# Find volume of solid produced

1. Jul 7, 2008

### c-murda

1. The problem statement, all variables and given/known data
y = x³ y= 0 and x = 1

and its revolved around the line x = 2

okay i have drawn the graph of y = x³ and other paramaters, but when i get ther area being rotated it produces a hollow center. how do i go about finding the volume?

would it be a washers i don't understand the gap between the line x = 2 and x = 1?

2. Jul 7, 2008

### Defennder

Yes there will be a hollow centre. The solid of revolution generated looks like a ring. You can use the washer method of calculating volume. First find the volume generated rotating f(y) (express y=f(x) in terms of x first) around x=2 for the region defined within the range of y=0 and y=1. When that is done, subtract the unwanted volume which corresponds to the hollow centre beteen the lines x=2 and x=1. You don't have to use calculus to do this one, just use the formula for volume of cylinder.

3. Jul 7, 2008

### c-murda

thanks!
using washer and shell

V=∫2pi(2-x)x³dx

i got 3pi / 5

4. Jul 8, 2008

### dynamicsolo

So you chose the shell method. I agree with your result.

To use the disk/washer method, the infinitesimal slices are taken perpendicular to the rotation axis, so we would have to integrate in the y-direction. The limits of integration would now be y = 0 to y = 1. The "washers" would be disks with an outer radius following the y = x^3 curve and an inner radius for the hole following the x = 1 curve. So the washers have holes of constant radius 2 - 1 unit. The outer radius curve will have to be inverted into x = y^(1/3); since it is to the left of x = 2, the outer radius will be 2 - y^(1/3).

This makes the infinitesimal volume of a washer

$$dV = [\pi(r_{outer})^2 - \pi(r_{inner})^2] dy = [\pi(2 -y^{1/3})^2 - \pi(1)^2] dy$$

The volume integral is then

$$V = \pi \int_{0}^{1} 4 - 4y^{1/3}+ y^{2/3} - 1 dy$$ ,

which also gets you $$V = \frac{3\pi}{5}$$. But shells is definitely the easier method for this one...

5. Jul 8, 2008

### HallsofIvy

Staff Emeritus
Another way to think about it (but exactly the same as the washer method) is this:

1. Suppose you rotate the area above the line y= 0, from x= 0 to x= 1, about the line y= 2. What volume would it have? (That's simply a cylinder- area= $\pi r^2 h$.)

2. Suppose you rotate the area above the curve y= x3, from x= 0 to x= 1, about the line y= 2. What volume would that have?

Now subtract the two.

6. Jul 10, 2008

### c-murda

so do disk and washer rotate perpendicular to rotation of axis and shell rotates paralell

7. Jul 10, 2008

### dynamicsolo

That's correct, and that is why the variable of integration depends on which method is used.