# Solve $\iiint_\limits{V}\sqrt{x^2+y^2+z^2}\;\mathrm{d}x\;\mathrm{d}y\;\mathrm{d}z$ using spherical coordinates

I just need more eyes to check if I got idea right:

The integral is:

$$\iiint_\limits{V}\sqrt{x^2+y^2+z^2}\;\mathrm{d}x\;\mathrm{d}y\;\mathrm{d}z$$ where $V$ is restricted by surface $x^2+y^2+z^2=z$

So coordinates are:

$$\begin{cases}x = \rho\cos\varphi\sin\theta \\ y = \rho\sin\varphi\sin\theta \\ z = \rho\cos\theta \\ |J| = \rho^2\sin\theta \end{cases}$$

and limits:

$$\begin{cases}\rho \ge 0 \\ 0 \le \varphi \le 2\pi \\ 0 \le \theta \le \color{red}{\frac{\pi}{2}}\end{cases}$$

Therefore simplyfing undersquare expression first:

$x^2+y^2+z^2 = \rho^2\cos^2\varphi\sin^2\theta + \rho^2\sin^2\varphi\sin^2\theta+\rho^2\cos^2\theta = \ldots = \rho^2$

Therefore:

$$\iiint_\limits{V}\sqrt{\rho^2} \cdot \rho^2\sin\theta \; \mathrm{d}\rho \; \mathrm{d}\varphi \; \mathrm{d}\theta = \iiint_\limits{V} \rho^3\sin\theta \; \mathrm{d}\rho \; \mathrm{d}\varphi \; \mathrm{d}\theta$$

as a result we get three integrals:

$$\int_0^1 \rho^3\sin\theta \; \mathrm{d}\rho \int_0^{2\pi} \mathrm{d}\varphi \int_0^\pi \mathrm{d}\theta = \ldots$$

Last three integrals are easy to calculate, questions:

1.) Did I find limits right ?

2.) Do I understand correctly that surface (sphere in this case) is usually given to find appropriate substitutions limits?

• If you take $\;\rho\ge\;$ , it means the upper limit is $\;\infty\;$ . This doesn't look correct...are you sure of the sphere that delimits $\;V/;$ ? – DonAntonio May 29 '17 at 19:13
• @DonAntonio, I am sure that written $V$ is correct, but not sure about limits (that's why I am actually asking) – M.Mass May 29 '17 at 19:15
• @M Ok, thanks.. – DonAntonio May 29 '17 at 19:16
• Do you have the final answer to this? – DonAntonio May 29 '17 at 19:39
• @DonAntonio, the correct one is $\frac{\pi}{10}$ But I got wrong one (posibly die to limits) – M.Mass May 29 '17 at 19:55

It is clear the azimut is $\;0\le\phi\le2\pi\;$ and $\;0\le\theta\le\pi/2\;$ ( as the radius "sweeps" the whole part of the solid over the plane $\;z=0\;$) . Now, we have on our solid that

$$\rho^2\cos^2\phi\sin^2\theta+\rho^2\sin^2\phi\sin^2\theta+\left(\rho\cos\theta-\frac12\right)^2\le\frac14\iff$$

$$\rho^2\sin^2\theta+\rho^2\cos^2\theta-\rho\cos\theta+\frac14\le\frac14\iff \rho(\rho-\cos\theta)\le0\implies0\le\rho\le\cos\theta$$

and from here your integral is

$$\int_0^{2\pi}\int_0^{\pi/2}\int_0^{\cos\theta}\rho^3\sin\theta\,d\rho\,d\theta\,d\phi=\frac\pi2\int_0^{\pi/2}\cos^4\theta\sin\theta\,d\theta=\left.-\frac\pi2\frac15\cos^5\theta\right|_0^{\pi/2}=\frac\pi{10}$$

No, the limits of integration that you have written correspond to the ball $x^2+y^2+z^2 \le 1$ which is centered at the origin and has radius $1$.

But the region $V$ that you are given is actually a ball centered at $(0,0,1/2)$ with radius $1/2$, since the inequality $x^2+y^2+z^2 \le z$ can be rewritten as $x^2+y^2+(z-1/2)^2 \le (1/2)^2$.

So your limits are not correct.

See this answer to a similar question for hints on how to proceed instead.

• I think the integration is on the sphere $\;x^2+y^2+z^2=z\iff x^2+y^2+\left(z-\frac12\right)^2=\frac14\;$ ... – DonAntonio May 29 '17 at 19:09
• @DonAntonio: I don't think I understand what you mean. It's a triple integral, so it's over a three-dimensional body, namely the ball bounded by that sphere (which is what I wrote). – Hans Lundmark May 29 '17 at 19:11
• The question has the first equality: $\;x^2+y^2+z^2=\color{red}z\;$, not $\;1\;$ ... – DonAntonio May 29 '17 at 19:11
• @DonAntonio: Yes, that's of course why I said “no”, meaning “the limits of integration that you have written are not correct” (as an answer to question number 1). – Hans Lundmark May 29 '17 at 19:12
• I've no idea what you're talking about: I haven't written any integration limits at all... I'm only saying that this question has no unit sphere as integration solid. – DonAntonio May 29 '17 at 19:13