# Finding Volume

1. Nov 30, 2003

### gigi9

Can someone plz help me w/ these problems below:
Find the volume generated by revolving the area abounded by x=y^2 and x=4 about
a)the line y=2
b)the line x= -1
***I tried to write out the integral not sure if it's correct:
a) V=pi* int. of (sqrt(x)^2-(2-sqrt(x))^2) dx
**integral form 0->4 ??
b) V=pi* int. from -2->2 of (4-y^2)-1^2 dy ??
If the integrals is wrong plz fix it for me...Thanks!

2. Dec 1, 2003

### HallsofIvy

Staff Emeritus
For both of these think of a "cross section" as being the area between two circles- a "washer".

For (a) (rotated around y=2), you are measuring the radius in the y direction from 2 to y= &radic;(x) and from 2 to y= -&radic(x): that is, the two circles have radius 2-x1/2 and 2+x1/2. The larger circle has area &pi;(2+x1/2)2= &pi;(4+ 2x1/2+x). The smaller circle has radius &pi;(2-x1/2)2= &pi;(4- 2x1/2+x).
The difference between those two areas is 4&pi; x1/2. That's what you need to integrate.

For (b) (rotated around x= -1), the radius is measured in the x direction from -1 to x= y2 and from -1 to x= 4. The two circles have radius y2+1 and 4+1= 5. The area of the larger circle is 25&pi; and the area of the smaller circle is &pi;(y2+1)2= &pi(y4+ 2y2+ 1).
The difference between those two areas is &pi;(24- y4- 2y2). That's what you need to integrate.

3. Dec 1, 2003

### PrudensOptimus

I am new to volumes too, what is this cross section?

And can all general objects' volumes be regarded as Washer and another category?(forgot what the other one was).