I am trying to solve the following integral:
$$\int_{-1}^{1}\int_0^{\sqrt{1 - y^2}} \int_{x^2 + y^2}^{\sqrt{x^2 + y^2}} xyz \, dz \, dx \, dy$$
I think that the best approach is to use cylindrical coordinates, But I'm now sure how to do that since this is one of the first exercises I make with cylindrical coordinates. How can I solve this?
Remarks:
You would first need to change the variables of integration to cylindrical coordinates using $ \ dx \ dy \ dz = r \ dr \ d\theta \ dz $.
Then you would also need to convert the function inside the integral to cylindrical coordinates using $x = r \cos\theta$, $y=r\sin\theta$, and $z = z$.
Finally, you will need to figure out the bounds for $r$, $\theta$, and $z$ in cylindrical coordinates for the region you are integrating over.
Solution:
First note that $x = r \cos\theta$, $y=r\sin\theta$, and $z = z$.
$I = \int_{y=-1}^{y=1} \int_{x=0}^{x=\sqrt{1-y^2}} \int_{z= x^2+y^2}^{z=\sqrt{x^2+y^2}} xyz \ dz \ dx \ dy$
Let's try to simplify the bounds of the integral using the conversion from cartesian to cylindrical coordinates.
$\int_{y=-1}^{y=1} \int_{x=0}^{x=\sqrt{1-y^2}} \int_{z=r^2}^{z=r}$.
Now, the bounds $\int_{y=-1}^{y=1} \int_{x=0}^{x=\sqrt{1-y^2}}$ strongly hint that the region they enclose is a semicircle in the x-y plane with equation $x = \sqrt{1-y^2}$ for $-1 \le y \le 1$. In cylindrical/polar coordinates this is $0 \le r \le 1$ and $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$.
So, overall the bounds become $\int_{\theta=-\frac{\pi}{2}}^{\theta=\frac{\pi}{2}} \int_{r=0}^{r=1} \int_{z=r^2}^{z=r}$.
Finally, we need to convert the argument of the integral into cylindrical coordinates using $x = r \cos\theta$, $y=r\sin\theta$, and $z = z$ and $ \ dx \ dy \ dz = r \ dr \ d\theta \ dz$.
So $I = \int_{-1}^1 \int_{0}^{\sqrt{1-y^2}} \int_{x^2+y^2}^{\sqrt{x^2+y^2}} xyz \ dz \ dx \ dy = \int_{\theta=-\frac{\pi}{2}}^{\theta=\frac{\pi}{2}} \int_{r=0}^{r=1} \int_{z=r^2}^{z=r} r^3z \cos\theta \sin\theta \ dz \ dr \ d\theta$.
So the integral simplifies to $I = \int_{\theta=-\frac{\pi}{2}}^{\theta=\frac{\pi}{2}} \int_{r=0}^{r=1} \int_{z=r^2}^{z=r} r^3z \cos\theta \sin\theta \ dz \ dr \ d\theta$