Calculate $\iiint_D (x^2+y^2+z^2) \; dx dy dz$ where $D = \{(x,y,z) \in \mathbb{R}^3 \,|\, x^2+y^2+z^2 \leq 1, \, x^2+y^2 \leq z^2, \, z \geq 0 \}$

Question

Calculate $\int \int \int_D (x^2+y^2+z^2) \; dx dy dz$, where $D = \{(x,y,z) \in \mathbb{R}^3 \,|\, x^2+y^2+z^2 \leq 1, \, x^2+y^2 > \leq z^2, \, z \geq 0 \}$

I did it this way. I just want to make sure it is correct. Because while I was solving it, I had some doubts about $\rho$ limits when chaging to spherical coordinates.

$\int \int \int_D (x^2+y^2+z^2) \; dx dy dz = \int_0^{2 \pi} \int_0^{\frac{\pi}{4}} \int_0^1 \rho^4 \cdot \sin(\varphi) \; d \theta d \varphi d \rho = 2 \pi \cdot \frac{\left( 1-\frac{1}{\sqrt{2}}\right)}{5}$

Answer 1

Everything is correct. Everything should be clear from this picture

This picture represents the region over which we are integrating. It is now clear that the limits for r coordinates are (0,1), limits for $\phi$ coordinates are $(0,\frac{\pi}{4})$, while the integral is completely independent of $\theta$. So:

$$ \int \int \int_D (x^2+y^2+z^2) \; dx dy dz =2\pi \int_0^1 \left( \int_0^{\pi/4} \sin(\theta) d\theta \right) r^4 dr = \frac{2\pi}{5} \left(1-\frac{1}{\sqrt{2}} \right) $$ as the OP already obtained.

Answer 2

An alternative approach is using polar coordinates for $x$ and $y$. Advantage - no trig., disadvantage - messy arithmetic.

$I=2\pi \int_0^1 \int _0^{f(z)} (r^2+z^2)rdrdz$, where $f(z)=min(z,\sqrt{1-z^2}$

The $z$ integral is split into two parts at $z_0=\frac{1}{\sqrt{2}}$, with $f(z)=z$ for the first part and $f(z)=\sqrt{1-z^2}$ for the second.

Integrate $r$ and get $I=2\pi(\int_0^{z_0}\frac{3z^4}{4}dz+\int_{z_0}^1\frac{1-z^4}{4}dz)$

Finally after $z$ integration $I=\frac{2\pi}{5}(1-\frac{1}{\sqrt{2}})$

Note: the first integral is that over the cone and the second is that over the dome in the picture above (RedGiant).