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I know how to integrate functions of the form $$ \iiint_\Omega f(r,\theta, \phi) \,dV,$$ where $dV = r^2 \sin \phi \,dr \,d \theta \,d \phi$ and $x=r \sin \phi \cos \theta$, $y=r \sin \phi \sin \theta$, $z=r \cos \phi$, but I don't know how to integrate

$$\iiint_{\Omega}{} \frac{1}{(1+z)^2} \,dx \,dy \,dz$$

How do I change $dx dy dz$ into something that's integrable with spherical polar coordinates? I differentiated $x=r \sin \phi \cos \theta$, $y=r \sin \phi \sin \theta$, $z=r \cos \phi$, each with respect to $r$, $\theta$, $\phi$, but couldn't find a connection. I appreciate your help.

Edit: $\Omega$ is defined by $\Omega := \left\{(x,y,z) \ | \ z \ge 0, x^2 +y^2 \le z^2 \le 1 - x^2 - y^2 \right\}$. I think this is what $\Omega$ looks like, the top half of the sphere with radius $1$.

enter image description here

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Does the assignment tell you to use spherical coordinates? If not, look at the Integral $\int \frac{1}{(1+z)^2}\; \text{d}x$. The Integrand is constant with respect to x. Same goes for the Integral over y. Only the last Integral has to be calculated – CBenni Jun 2 '13 at 13:11
@CBenni It's a past exam paper, and it asks to use spherical coordinates. Thanks for the observation though. If I could use that, it would've been much easier! – user4167 Jun 2 '13 at 13:13
@BabakS. Yes, $\Omega := \left\{(x,y,z) \ | \ z \ge 0, x^2 +y^2 \le z^2 \le 1 - x^2 - y^2 \right\}$. I know the constraint $z \ge 0$ and $z^2 \le 1 -x^2 - y^2$, but not sure what $x^2 +y^2 \le z^2$ represents though. – user4167 Jun 2 '13 at 13:55
@user4167: Have tried to draw the desired area? Did you get the limits? It seems easy to find the correct triple integral. – Babak S. Jun 2 '13 at 14:08
@BabakS. Yes, I just uploaded the picture. I'm not sure, but I think it is the top half of the sphere with radius $1$. I could do the first integration, but first I need to make sure that the limits are correct. Could you please check? Thank you. – user4167 Jun 2 '13 at 14:31
up vote 2 down vote accepted

The limits as you see are as follows:


enter image description here

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Thank you for the picture! – user4167 Jun 2 '13 at 14:46
You're Welcome. You did it by yourself before. :-) – Babak S. Jun 2 '13 at 14:51
Nice picture, indeed! +1 – amWhy Jun 2 '13 at 15:00
@amWhy: thanks Amy. – Babak S. Jun 2 '13 at 15:01
@amWhy: Red-correcting the assignments. 70 sheets. I wanna cry... – Babak S. Jun 2 '13 at 15:04

$\mathrm{d} V=\mathrm{d} x\mathrm{d} y\mathrm{d} z$

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Is that it? I thought it would be more complicated than that. – user4167 Jun 2 '13 at 13:11
@user4167 In your example, the flow is $dx\,dy\,dz=dV=r^2 \sin\phi \,dr\,d\theta\,d\phi$. – ˈjuː.zɚ79365 Jun 2 '13 at 13:24
@user79365 What do you mean by flow? Do you mind explaining further? Thank you. – user4167 Jun 2 '13 at 13:27
@user4167 Maybe not the best expression: I meant the process of computations. You are given an integral with $dx\,dy\,dz$. This is the same as $dV$. And $dV$ is the same as $r^2 \sin\phi \,dr\,d\theta\,d\phi$. And you know how to integrate with $r^2 \sin\phi \,dr\,d\theta\,d\phi$. Of course, $(1+z)^2$ in the denominator will be $(1+r\cos\phi)^2$. – ˈjuː.zɚ79365 Jun 2 '13 at 13:29
@user4167 Try this: Computing the Volume Element – ˈjuː.zɚ79365 Jun 2 '13 at 13:36

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