Calculus I

Preface Here 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 you’ve reached the level of working the harder problems then you will probably already understand the basics fairly well and won’t 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 a popup 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 haven’t been removed here. These issues may make the solutions a little difficult to follow at times, but they should still be readable.

Volumes of Solids of Revolution / Method of Rings 1. Use the method of disks/rings to determine the volume of the solid obtained by rotating the region 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 let’s 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 we’ll 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.

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

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 aren’t a very visual person. 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 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).

© 2007 Paul Dawkins 2 http://tutorial.math.lamar.edu/terms.aspx

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 of y. This in turn means that we’ll need to rewrite the equation of the boundary curve to get into terms of y. Here is another sketch of a representative disk with all of the various quantities we need put into it.

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

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 the y-axis to the curve defining the edge of the solid. In other words,

2Radius y= The area of the disk is then,

( ) ( ) ( )22 2 4RadiusA y y yπ π π= = =

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 yπ π π= = =∫

2. Use the method of disks/rings to determine the volume of the solid obtained by rotating the region bounded by 27y x= − , 2x = − , 2x = and the x-axis about the x-axis. Hint : Start with sketching the bounded region. Step 1 We need to start the problem somewhere so let’s 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 we’ll 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.

© 2007 Paul Dawkins 4 http://tutorial.math.lamar.edu/terms.aspx

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 aren’t a very visual person. 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 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).

© 2007 Paul Dawkins 5 http://tutorial.math.lamar.edu/terms.aspx

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 of x. Therefore the equation of the curve will need to be in terms of x (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 the x-axis to the curve defining the edge of the solid. In other words,

© 2007 Paul Dawkins 6 http://tutorial.math.lamar.edu/terms.aspx

Calculus I

2Radius 7 x= − The area of the disk is then,

( ) ( ) ( ) ( )22 2 2 4Radius 7 49 14A x x x xπ π π= = − = − +

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 2x = − and the “last” disk would occur at 2x = . Our limits are then : 2 2x− ≤ ≤ . The volume is then,

( ) ( ) 22 2 4 3 5 201214 13 5 152 2

49 14 49V x x dx x x xπ π π− −

= − + = − + =∫

3. Use the method of disks/rings to determine the volume of the solid obtained by rotating the region bounded by 2 6 10x y y= − + and 5x = about the y-axis. Hint : Start with sketching the bounded region. Step 1 We need to start the problem somewhere so let’s 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 we’ll 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.

© 2007 Paul Dawkins 7 http://tutorial.math.lamar.edu/terms.aspx

Calculus I

Here is the work used to determine the intersection points (we’ll need these later).

( )( ) ( ) ( )

2

2

6 10 56 5 0

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

y yy y

y y y y

− + =

− + =

− − = ⇒ = = ⇒

Hint : Give a good attempt at sketching what the solid of revolution looks like and sketch in a representative ring. Note that this can be a difficult thing to do especially if you aren’t a very visual person. 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 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 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 the edges of the of the back half of the solid).

© 2007 Paul Dawkins 8 http://tutorial.math.lamar.edu/terms.aspx

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 of y. Therefore the equation of the curves will need to be in terms of y (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,

© 2007 Paul Dawkins 9 http://tutorial.math.lamar.edu/terms.aspx

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

π

π π

= − = − − + = − + − + −

Step 4 The 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,

( )

( )

5 2 3 4

152 3 4 556 10881

3 5 151

75 120 56 12

75 60 3

V y y y y dy

y y y y y

π

π π

= − + − + −

= − + − + − =

∫

4. Use the method of disks/rings to determine the volume of the solid obtained by rotating the region bounded by 22y x= and 3y x= about the x-axis. Hint : Start with sketching the bounded region. Step 1 We need to start the problem somewhere so let’s 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 we’ll 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.

© 2007 Paul Dawkins 10 http://tutorial.math.lamar.edu/terms.aspx

Calculus I

Here is the work used to determine the intersection points (we’ll need these later).

( ) ( ) ( )

3 2

3 2

2

22 0

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

x xx x

x x x x

=

− =

− = ⇒ = = ⇒

Hint : Give a good attempt at sketching what the solid of revolution looks like and sketch in a representative ring. Note that this can be a difficult thing to do especially if you aren’t a very visual person. 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 revolution. So, do the best you can at getting these sketches. Step 2 Here is a sketch of the solid of revolution.

© 2007 Paul Dawkins 11 http://tutorial.math.lamar.edu/terms.aspx

Calculus I

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

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 x-axis we know that the area formula will need to be in terms of x. Therefore the equation of the curves will need to be in terms of x (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 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.

© 2007 Paul Dawkins 12 http://tutorial.math.lamar.edu/terms.aspx

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= = The area of the ring is then,

( ) ( ) ( )

( ) ( ) ( )

2 2

2 22 3 4 6

Outer Radius Inner Radius

2 4

A x

x x x x

π

π π

= − = − = −

Step 4 The 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 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 xπ π π= − = − =∫

5. Use the method of disks/rings to determine the volume of the solid obtained by rotating the region bounded by 26 xy −= e and 26 4 2y x x= + − between 0x = and 1x = about the line

2y = − . Hint : Start with sketching the bounded region. Step 1 We need to start the problem somewhere so let’s 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 we’ll 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.

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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. We’ll 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. Hint : Give a good attempt at sketching what the solid of revolution looks like and sketch in a representative ring. Note that this can be a difficult thing to do especially if you aren’t a very visual person. 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 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 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

© 2007 Paul Dawkins 14 http://tutorial.math.lamar.edu/terms.aspx

Calculus I

representative ring 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).

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 a horizontal axis (i.e. parallel to the x-axis) we know that the area formula will need to be in terms of x. Therefore the equations of the curves will need to be in terms of x (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.

© 2007 Paul Dawkins 15 http://tutorial.math.lamar.edu/terms.aspx

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 of 26 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 of 26 4 2x x+ − ). 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 x−= + = + + − = + −e The area of the ring is then,

( ) ( ) ( )

( ) ( )( )

2 2

2 22 2

2 3 4 2 4

Outer Radius Inner Radius

8 4 2 2 6

60 64 16 16 4 24 36

x

x x

A x

x x

x x x x

π

π

π

−

− −

= − = + − − +

= + − − + − −

e

e e

Step 4 The final step is to then set up the integral for the volume and evaluate it. 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 1x≤ ≤ . The volume is then,

( )

( ) ( )

1 2 3 4 2 4

012 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

π

π π

− −

− − − −

= + − − + − −

= + − − + + + = + +

∫ e e

e e e e

6. Use the method of disks/rings to determine the volume of the solid obtained by rotating the region bounded by 210 6y x x= − + , 210 6y x x= − + − , 1x = and 5x = about the line 8y = . Hint : Start with sketching the bounded region.

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Calculus I

Step 1 We need to start the problem somewhere so let’s 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 we’ll 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.

Hint : Give a good attempt at sketching what the solid of revolution looks like and sketch in a representative ring. Note that this can be a difficult thing to do especially if you aren’t a very visual person. 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 revolution. So, do the best you can at getting these sketches. Step 2 Here is a sketch of the solid of revolution.

© 2007 Paul Dawkins 17 http://tutorial.math.lamar.edu/terms.aspx

Calculus I

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

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 a horizontal axis (i.e. parallel to the x-axis) we know that the area formula will need to be in terms of x. Therefore the equations of the curves will need to be in terms of x (which in this case they already are).

© 2007 Paul Dawkins 18 http://tutorial.math.lamar.edu/terms.aspx

Calculus I

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

210 6y 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 6x 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 a positive 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 6x x x x= − − + − = − + The area of the ring is then,

© 2007 Paul Dawkins 19 http://tutorial.math.lamar.edu/terms.aspx

Calculus I

( ) ( ) ( )

( ) ( ) ( )

2 2

2 22 2 2

Outer Radius Inner Radius

18 6 2 6 320 192 32

A x

x x x x x x

π

π π

= − = − + − − + − = − +

Step 4 The final step is to then set up the integral for the volume and evaluate it. From the graph from Step 1 we can see that the “first” ring in the solid would occur at 1x = 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 xπ π π= − + = − + =∫

7. Use the method of disks/rings to determine the volume of the solid obtained by rotating the region bounded by 2 4x y= − and 6 3x y= − about the line 24x = . Hint : Start with sketching the bounded region. Step 1 We need to start the problem somewhere so let’s 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 we’ll 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.

© 2007 Paul Dawkins 20 http://tutorial.math.lamar.edu/terms.aspx

Calculus I

To get the intersection points shown on the sketch all we need to do is set the two equations equal and solve (we’ll need these in a bit).

( )( ) ( ) ( )

2

2

4 6 33 10 0

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

y yy y

y y y y

− = −

+ − =

+ − = ⇒ = − = ⇒ −

Hint : Give a good attempt at sketching what the solid of revolution looks like and sketch in a representative ring. Note that this can be a difficult thing to do especially if you aren’t a very visual person. 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 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 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

© 2007 Paul Dawkins 21 http://tutorial.math.lamar.edu/terms.aspx

Calculus I

representative ring 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).

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 a vertical axis (i.e. parallel to the y-axis) we know that the area formula will need to be in terms of y. Therefore the equation of the curves will need to be in terms of y (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.

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

© 2007 Paul Dawkins 22 http://tutorial.math.lamar.edu/terms.aspx

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. Let’s 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 negative

function 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 just 2 4y − . We don’t 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 the y-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 won’t need to do the second as the formula will be the same.

The area of the ring is then,

( ) ( ) ( )

( ) ( ) ( )

2 2

2 22 2 4

Outer Radius Inner Radius

28 18 3 460 108 65

A x

y y y y y

π

π π

= − = − − + = − − +

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

© 2007 Paul Dawkins 23 http://tutorial.math.lamar.edu/terms.aspx

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 yπ π π− −

= − − + = − − + =∫

8. Use the method of disks/rings to determine the volume of the solid obtained by rotating the region bounded by 2 1y x= + , 4x = and 3y = about the line 4x = − . Hint : Start with sketching the bounded region. Step 1 We need to start the problem somewhere so let’s 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 we’ll 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.

Hint : Give a good attempt at sketching what the solid of revolution looks like and sketch in a representative ring. Note that this can be a difficult thing to do especially if you aren’t a very visual person. 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 revolution. So, do the best you can at getting these sketches.

© 2007 Paul Dawkins 24 http://tutorial.math.lamar.edu/terms.aspx

Calculus I

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

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

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 a vertical axis (i.e. parallel to the y-axis) we know that the area formula will need to be in terms of y. Therefore the equations of the curves will need to be in terms of y and so we’ll need to rewrite the equation of the line to be in terms of y. Here is another sketch of a representative ring with all of the various quantities we need put into it.

© 2007 Paul Dawkins 25 http://tutorial.math.lamar.edu/terms.aspx

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 ( )1

2 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 1

2 2 2Inner Radius 4 1 Outer Radius 4 4 8y y= + − = + = + = The area of the ring is then,

( ) ( ) ( )

( ) ( ) ( )

2 2

22 27 207 71 12 2 4 2 4

Outer Radius Inner Radius

8

A x

y y y

π

π π

= − = − + = − −

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

© 2007 Paul Dawkins 26 http://tutorial.math.lamar.edu/terms.aspx

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 yπ π π= − − = − − =∫

© 2007 Paul Dawkins 27 http://tutorial.math.lamar.edu/terms.aspx