1. ## volume rev.

The region R is bounded by y= -(x-2)^2 and y=-9. There is a number k, k < -9, such that when R is revolved about the line y=k, the resulting solid has the same volume as the when the region R is revolved about the x-axis. Set up, but don't solve, the integral expression that can be used to find k. I am having trouble with the k integrand part. help would be appreciated. Thanks!!

2. The volume when revolved about the x-axis is

${\pi}\int_{-1}^{5}\left[(-9)^{2}-(-(x-2)^{2})^{2}\right]dx=\frac{1944\pi}{5}$

Now, can you come up with an integral where you revolve it around y=k.

Set it up, integrate, then set equal to the above result and solve for k.

It is the same when y = -54/5 = -10.8