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I need to compute: $\displaystyle\int \int \int z dxdydz$

over the domain: $\left\{x^2+y^2+z^2\leqslant 16,z\geqslant 0\right\}$

Im trying to use spherical coords as:

\begin{equation} \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{4} r \cos(\theta )r^2\sin(\varphi ) \;dr\,d\theta \,d\varphi \end{equation}

It gives me $0$ as result. But i know that is wrong, the same formula computed with Mathematica software returns $64 \pi$.

{x, y, z} = 
  r {Cos[ϕ] Sin[θ], Sin[ϕ] Sin[θ], Cos[θ]};

Integrate[
 z Abs[Det[D[{x, y, z}, {{r, θ, ϕ}}]]], 
   {r, 0, 4}, {θ, 0, π/2}, {ϕ, 0, 2 Pi}]

My procedure is:

$$\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{4} r \cos(\theta )r^2\sin(\varphi ) \;dr\,d\theta \,d\varphi=\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{4} r^3 \cos(\theta )\sin(\varphi ) \;dr\,d\theta \,d\varphi=$$ $$=\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}}64 \cos(\theta ) \sin(\varphi ) d\theta d\varphi = \int_{0}^{2\pi} -64 \sin(\varphi ) d\varphi = 64 \cos(2\pi ) - 64 \cos(0)=0$$

EDIT: I see the name of the angles can be confusing, so this is mi current notation:

$x=r \sin(\theta ) \cos (\varphi )$

$y=r \sin(\theta ) \sin (\varphi )$

$z=r \cos(\theta)$

$\varphi$ should be the azimuthal angle, and $\theta$ the polar angle

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  • 1
    $\begingroup$ This will be zero because sine is integrated over its period. Could you please designate the task more precisely? $\endgroup$
    – Тимофей Ломоносов
    Jun 28, 2014 at 4:53
  • $\begingroup$ Even WolframAlpha agrees with the result of 0, are you sure that's supposed to be $\sin{\varphi}$ and not $\sin{\theta}$? $\endgroup$
    – Silynn
    Jun 28, 2014 at 4:53
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    $\begingroup$ You appear to be using $ \ \theta \ $ as the "polar angle" and $ \ \phi \ $ as the "azimuthal angle". So the factor $ \ \sin \ \phi \ $ ought to be $ \ \sin \ \theta \ $ . (After your edit: Yes, if you have $ \ z \ $ as $ \ r \ \cos \ \theta \ $ , then that factor with sine is incorrect and you will get a non-zero result if you are integrating over a hemisphere. [Were you to integrate over the full sphere, the integral would be zero again...] ) $\endgroup$
    – colormegone
    Jun 28, 2014 at 4:55
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    $\begingroup$ Yes. The integral is "separable" as $$ \int_{0}^{2\pi} d\varphi \ \int_{0}^{\frac{\pi}{2}} \cos(\theta )\sin(\theta) \ d\theta \ \int_{0}^{4} r^3 \ dr\ \ \ . $$ You will now obtain $ \ 64 \pi \ $ . $\endgroup$
    – colormegone
    Jun 28, 2014 at 5:33
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    $\begingroup$ It would seem so -- this happens to people a lot because of the differing notations for the spherical angles. So beware... $\endgroup$
    – colormegone
    Jun 28, 2014 at 5:43

1 Answer 1

Reset to default
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The problem is in (1) where $\sin\varphi$ should be $\sin\theta$.

$$\int_{0}^{2\pi}\sin\theta d\varphi \int_{0}^{\frac{\pi}{2}}d\theta \int_{0}^{4}r^2dr (r \cos\theta ) $$

$$=\int_{0}^{2\pi} d\varphi \int_{0}^{\frac{\pi}{2}}\sin\theta \cos\theta d\theta \int_{0}^{4}r^3dr =2\pi*(1/2)* 64=64\pi$$

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