Trying to swap to polar and solve the following double integral, but I am not getting the same answer.
$ \displaystyle\int_{-3/4}^{3/4} \int_{-\sqrt{3/4-x^2}}^{\sqrt{3/4-x^2}} 1/2\left(3-4\sqrt{x^2+y^2}\right) dy dx $
Here is the conversion into polar: $$ \int_0^{2\pi} \int_0^{3/4} rrdrd\theta = 9\pi/32 $$ Not sure if this is correct?
The above are an intermediate step which may be incorrect. The original problem tries to find the volume bounded between the following:
$cone: z = 2 - \sqrt{x^2+y^2}$, top of hyperboloid: $z= \sqrt{1+x^2+y^2}$
$cylinder: (x-1)^2+y^2 = 1$, plane $z=0$, cone $z= \sqrt{x^2+y^2}$
The region of integration in your Cartesian coordinate double integral does not correspond to the one in your polar coordinate double integral. The latter—which I suspect is the one you actually want—is the disc, $\ r\le\frac{3}{4}\ $ in polar coordinates, or $\ \sqrt{x^2+y^2}\le\frac{3}{4}\ $ in Cartesian coordinates. However, if that is the region you want to integrate over, then the limits on your inner integral in Cartesian coordinates aren't correct. They should be $\ -\sqrt{\frac{9}{16}-x^2}\ $ and $\ \sqrt{\frac{9}{16}-x^2}\ $.