The region enclosed by the x-axis and the parabola

23f x x x is revolved about the line x = –1 to

generate the shape of a cake. What is the volume of the cake?

DO NOW

THE SHELL METHOD

Section 7.3d

The region enclosed by the x-axis and the parabola

23y f x x x is revolved about the line x = –1 togenerate the shape of a cake. What is the volume of the cake?

x = –1

Let’s first try to integrate withrespect to y Washers!!!

To find the inner and outer radii of our washer, we would needto get the equation of the original parabola in terms of x...not an easy task…

The region enclosed by the x-axis and the parabola

23y f x x x is revolved about the line x = –1 togenerate the shape of a cake. What is the volume of the cake?

Instead of slicing horizontally,we will cut a series of cylindricalslices by cutting straight downall the way around the solid.

The radii of the cylinders gradually increase, and the heights ofthe cylinders follow the contour of the parabola: smaller to larger,then back to smaller.

When we unroll one of these shells, we have a rectangularprism, with height equal to the height of the shell, lengthequal to the circumference of the shell, and thickness equalto .x

The region enclosed by the x-axis and the parabola

23y f x x x is revolved about the line x = –1 togenerate the shape of a cake. What is the volume of the cake?

Volume of each shell:

2 rh x Here, we have

1r x 23h x x

2,3x x x

To find the total volume of the solid, integrate with respect to x:

3 2

02 1 3V x x x dx 3 2 3

02 3 2x x x dx

32 3 4

0

3 2 12

2 3 4x x x

45

2

The region bounded by the curve , the x-axis, and theline x = 4 is revolved about the x-axis to generate a solid. Findthe volume of the solid.

y x

Let’s solve this problem first by integrating with respect to x:

2A x rCross section area:

2x x 4,2

4

0V xdxVolume:

42

02

x

8

The region bounded by the curve , the x-axis, and theline x = 4 is revolved about the x-axis to generate a solid. Findthe volume of the solid.

y x

Now let’s integrate with respect to y using our new technique:

yRadius of each shell:

4,2 2 ,y y

2x y

24 yHeight of each shell:

Limits of integration: 0 to 2

dyThickness of each shell:

The region bounded by the curve , the x-axis, and theline x = 4 is revolved about the x-axis to generate a solid. Findthe volume of the solid.

y x

Now let’s integrate with respect to y using our new technique:

2 2

02 4V y y dy

Volume: 4,2 2 ,y y

2x y

2 3

02 4y y dy

242

0

2 24

yy

8

Let’s redo a problem from a previous class, this time using shells…

Find the volume of the solid generated by revolving the regionbounded above by the square root function and below by theidentity function about the y-axis.

Cross section area:

22 2A y y y

Volume:

1 2 4

0V y y dy

13 5

03 5

y y

2

15

2 4y y

1,12x y

x y

Our previous solution

Let’s redo a problem from a previous class, this time using shells…

Find the volume of the solid generated by revolving the regionbounded above by the square root function and below by theidentity function about the y-axis.

Shell radius: r x

Volume:

1

02V x x x dx 1 3 2 2

02 x x dx

2

15

1,1

y x

y xShell height: h x x Shell thickness: dx

15 2 3

0

2 12

5 3x x

Let’s redo a problem from a previous class, this time using shells…

Find the volume of the solid generated by revolving the triangularregion bounded by the lines y = 2x, y = 0, and x = 1 about theline x = 2.

2x

R

r

2

212 12

A y y

Cross section area:

213 2

4y y

42

03 2

4

yV y dy

Volume: 232

0

312

yy y

8

3

Our previous solution

Let’s redo a problem from a previous class, this time using shells…

Find the volume of the solid generated by revolving the triangularregion bounded by the lines y = 2x, y = 0, and x = 1 about theline x = 2.

2x Shell radius: 2 xShell height: 2xShell thickness: dx

Volume:

1

02 2 2V x x dx 1 2

04 2x x dx

132

0

43

xx

8

3