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Let $r>0$ and $S_r=\{(x,y,z)\in\mathbb{R}^3|x^2+y^2+z^2=r^2\}$.

Show that $$\int_{S_r}\cos x\cos y\cos z\ dS=\int_{S_r}\cos (x+y+z)\ dS$$ and find the value of the integrals.

I know that we can parametrize $S_r$ and try to solve the integrals by hand or try to expand the integrands using trigonometric identities, but I had no success in doing so.

I am asking for a hint on how to do this. Moreover, is there some way to handle with integrals over spheres that involve "cosines and sines" (before parametrization). What I mean by this is that when the integrand does not involve "cosines and sines", parametrization often helps but in a situation like the above I don't know what to do.

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  • $\begingroup$ Could you perhaps do something with the divergence theorem? Make each integrand on the form $\vec F(x, y, z)\cdot \vec n$ where $\vec n$ is the unit normal vector of the sphere and $\vec F$ is a differentiable vector field defined on all of $\Bbb R^3$. If you're clever enough with what fields you choose, that may make the calculation relatively simple. $\endgroup$
    – Arthur
    Feb 7, 2018 at 9:16
  • $\begingroup$ @Arthur But the unit normal vector is $\vec{n}(x,y,z)=\frac{1}{r}(x,y,z)$. So, how am I supposed to 'rewrite' $\cos(x+y+z)$ in terms of a flux without dividing by $x$ for example (which makes the field not differentiable). $\endgroup$ Feb 7, 2018 at 9:50
  • $\begingroup$ If I had known that, I would've written it as an answer instead of a comment. There might not be a nice $\vec F$, and then my idea doesn't work. $\endgroup$
    – Arthur
    Feb 7, 2018 at 9:52

2 Answers 2

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Note that $$\eqalign{&\cos x\cos y\cos z\cr &\qquad={1\over4}\bigl(\cos(x+y+z)+\cos(x+y-z)+\cos(y+z-x)+\cos(z+x-y)\bigr)\ .\cr}$$ Both statements then follow from the observation that $$\int_{S_r}\cos\bigl({\bf u}\cdot{\bf x})\>dS=\int_{-r}^r\cos\bigl(|{\bf u}|\>z)\>2\pi r\>dz={4\pi r\over\sqrt{3}}\>\sin\bigl(\sqrt{3}\, r\bigr)\ .$$ Here we have used that the area of a sphere between two parallel planes is equal to the area cut out by these planes from the cylinder tangent to the sphere along the corresponding equator.

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As a partial answer I can show the first equation is valid:

On the upper surface of the sphere, $$\vec r=\langle x,y,z\rangle=\langle x,y,\sqrt{r^2-x^2-y^2}\rangle$$ So $$d\vec r=\langle 1,0,\frac{-x}{\sqrt{r^2-x^2-y^2}}\rangle dx+\langle 0,1,\frac{-y}{\sqrt{r^2-x^2-y^2}}\rangle dx$$ And then $$\begin{align}d\vec A&=\pm\langle 1,0,\frac{-x}{\sqrt{r^2-x^2-y^2}}\rangle dx\times\langle 0,1,\frac{-y}{\sqrt{r^2-x^2-y^2}}\rangle dx\\ &=\pm\langle\frac{x}{\sqrt{r^2-x^2-y^2}},\frac{y}{\sqrt{r^2-x^2-y^2}},1\rangle dx\,dy\\ &=\langle\frac{x}{\sqrt{r^2-x^2-y^2}},\frac{y}{\sqrt{r^2-x^2-y^2}},1\rangle dx\,dy\end{align}$$ Because the outward normal points up here. Then $$d^2A=\left|\left|d^2\vec A\right|\right|=\frac{r\,dx\,dy}{\sqrt{r^2-x^2-y^2}}$$ On the lower surface of the sphere, $$\vec r=\langle x,y,z\rangle=\langle x,y,-\sqrt{r^2-x^2-y^2}\rangle$$ And we can work out $$d^2A=\left|\left|d^2\vec A\right|\right|=\frac{r\,dx\,dy}{\sqrt{r^2-x^2-y^2}}$$ Just as it was on the upper surface.

Start with the right hand side: $$\begin{align}\int\int_{S_r}\cos(x+y+z)d^2A&=\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\left[\cos\left(x+y-\sqrt{r^2-x^2-y^2}\right)\right.\\ &+\left.\cos\left(x+y+\sqrt{r^2-x^2-y^2}\right)\right]\frac{r\,dy}{\sqrt{r^2-x^2-y^2}}dx\end{align}$$ The outer $(x)$ integral has a domain that is a circle parallel to the $yz$-plane and the inner integral just needs the two points that have the same $x$ and $y$. Now we can expand the cosines using the angle sum formula to get $$\begin{align}\int\int_{S_r}\cos(x+y+z)d^2A&=\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}2\cos(x+y)\cos\left(\sqrt{r^2-x^2-y^2}\right)\frac{r\,dy}{\sqrt{r^2-x^2-y^2}}dx\\ &=\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\cos(x+y)\cos\left(-\sqrt{r^2-x^2-y^2}\right)\frac{r\,dy}{\sqrt{r^2-x^2-y^2}}dx\\ &+\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\cos(x+y)\cos\left(\sqrt{r^2-x^2-y^2}\right)\frac{r\,dy}{\sqrt{r^2-x^2-y^2}}dx\\ &=\int\int_{S_r}\cos(x+y)\cos z\,d^2A\end{align}$$ Let's set up the same integral with $x$ outer, $z$ middle and $y$ inner: $$\begin{align}\int\int_{S_r}\cos(x+y)\cos z\,d^2A&=\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\left[\cos\left(x-\sqrt{r^2-x^2-z^2}\right)\cos z\right.\\ &+\left.\cos\left(x+\sqrt{r^2-x^2-z^2}\right)\cos z\right]\frac{r\,dz}{\sqrt{r^2-x^2-z^2}}dx\\ &=\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}2\cos x\cos\left(\sqrt{r^2-x^2-z^2}\right)\cos z\frac{r\,dz}{\sqrt{r^2-x^2-z^2}}dx\\ &=\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\cos x\cos\left(-\sqrt{r^2-x^2-z^2}\right)\cos z\frac{r\,dz}{\sqrt{r^2-x^2-z^2}}dx\\ &+\int_{-r}^r\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\cos x\cos\left(\sqrt{r^2-x^2-z^2}\right)\cos z\frac{r\,dz}{\sqrt{r^2-x^2-z^2}}dx\\ &=\int\int_{S_r}\cos x\cos y\cos z\,d^2A\end{align}$$ EDIT: Now we have to actually finish up the integral. We start out with the right hand side and transform to $$\begin{align}u&=\frac1{\sqrt3}(x+y+z)\\ v&=\frac1{\sqrt6}(-2x+y+z)\\ w&=\frac1{\sqrt2}(z-y)\end{align}$$ This is just a rigid rotation about the origin, so $$\begin{align}\int\int_{S_r}\cos(x+y+x)d^2A&=\int\int_{S_r}\cos\left(\sqrt3u\right)d^2A\\ &=\int_{-r}^r\int_{-\sqrt{r^2-u^2}}^{\sqrt{r^2-u^2}}2\cos\left(\sqrt3u\right)\frac{r\,dv}{\sqrt{r^2-u^2-v^2}}du\\ &=\int_{-r}^r\int_{-\pi/2}^{\pi/2}2\cos\left(\sqrt3u\right)\frac{r\sqrt{r^2-u^2}\cos\theta\,d\theta}{\sqrt{r^2-u^2}\cos\theta}du\\ &=2\pi r\int_{-r}^r\cos\left(\sqrt3u\right)du\\ &=\frac{2\pi r}{\sqrt3}\left.\sin\left(\sqrt3u\right)\right|_{-r}^r\\ &=\frac{4\pi r}{\sqrt3}\sin\left(\sqrt3r\right)\end{align}$$ EDIT: Pity that factor of $\frac r{\sqrt{r^2-u^2-v^2}}$ in the area element destroyed my beautiful solution with the Bessel functions and left us with this prosaic answer.

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