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I need to calculate $$\iint_D \sqrt{x^2+y^2+z^2} dx dy dz$$ where $D=\{ (x,y,z):x^2+y^2+z^2\leq z\}$ .

After substituting $x=r\cos\theta\sin\phi , y=r\sin\theta\sin\phi , z=r\cos\phi $ into the inequality $x^2+y^2+z^2\leq z$, I received that $0\leq r \leq\cos \phi$ so as far as I understand this $\phi$ should be in $[-\frac{\pi}{2} , \frac{\pi}{2}] $ . The problem is that when I calculate it with these boundaries I get the integral is zero, and when I calculate it for $\phi \in [0 , \frac{\pi}{2}] $ and multiply by $2$, I get $\frac{\pi}{10}$.

So:

Why is it not correct to take $\phi \in [-\frac{\pi}{2} , \frac{\pi}{2}] $?

Thanks in advance.

user91500
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homogenity
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1 Answers1

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Note that $\phi$ is commonly defined in $[0,\pi]$, not $[-\pi/2,\pi/2]$. Therefore, since $\cos \phi >0$, your integral should be only over $[0,\pi/2]$: $$2\pi \int\left(\int_0^{\cos\phi} r^3 \sin \phi dr \right)d \phi = \pi/ 10$$

Nathaniel Bubis
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