swap integral to polar

Question

Trying to swap to polar and solve the following double integral, but I am not getting the same answer.
$
displaystyleint_{-3/4}^{3/4} int_{-sqrt{3/4-x^2}}^{sqrt{3/4-x^2}} 1/2left(3-4sqrt{x^2+y^2}right) dy dx
$
Here is the conversion into polar:
$$
int_0^{2pi} int_0^{3/4} rrdrdtheta = 9pi/32
$$
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 $z= sqrt{x^2+y^2}$

lonza leggiera · Answer

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, $ rlefrac{3}{4} $ in polar coordinates, or $ sqrt{x^2+y^2}lefrac{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} $.

Answered by lonza leggiera on January 1, 2021

swap integral to polar

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