# Spherical coordinate integration help -- spheres.

$$L = \iiint_{D} \frac{1}{(x^2+y^2+z^2)^\frac{3}{2}}dxdydz$$ where D is a part of a sphere $$x^2+y^2+z^2 \leq 1$$ that lies in the first octant between the planes $$z=\frac{1}{2}$$ and $$z=\frac{1}{\sqrt 2}$$

We are asked to find the solution to this, in spherical coordinates and I have found a expression for $$L$$ but I'm not sure how to go about completing the solution.

My thoughts were to decompose the integral and use spherical coordinates and then express $$L$$ as the following.

$$L = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{4}} \int_{\frac{1}{2 \cos \phi}}^{\frac{1}{\sqrt 2 \cos \phi}} \rho^{1/2} \sin \phi d \rho d\phi d \theta + \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{4}} \int_{\frac{1}{2 \cos \phi}}^{1} \rho^{1/2} \sin \phi d \rho d\phi d \theta$$

Using geometry and my question is, is this correct? - I based it off a sketch off $$z$$ against $$r$$ and then saw we have to decompose this accordingly ^ and then I don't know how to complete the integral i.e. find out what the value of $$L$$ is.

Could someone help me with this?

• If you do not insist on spherical coordinates way, then more easy seems to consider plane $OYZ$ and calculating volume of revolving over $OZ$ figure. Jan 19, 2021 at 11:38
• The integrand is supposed to be $\rho^{-1}\sin\phi$ Jan 20, 2021 at 8:21

Asserting that $$(x,y,z)$$ belongs to the first octant means that $$x,y,z\geqslant0$$. Actually, asserting that $$z\geqslant0$$ is useless here, since we know that $$z\in\left[\frac12,\frac1{\sqrt2}\right]$$. So, we have

• $$x=\rho\cos(\theta)\sin(\phi)\geqslant0$$, meaning that $$\theta\in\left[-\frac\pi2,\frac\pi2\right]$$;
• $$y=\rho\sin(\theta)\sin(\phi)\geqslant0$$, meaning that $$\theta\in[0,\pi]$$;
• $$z=\rho\cos(\phi)\in\left[\frac12,\frac1{\sqrt2}\right]$$, meaning that $$\phi\in\left[0,\frac\pi2\right]$$, and that $$\frac1{2\cos(\phi)}\leqslant\rho\leqslant\frac1{\sqrt2\cos(\phi)}$$. But we also want to have $$\rho\leqslant1$$. So, if $$\phi\in\left[0,\frac\pi4\right]$$, $$\frac1{2\cos(\phi)}\leqslant\rho\leqslant\frac1{\sqrt2\cos(\phi)}$$, and if $$\phi\in\left[\frac\pi4,\frac\pi3\right]$$, $$\frac1{2\cos(\phi)}\leqslant\rho\leqslant1$$

So, the integral should be$$\begin{multline}\int_0^{\pi/2}\int_0^{\pi/4}\int_{1/(2\cos\phi)}^{1/\left(\sqrt2\cos\phi\right)}\rho^{1/2}\sin(\phi)\,\mathrm d\rho\,\mathrm d\phi\,\mathrm d\theta+\\+\int_0^{\pi/2}\int_{\pi/4}^{\pi/3}\int_{1/(2\cos\phi)}^1\rho^{1/2}\sin(\phi)\,\mathrm d\rho\,\mathrm d\phi\,\mathrm d\theta.\end{multline}$$

• But shouldn't we also enforce the condition $\rho \leq 1$? I don't see how this does that. Jan 19, 2021 at 12:54
• When you sketch it out, we have to use the $\rho = 1$ into account, which makes the integral into two parts?
– user876001
Jan 19, 2021 at 13:06
• @DavidK You are right, of course. I've edited my answer. Jan 19, 2021 at 13:10
• Why is it $\phi \in [\frac{\pi}{6}, \frac{\pi}{4}]$ rather than $\phi \in [\frac{\pi}{3}, \frac{\pi}{4}]$?
– user876001
Jan 19, 2021 at 15:12
• @Matt You're right, except that it should be $\left[\frac\pi4,\frac\pi3\right]$. Jan 19, 2021 at 15:17

I see two errors.

One error is that you let $$\phi$$ cover the same range of angles in both integrals. You have one integral that covers all the region in the first octant within the solid cone $$z^2 \geq x^2 + y^2$$ and between the planes $$z=\frac12$$ and $$z =\frac1{\sqrt2},$$ which covers part of the region you are trying to integrate. This is fine. But the second integral covers the region in the first octant within the same cone, above the plane $$z=\frac12$$ and within the sphere. This provides double coverage on the region you already integrated and includes a region above the plane $$z =\frac1{\sqrt2},$$ which should have been excluded.

The other error appears to be in converting the expression $$(x^2 + y^2 + z^2)^{3/2}.$$ Remember that in the conversion to spherical coordinates, $$\rho = (x^2 + y^2 + z^2)^{1/2}.$$ You seem to have taken $$\rho$$ to be equal to $$x^2 + y^2 + z^2$$ instead.

• That makes a lot of sense thanks
– user876001
Jan 19, 2021 at 15:12