Volume enclosed by paraboloid and plane

Question

The volume, $V$, enclosed by paraboloid $z=x^2 + y^2$ and the plane $z=3-2y$ can be expressed as a triple integral. Determine the limits describing the enclosed volume. By evaluating the integral, show the volume is $8\pi$.$\\$

There is then a hint that $\cos^4\theta = \frac{1}{8}\cos4\theta + \frac{1}{2}\cos2\theta + \frac{3}{8}$.

I worked out the limits in Cartesian as $V=\int_{y=-3}^{y=1}{\int_{x=-\sqrt{4-(y+1)^2}}^{x=\sqrt{4-(y+1)^2}}{\int_{z=x^2+y^2}^{z=3-2y}dz}dx}dy$

This looks pretty hard to compute though, so I then made the polar limits as $V=\int_{R=0}^{R=2}{\int_{\theta=0}^{\theta=2\pi}{\int_{z=R^2}^{z=3-2Rsin\theta}R dz}d\theta}dR$,

but when I compute this I don't get $8\pi$. I'm wondering if this is me making a calculation error, or if the limits are incorrect. I also never used the hint, so was wondering if anyone could shine a light on when that comes in handy.

Thanks!

Answer 1

If u integrate what u got in cartesian coordinates, u get the answer

Given integral = $\int_{y=-3}^1(1-\frac{2}{3}y-\frac{y^2}{3})4 \sqrt {4-(y+1)^2} dy $

Now put $y+1 = 2 \sin t$, we get

given integral = $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(\frac{4}{3}-\frac{8}{9} \sin t - \frac{4}{9} \sin^2 t)16 \cos^2 t dt \\ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{64}{3} \cos^4 t dt = 8 \pi$

Answer 2

Consider that $$(3 - 2 y) - (x^2 + y^2) = 4 - x^2 - (y + 1)^2,$$ therefore we're integrating over a disk of radius $2$. In cylindrical coordinates $(x, y, z) = (r \cos t, -1 + r \sin t, z)$, the difference between the $z$ coordinates of the surfaces is $4 - r^2$, and $$V = \int_0^{2 \pi} \int_0^2 (4 - r^2) r dr dt = 8 \pi.$$