Calculate the triple integral: $\iiint_W(x^2+y^2)z\ dx\,dy\,dz$

Question

I am trying to solve the following triple integral:

$$ \iiint_W(x^2+y^2)z\,dx\,dy\,dz \\ W=\{(x,y,z) \in \mathbb{R}: x^2+y^2+z^2 \le 9; x^2+y^2\le 1; x \ge 0; y \ge 0; z \ge 0\} $$

From which I know that there are two surfaces:

$$ x^2+y^2+z^3 = 9 \rightarrow \text{Sphere of radius 3} \\ x^2+y^2=1 \rightarrow \text{Cylinder with a bade radius of 1} $$

Converting this to Cylindrical Coordinates:

$$ \rho^2+z^2 \le 9 \implies \rho^2 \le 9-z^2 \implies \rho \le \sqrt{9-z^2} \\ \rho^2\le 1 $$

Where do I go from here?

Answer 1

Hint:

from $\rho^2\le 1$ you have the limits for $\rho$, that is : $0\le \rho \le 1$

so you can find the limits for $z$ that is: $0\le z \le \sqrt{9-\rho^2}$

Finally, from the condition that $x,y$ are positive, you have the limits $0\le \theta \le \frac{\pi}{2}$

Can you do from this?

Answer 2

The integral is getting calculated in 1st octant therefore: $$\iiint_W(x^2+y^2)z\,dx\,dy\,dz=\iiint_W R^5\sin^{3}\theta\cos\theta dR d\theta d \phi$$also we know that for $0<\theta<\theta_0$ where $\theta_0=\sin^{-1}\frac{1}{3}$ we have $0<R<3$ and for $\frac{\pi}{2}>\theta>\theta_0$ we have $0<R<\frac{1}{\sin\theta}$. So we can split the integral by two:$$I=\int_{0}^{\frac{\pi}{2}}\int_{\theta_0}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sin\theta}} R^5\sin^{3}\theta\cos\theta dR d\theta d \phi+\int_{0}^{\frac{\pi}{2}}\int_{0}^{\theta_0}\int_{0}^{3} R^5\sin^{3}\theta\cos\theta dR d\theta d \phi$$$$=\frac{1}{6}\int_{0}^{\frac{\pi}{2}}\int_{\theta_0}^{\frac{\pi}{2}}\frac{1}{\sin^{3}\theta}\cos\theta d\theta d \phi+\frac{243}{2}\int_{0}^{\frac{\pi}{2}}\int_{0}^{\theta_0}\sin^{3}\theta\cos\theta d\theta d \phi=\frac{3\pi}{16}+\frac{\pi}{3}=\frac{25\pi}{48}$$