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

Question

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

anonymous · Answer

GT as in Georgia Tech?

anonymous · Answer

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

rogue · Answer

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.
$$V = \pi \int\limits_{-2}^{2} (x ^{2} - 4)^{2} dx$$

turingtest · Answer

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.

anonymous · Answer

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

anonymous · Answer

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

turingtest · Answer

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

turingtest · Answer

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