Question

rewrite the integral in spherical and cylindrical coordinates

Answer 1

\[\int\limits_{-3}^{3}\int\limits_{-\sqrt{9-y^2}}^{0}\int\limits_{\sqrt{x^2+y^2}}^{\sqrt{18-x^2-y^2}} xy dz dx dy\]

Answer 2

The points on the \(y\) axis indicate the limits for \(y\), of course, with \(-3\le y\le3\). The limits on \(x\) indicate that it is ranging from the lower half of the circle (marked with arrows) in the \(x\)-\(y\) plane (which has radius 3) to the \(y\) axis (the line \(x=0\)). The third set of limits indicate that \(z\) ranges from the greenish surface of the cone to the hemisphere. The point here is that we're restricted to the third and fourth quadrants of the \(x\)-\(y\) plane, and the octants of the \(x\)-\(y\)-\(z\) space above these quadrants.

Answer 3

The conversion to cylindrical is the simpler one. The range on \(z\) stays the same.

Here \(r\) is ranging from the origin to the edge of the white circle, and so \(0\le r\le3\).

\(\theta\) is restricted to the third and fourth quadrants, which would indicate \(\pi\le \theta\le2\pi\), but due to symmetry we can use \(0\le \theta\le\pi\) (or \(-\dfrac{\pi}{2}\le\theta\le\dfrac{\pi}{2}\), or any interval of length \(\pi\) for that matter).

Converting to cylindrical, you have
\[\begin{cases}
x^2+y^2=r^2\\
x=r\cos\theta\\
y=r\sin\theta\\
z=z\\
dx~dy~dz=r~dr~d\theta
\end{cases}\]
and so the integral is
\[\int_0^{\pi}\int_0^3\int_{\sqrt{r^2}}^{\sqrt{18-r^2}}r^3\sin\theta\cos\theta~dz~dr~d\theta=\int_0^{\pi}\int_0^3\int_{r}^{\sqrt{18-r^2}}r^3\sin\theta\cos\theta~dz~dr~d\theta\]

Answer 4

In cylindrical, you have \(\theta\) ranging over the same interval. The other angle, \(\phi\), starts at the \(z\) axis and ends at the intersection of the cone and hemisphere, which occurs at half a right angle, so \(0\le\phi\le\dfrac{\pi}{4}\).

\(\rho\) ranges from the origin to the surface of the sphere, which has radius \(\sqrt{18}\).

Converting to spherical, you have
\[\begin{cases}
x^2+y^2+z^2=\rho^2\\x=\rho\sin\phi\cos\theta\\
y=\rho\sin\phi\sin\theta\\
z=\rho\cos\phi\\
dx~dy~dz=\rho^2\sin\phi~d\rho~d\theta~d\phi
\end{cases}\]
So you have
\[\int_0^\pi\int_0^{\pi/4}\int_0^\sqrt{18}\rho^4\sin^3\phi\sin\theta\cos\theta ~d\rho~d\phi~d\theta \]

Answer 5

thank you so much @SithsAndGiggles everything is so clear now.