1. ## finding the volume

im supposed to find the volume of the solid that results when the region enclosed by y = x^(1/2), y = 0 and x = 9. the region is revolved about the y = 3 line?

well my intergral that i got for this problem is.

V = pi*(the intergral of) (9 - y^(2))^(2) can someone tell me if the set up is right? basically i took the length of the strip to be my radius and then squared it times pi? thanks in advance...

2. Since you are integrating with respect to y and the rotation is about a horizontal line (y=3), you should be using the cylindrical shells method.

Alternatively, you could leave the function defined as $\displaystyle y=\sqrt{x}$ and integrate with respect to x, and use the "discs" (aka "washers") method.

3. so is my setup right? v equals the intergral of y*(9-y squared) with limits running from 0 to 3?

4. Originally Posted by slapmaxwell1
so is my setup right? v equals the intergral of y*(9-y squared) with limits running from 0 to 3?
Since you are revolving about the line y=3, the radius will be 3-y, not y.

So:

$\displaystyle V = 2\pi \int_0^3 (3-y)(9-y^2)\,dy$