Finding the volume of a solid revolution (shell method)

Cottontails

Let f(x)=9-x^2. Let A be the area enclosed by the graph y=f(x) and the region y>=0.
Suppose A is rotated around the vertical line x=7 to form a solid revolution S.
So, using the shell method, I was able to find the indefinite integral used.
I found the shell radius to be (7-x) and the shell height to be (9-x^2)
Therefore, the volume is 2∏∫(7-x)(9-x^2)dx = 2∏∫(63-9x-7x^2+x^3)Δx
However, I am just unsure for what the interval should be. For whether it should be [-3,3] or [0,7]. So, which one is it?
I also tried using the disc method, with π ∫ [7² - (9 - y)] dy from 0 to 9 and got the answer 801∏/2. Is that right?
Any help is appreciated. Thanks.

elvishatcher

To find your limits, think about the description of the area A. It is the area enclosed by the function and the region y >= 0. Picture this area on a graph (make a sketch if it helps), and it should be pretty clear that the integral should go from x=-3 to x=3. Using the other method, your formulas for your radii aren't quite right. Your outer radius should be [tex]7+\sqrt{9-y}[/tex] and your inner radius should be [tex]7-\sqrt{9-y}[/tex]. Again, picture the situation on a graph (for this one, it might be especially helpful to actually draw a quick sketch) and you should see why you get that.