Compute $\iiint x+y+z$ over the region inside $x^2+y^2+z^2 \le 1$ in the fist octant I feel this should be an easy question, but I seem to be struggling with it. So, I started by finding my bounds of integration. In this case, I get $0 \le x \le 1$, $0 \le y \le \sqrt{1-x^2}$ and $0 \le z \le \sqrt{1-x^2-y^2}$. Then, upon integrating, I get:
\begin{align}
= {} & \int_0^1 \int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}(x+y+z) \, dz \, dy \, dx \\[8pt]
= {} &  \int_0^1 \int_0^{\sqrt{1-x^2}}(\sqrt{1-x^2-y^2})(x+y) - \frac{1-x^2-y^2}{2} \, dy \, dx
\end{align}
From here the integration got pretty hairy, and using an online calculator the next inner integral with respect to $y$ resulted in imaginary numbers, which seems way too complex for a final answer of $\frac{3\pi}{16}$. My guess is my bounds of integration are wrong, but I'm not sure why or what the right ones should be.
Thanks!
 A: In spherical coordinates, that amounts to computing$$\int_0^{\pi/2}\int_0^{\pi/2}\int_0^1\bigl(\rho\cos(\theta)\sin(\varphi)+\rho\sin(\theta)\sin(\varphi)+\rho\cos(\varphi)\bigr)\rho^2\sin(\varphi)\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta.$$Can you take it from here?You can also do it in cylindrical coordinates:$$\int_0^{\pi/2}\int_0^1\int_0^{\sqrt{1-z^2}}\bigl(r\cos(\theta)+r\sin(\theta)+z\bigr)r\,\mathrm dr\,\mathrm dz\,\mathrm d\theta.$$
A: If $E$ is the region $x^2 + y^2 + z^2 \leq 1$ in first octant. Please note that due to symmetry,
$$I = \iiint_E (x+y+z) \ dV = 3 \iiint_E z \ dV$$
In spherical coordinates,
$x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi, z = \rho \cos\phi$
So $x^2 + y^2 + z^2 \leq 1 \implies \rho \leq 1$
As we are in first octant, $0 \leq \phi \leq \pi/2, 0 \leq \theta \leq \pi/2$
So, $ \displaystyle I = 3 \int_0^{\pi/2} \int_0^{\pi/2} \int_0^1 \rho^3 \cos\phi \sin\phi \ d\rho \ d\phi \ d\theta = \frac{3\pi}{16}$
Edit:
In polar coordinates,
$$x = r \cos\theta, y = r \sin\theta$$
$x^2 + y^2 + z^2 \leq 1 \implies z \leq \sqrt{1-r^2}$ in first octant.
Also, $0 \leq r \leq 1$ and $0 \leq \theta \leq \pi/2$.
So, integral is $I = 3 \displaystyle \int_0^{\pi/2} \int_0^1 \int_0^{\sqrt{1-r^2}} r \ z \ dz \ dr \ d\theta$
which is a straightforward integral.
