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"Integrate $G(x,y,z) = x^2$ over the unit sphere $x^2 + y^2 + z^2 = 1.$"

Alright, so using the formula

$\int \int_R x^2 \sqrt{ 1+ \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}} dA$

and working it down, I arrive at the integral:

$\int \int_R \frac{x^2}{z} dA$

When I try to convert this to polar coordinates, it becomes very messy, and when I look at the given solution they convert it to spherical coordinates. This is very odd to me, as I was under the impression that spherical coordinates were only used in a triple integral. The "official" conversion results in:

$\int_o^{2\pi} \int_o^{\pi} sin^3 (\phi) cos^2(\theta)d\phi d\theta$

And I have no idea how they arrived at that. If anyone could point me in the right direction here, I'd greatly appreciate it.

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    The exact wording is "integrate the given function over the given surface," and the section is titled Surface Integrals, so pretty sure that it is indeed the sphere surface. – 2017-02-02

2 Answers 2

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Usually spherical coordinates are for volumes. But if you are on the surface of the sphere, you set $\rho = 1$

$x = \cos\theta \sin \phi\\ y = \sin\theta \sin \phi\\ z = \cos \phi$

$ dy\;dx= \| (\frac {dx}{d\phi},\frac {dy}{d\phi}, \frac {dz}{d\phi}) \times (\frac {dx}{d\theta},\frac {dy}{d\theta}, \frac {dz}{d\theta})\|d\phi\;d\theta$

$dy\;dx = \sin\phi \;d\phi\;d\theta$

$\iint (\cos\theta \sin \phi)^2 \sin\phi \;d\phi\;d\theta$

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    Ohhhhhh, I see, thank you! Both good answers, but yours is more straightforward. – 2017-02-02
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Parametrize with spherical coordinates

Note $\rho \sin (\phi)=r$ and $\rho \cos (\phi)=z$. This gives that,

$$\vec r(\phi,\theta)=\langle 1\sin (\phi) \cos (\theta), 1\sin (\phi) \sin (\phi), 1 \cos (\phi) \rangle$$

Now we need to compute $|r_\phi \times r_\theta|$. Lucky for me already know if there was a $\rho$ instead of $1$ above I would get $\rho^2 \sin (\phi)$ because that is the Jacobian associated with a change to spherical coordinates. So we get $1^2 \sin (\phi)$. Hence $dS=1^2 \sin (\phi) dA$.

So then

$$\iint_{S} x^2 dS=\iint (\sin (\phi) \cos (\theta))^2 1^2 \sin (\phi) dA$$

Of course $\theta \in [0,2\pi]$ and $\phi \in [0,\pi]$.