determine the volume of the solid obtained by rotating theregion bounded by y x= , 3y = and the y-axis about the y-axis.

Hint : Start with sketching the bounded region.

Step 1

We need to start the problem somewhere so lets start simple.

Knowing what the bounded region looks like will definitely help for most of these types of

problems since we need to know how all the curves relate to each other when we go to set up the

area formula and well need limits for the integral which the graph will often help with.

Here is a sketch of the bounded region with the axis of rotation shown.
http://tutorial.math.lamar.edu/terms.aspxhttp://tutorial.math.lamar.edu/terms.aspx


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Calculus I


Hint : Give a good attempt at sketching what the solid of revolution looks like and sketch in a

representative disk.

Note that this can be a difficult thing to do especially if you arent a very visual person.

However, having a representative disk can be of great help when we go to write down the areaformula. Also, getting the representative disk can be difficult without a sketch of the solid of

revolution. So, do the best you can at getting these sketches.

Step 2

Here is a sketch of the solid of revolution.

Here are a couple of sketches of a representative disk. The image on the left shows a

representative disk with the front half of the solid cut away and the image on the right shows a

representative disk with a wire frame of the back half of the solid (i.e. the curves representing

the edges of the of the back half of the solid).


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Calculus I


Hint : Determine a formula for the area of the disk.

Step 3

We now need to find a formula for the area of the disk. Because we are using disks that are

centered on the y-axis we know that the area formula will need to be in terms ofy. This in turn

means that well need to rewrite the equation of the boundary curve to get into terms ofy.

Here is another sketch of a representative disk with all of the various quantities we need put into

it.


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Calculus I


As we can see from the sketch the disk is centered on the y-axis and placed at some y. The radius

of the disk is the distance from they-axis to the curve defining the edge of the solid. In other

words,

2Radius y=

The area of the disk is then,

( ) ( ) ( )22 2 4Radius A y y yp p p= = =

Step 4

The final step is to then set up the integral for the volume and evaluate it.

For the limits on the integral we can see that the first disk in the solid would occur at 0y = and

the last disk would occur at 3y = . Our limits are then : 0 3y .

The volume is then,

3 34 5 2431

5 500V y dy yp p p= = =

2. Use the method of disks/rings to determine the volume of the solid obtained by rotating the

region bounded by

2

7y x= -

, 2x= -

, 2x=

and the x-axis about the x-axis.


Step 1







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Calculus I



representative disk.


However, having a representative disk can be of great help when we go to write down the area

formula. Also, getting the representative disk can be difficult without a sketch of the solid of


Step 2


Here are a couple of sketches of a representative disk. The image on the left shows a

representative disk with the front half of the solid cut away and the image on the right shows a

representative disk with a wire frame of the back half of the solid (i.e. the curves representing



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Calculus I


Hint : Determine a formula for the area of the disk.

Step 3

We now need to find a formula for the area of the disk. Because we are using disks that are

centered on the x-axis we know that the area formula will need to be in terms ofx. Therefore the

equation of the curve will need to be in terms ofx (which in this case it already is).

Here is another sketch of a representative disk with all of the various quantities we need put into

it.

As we can see from the sketch the disk is centered on the x-axis and placed at some x. The radius

of the disk is the distance from thex-axis to the curve defining the edge of the solid. In other

words,


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Calculus I


2Radius 7= -

The area of the disk is then,

( ) ( ) ( ) ( )

22 2 2 4Radius 7 49 14 A x x x xp p p= = - = - +

Step 4


For the limits on the integral we can see that the first disk in the solid would occur at 2x = -

and the last disk would occur at 2x = . Our limits are then : 2 2x- .

The volume is then,

( ) ( )22

2 4 3 5201214 13 5 152 249 14 49V x x dx x x xp p p- -= - + = - + =


region bounded by2 6 10 x y y= - + and 5x = about the y-axis.


Step 1







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Calculus I


Here is the work used to determine the intersection points (well need these later).

( ) ( ) ( ) ( )

2

2

6 10 5

6 5 0

5 1 0 1, 5 5,1 & 5,5

y y

y y

y y y y

- + =

- + =

- - = = =


representative ring.


However, having a representative ring can be of great help when we go to write down the area

formula. Also, getting the representative ring can be difficult without a sketch of the solid of


Step 2


Here are a couple of sketches of a representative ring. The image on the left shows a

representative ring with the front half of the solid cut away and the image on the right shows a

representative ring with a wire frame of the back half of the solid (i.e. the curves representing



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Calculus I


Hint : Determine a formula for the area of the ring.

Step 3

We now need to find a formula for the area of the ring. Because we are using rings that are

centered on the y-axis we know that the area formula will need to be in terms ofy. Therefore the

equation of the curves will need to be in terms ofy (which in this case they already are).

Here is another sketch of a representative ring with all of the various quantities we need put into

it.

As we can see from the sketch the ring is centered on the y-axis and placed at some y. The inner

radius of the ring is the distance from the y-axis to the curve defining the inner edge of the solid.

The outer radius of the ring is the distance from the y-axis to the curve defining the outer edge of

the solid. In other words,

2Inner Radius 6 10 Outer Radius 5y y= - + =

The area of the ring is then,


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Calculus I


( ) ( ) ( )

( ) ( ) ( )

2 2

22 2 2 3 4

Outer Radius Inner Radius

5 6 10 75 120 56 12

A x

y y y y y y

p

p p

= -

= - - + = - + - + -

Step 4The final step is to then set up the integral for the volume and evaluate it.

From the intersection points shown in the graph from Step 1 we can see that the first ring in the

solid would occur at 1y = and the last ring would occur at 5y = . Our limits are then :

1 5y .

The volume is then,

( )

( )

52 3 4

15

2 3 4 456 108813 5 15

1

75 120 56 12

75 60 3

V y y y y dy

y y y y y

p

p p

= - + - + -

= - + - + - =


region bounded by22y x= and 3y x= about the x-axis.


Step 1







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Calculus I


Here is the work used to determine the intersection points (well need these later).

( ) ( ) ( )

3 2

3 2

2

2

2 0

2 0 0, 2 0,0 & 2,8

x x

x x

x x x x

=

- =

- = = =







Step 2Here is a sketch of the solid of revolution.


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Calculus I







Step 3


centered on the x-axis we know that the area formula will need to be in terms ofx. Therefore the

equation of the curves will need to be in terms ofx (which in this case they already are).

Here is another sketch of a representative ring with all of the various quantities we need put intoit.

As we can see from the sketch the ring is centered on the x-axis and placed at some x. The inner

radius of the ring is the distance from the x-axis to the curve defining the inner edge of the solid.


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Calculus I


The outer radius of the ring is the distance from the x-axis to the curve defining the outer edge of

the solid. In other words,

3 2Inner Radius Outer Radius 2x x= =


( ) ( ) ( )

( ) ( ) ( )

2 2

2 22 3 4 6


2 4

A x

x x x x

p

p p

= -

= - = -

Step 4


From the intersection points shown in the graph from Step 1 we can see that the first ring in thesolid would occur at 0x = and the last ring would occur at 2x = . Our limits are then :

0 2x .

The volume is then,

( ) ( )22

4 6 5 7 2564 15 7 350 0

4V x x dx x xp p p= - = - =


region bounded by26 xy -= e and 26 4 2 y x x= + - about the line 2y = - .


Step 1







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Calculus I


For the intersection point on the left a quick check by plugging 0x = into both equations shows

that the intersection point is in fact ( )0,6 as we might have guessed from the graph. Well be

needing this point in a bit. From the sketch of the region it is also clear that there is no

intersection point on the right.







Step 2





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Calculus I





Step 3


centered on a horizontal axis (i.e. parallel to the x-axis) we know that the area formula will need

to be in terms ofx. Therefore the equations of the curves will need to be in terms ofx (which in

this case they already are).


it.


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Calculus I


From the sketch we can see the ring is centered on the line 2y = - and placed at some x.

The inner radius of the ring is the distance from the axis of rotation to the x-axis (a distance of 2)

followed by the distance from the x-axis to the curve defining the inner edge of the solid (a

distance of26 x-e ).

Likewise, the outer radius of the ring is the distance from the axis of rotation to the x-axis (again,

a distance of 2) followed by the distance from the x-axis to the curve defining the outer edge of

the solid (a distance of26 4 2x+ - ).

So the inner and outer radii are,

2 2 2Inner Radius 2 6 Outer Radius 2 6 4 2 8 4 2x x x x-= + = + + - = + -e


( ) ( ) ( )

( ) ( )

( )

2 2

2 22 2

2 3 4 2 4


8 4 2 2 6

60 64 16 16 4 24 36

x

x x

A x

x x

x x x x

p

p

p

-

- -

= -

= + - - +

= + - - + - -

e

e e

Step 4


From the graph from Step 1 we can see that the first ring in the solid would occur at 0x = and

the last ring would occur at 1x = . Our limits are then : 0 1 .

The volume is then,

( )

( ) ( )

12 3 4 2 4

0

12 3 4 5 2 4 2 416 9374

3 5 150

60 64 16 16 4 24 36

60 32 4 12 9 12 9

x x

x x

V x x x x dx

x x x x x

p

p p

- -

- - - -

= + - - + - -

= + - - + + + = + +

e e

e e e e


region bounded by210 6 y x x= - + , 210 6 y x x= - + - , 1= and 5x = about the line 8y = .



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Calculus I


Step 1










formula. Also, getting the representative ring can be difficult without a sketch of the solid ofrevolution. So, do the best you can at getting these sketches.

Step 2



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Calculus I







Step 3


centered on a horizontal axis (i.e. parallel to the x-axis) we know that the area formula will need

to be in terms ofx. Therefore the equations of the curves will need to be in terms ofx (which in

this case they already are).


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Calculus I



it.

From the sketch we can see the ring is centered on the line 8y = and placed at some x.

The inner radius of the ring is then the distance from the axis of rotation to the curve defining the

inner edge of the solid. To determine a formula for this first notice that the axis of rotation is a

distance of 8 from the x-axis. Next, the curve defining the inner edge of the solid is a distance of210 6 y x x= - + from the x-axis. The inner radius is then the difference between these two

distances or,

( )2 2Inner Radius 8 10 6 2 6 x x x x= - - + = - + -

The outer radius is computed in a similar manner. It is the distance from the axis of rotation to

the x-axis (a distance of 8) and then it continues below the x-axis until it reaches the curve

defining the outer edge of the solid. So, we need to add these two distances but we need to be

careful because the lower function is in fact negative value and so the distance of the point on

the lower function from the x-axis is in fact : ( )210 6x x- - + - as is shown on the sketch. The

negative in front of the equation makes sure that the negative value of the function is turned into apositive quantity (which we need for our distance). The outer radius is then the sum of these two

distances or,

( )2 2Outer Radius 8 10 6 18 6 x x x= - - + - = - +



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Calculus I


( ) ( ) ( )

( ) ( ) ( )

2 2

2 22 2 2


18 6 2 6 320 192 32

A x

x x x x x x

p

p p

= -

= - + - - + - = - +

Step 4


From the graph from Step 1 we can see that the first ring in the solid would occur at 1= and

the last ring would occur at 5x = . Our limits are then : 1 5x .

The volume is then,

( ) ( )55

2 2 332 8963 31 1

320 192 32 320 96V x x dx x x xp p p= - + = - + =


region bounded by2 4x y= - and 6 3x y= - about the line 24x = .


Step 1







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Calculus I


To get the intersection points shown on the sketch all we need to do is set the two equations equal

and solve (well need these in a bit).

( ) ( ) ( ) ( )

2

2

4 6 3

3 10 0

5 2 0 5, 2 5, 21 & 2,0

y y

y y

y y y y

- = -+ - =

+ - = = - = -







Step 2





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Calculus I





Step 3


centered on a vertical axis (i.e. parallel to the y-axis) we know that the area formula will need to

be in terms ofy. Therefore the equation of the curves will need to be in terms ofy (which in this

case they already are).


it.

From the sketch we can see the ring is centered on the line 24x = and placed at some y.

The inner radius of the ring is then the distance from the axis of rotation to the curve defining the

inner edge of the solid. To determine a formula for this first notice that the axis of rotation is a

distance of 24 from the y-axis. Next, the curve defining the inner edge of the solid is a distance of


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Calculus I


6 3x y= - from the y-axis. The inner radius is then the difference between these two distances

or,

( )Inner Radius 24 6 3 18 3y y= - - = +

The outer radius is computed in a similar manner but is a little trickier. In this case the curve

defining the outer edge of the solid occurs on both the left and right of the y-axis.

Lets first look at the case as shown in the sketch above. In this case the value of the function

defining the outer edge of the solid is to the left of the y-axis and so has a negative value. The

distance of this point from the y-axis is then ( )2 4y- - where the minus sign turns the negativefunction value into a positive value that we need for distance. The outer radius for this case is

then the sum of the distance of the axis of rotation to the y-axis (a distance of 24) and the distance

of the curve defining the outer edge to the y-axis (which we found above).

If the curve defining the outer edge of the solid is to the right of the y-axis then it will have a

positive value and so the distance of points on the curve and the y-axis is just2 4y - . We dont

need the minus sign in this case because the function value is already positive, which we need for

distance. The outer radius in this case is then the distance from the axis of rotation to they-axis

(a distance of 24) minus this new distance.

Nicely enough in either case the outer radius is then,

( )2 2Outer Radius 24 4 28y y= - - = -

Note that in cases like this where the curve defining an edge has both positive and negative values

the final equation of the radius (inner or outer depending on the problem) will be the same. You

just need to be careful in setting up the case you choose to look at. If you get the first case set up

correctly you wont need to do the second as the formula will be the same.


( ) ( ) ( )

( ) ( ) ( )

2 2

2 22 2 4


28 18 3 460 108 65

A x

y y y y y

p

p p

= -

= - - + = - - +

Step 4



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Calculus I


From the intersection points of the two curves we found in Step 1 we can see that the first ring

in the solid would occur at 5y = - and the last ring would occur at 2y = . Our limits are then

: 5 2y- .

The volume is then,

( ) ( )22

2 4 2 3 565 3155613 5 155 5

460 108 65 460 54V y y y dy y y y yp p p- -

= - - + = - - + =


region bounded by 2 1y x= + , 4x = and 3y = about the line 4x = - .


Step 1













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Calculus I


Step 2


Here are a couple of sketches of a representative ring. The image on the left shows arepresentative ring with the front half of the solid cut away and the image on the right shows a




Step 3


centered on a vertical axis (i.e. parallel to the y-axis) we know that the area formula will need to

be in terms ofy. Therefore the equations of the curves will need to be in terms ofy and so well

need to rewrite the equation of the line to be in terms ofy.


it.


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Calculus I


From the sketch we can see the ring is centered on the line 4x = - and placed at some y.

The inner radius of the ring is the distance from the axis of rotation to the y-axis (a distance of 4)

followed by the distance from the y-axis to the curve defining the inner edge of the solid (a

distance of ( )12 1y - ).

Likewise, the outer radius of the ring is the distance from the axis of rotation to the y-axis (again,

a distance of 4) followed by the distance from the y-axis to the curve defining the outer edge of

the solid (a distance of 4).

So the inner and outer radii are,

( ) 71 12 2 2Inner Radius 4 1 Outer Radius 4 4 8y y= + - = - = + =


( ) ( ) ( )

( ) ( ) ( )

2 2

22 27 207 71 12 2 4 2 4


8

A x

y y y

p

p p

= -

= - - = - -

Step 4


From the intersection points of the two curves we found in Step 1 we can see that the first ring

in the solid would occur at 3y = and the last ring would occur at 9y = . Our limits are then :

3 9y .


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Calculus I

The volume is then,

( ) ( )99

2 2 3207 7 207 71 14 2 4 4 4 123 3

126V y y dx y y yp p p= - - = - - =

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CalcI Volume Rings Solutions