Volumes Generated by Revolving the Area Bounded by x=y^2 and x=4

[gigi9]

Can someone please 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 please fix it for me...Thanks!

 

[HallsofIvy]

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= √(x) and from 2 to y= -&radic(x): that is, the two circles have radius 2-x^1/2 and 2+x^1/2. The larger circle has area π(2+x^1/2)²= π(4+ 2x^1/2+x). The smaller circle has radius π(2-x^1/2)²= π(4- 2x^1/2+x).
The difference between those two areas is 4π x^1/2. 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 25π and the area of the smaller circle is π(y²+1)²= &pi(y⁴+ 2y²+ 1).
The difference between those two areas is π(24- y⁴- 2y²). That's what you need to integrate.

 

[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).