1. ## Converting Coordinate Systems

Bit stuck on converting from Cartesian to Spherical:

Let $S$ be the spherical surface defined by $x^2 + y^2 + z^2 = a^2$, and let $\mathbf{F} = z^3 \mathbf{k}$.

Using SPs $(r,\theta,\phi)$, and with $\mathbf{\delta S} = a^2 \sin{\theta} \delta \theta \delta \phi \mathbf{\hat{r}}$

i) Show $\mathbf{F} \cdot \mathbf{\delta S} = a^5 \cos^4{\theta} \sin{\theta} \delta \theta \delta \phi$

ii) Calculate $\int_S \mathbf{F} \cdot \mathbf{dS}$.

i)Well converting $\mathbf{F}$ to Spherical, we get $(z^3,0,0)$, so the scalar product gives:

$z^3(a^2 \sin{\theta} \delta \theta \delta \phi)$.

Now we also know that $z = a \cos{\theta}$, either from using $z = a^2 - (x^2 + y^2)$, or from the definition of SPs.

So substituting this in gives:

$\cos^3{\theta} a^3(a^2 \sin{\theta} \delta \theta \delta \phi)$

Which simplifies to:

$a^5 \cos^3{\theta} \sin{\theta} \delta \theta \delta \phi$

I'm just at a bit of a loss where they get their extra $\cos{\theta}$ from?

ii) Integrating this gives us $a^5 \int^{2 \pi}_0 d\phi \int^{pi}_0 \cos^4{\theta} \sin{\theta} d\theta$

$\frac{-2 \pi a^5}{5} \left[\cos^5{\theta} \right]^{\pi}_0$

Which gives $\frac{4 \pi a^5}{5}$, would this be right?

Thanks in advance for any help

2. Would the first bit involving the extra $\cos{\theta}$ be because:

$\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos{\theta}$?

I think this would explain where they've got the extra $\cos{\theta}$ from?