## anonymous 4 years ago Find the volume of the solid generated by revolving the described region about the given axis: The region bounded above by the line y=4 , below by the curve y=x^2 , rotated along the following lines: y=4

1. anonymous

GT as in Georgia Tech?

2. anonymous

$\Pi \int\limits\limits\limits_{-2}^{2}(x^2-4)^2$

3. Rogue

If you think about it for it a bit, its going to be the volume of the solid generated by revolving $y = x ^{2} - 4$ about the x - axis. The equation crosses the x-axis positive and negative 2, so those are your limits of integration. This is a disk, which has circular partitions. If we add up the area of all the circular partitions, we can get the volume. $A = \pi r ^{2}$ The integral of area is volume. Your radius is x^2 - 4. So now just do the integration. $V = \pi \int\limits_{-2}^{2} (x ^{2} - 4)^{2} dx$

4. TuringTest

nash and rogue are right just to clean this up a little we can note that the integrand is even, so$\pi\int_{-2}^{2}(x^2-4)^2dx=2\pi\int_{0}^{2}(x^4-8x^2+16)dx$which is a very straightforward integration.

5. anonymous

Thank you guys, my final answer was 512/15pi, which was right!

6. anonymous

So what about with rotation y=8, what would the limits be?

7. TuringTest

same area bounded by y=x^2 and y=4 about y=8 ?

8. TuringTest

...or is it now the area bounded between y=x^2 and y=8 about y=8 ?