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:

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


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

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?

  • $\begingroup$ 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 $\;V/;$ ? $\endgroup$ – DonAntonio May 29 '17 at 19:13
  • $\begingroup$ @DonAntonio, I am sure that written $V$ is correct, but not sure about limits (that's why I am actually asking) $\endgroup$ – M.Mass May 29 '17 at 19:15
  • $\begingroup$ @M Ok, thanks.. $\endgroup$ – DonAntonio May 29 '17 at 19:16
  • $\begingroup$ Do you have the final answer to this? $\endgroup$ – DonAntonio May 29 '17 at 19:39
  • $\begingroup$ @DonAntonio, the correct one is $\frac{\pi}{10}$ But I got wrong one (posibly die to limits) $\endgroup$ – 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


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

and from here your integral is



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.

  • $\begingroup$ I think the integration is on the sphere $\;x^2+y^2+z^2=z\iff x^2+y^2+\left(z-\frac12\right)^2=\frac14\;$ ... $\endgroup$ – DonAntonio May 29 '17 at 19:09
  • $\begingroup$ @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). $\endgroup$ – Hans Lundmark May 29 '17 at 19:11
  • $\begingroup$ The question has the first equality: $\;x^2+y^2+z^2=\color{red}z\;$, not $\;1\;$ ... $\endgroup$ – DonAntonio May 29 '17 at 19:11
  • $\begingroup$ @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). $\endgroup$ – Hans Lundmark May 29 '17 at 19:12
  • $\begingroup$ 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. $\endgroup$ – DonAntonio May 29 '17 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.