Setting up spherical integrals

answered 2014-07-25 18:04:23 -0600

Gear Junky
278 ●2 ●7 ●13

Ok, so here's my whack at it while building off of Christina's input. Like Christina said $\phi$ is the angle that is measured from the $+z$-axis, having no correlation with the $xy$-plane. And also like Christina said, $\theta$ is measured off of the $+x$-axis when looking onto the $xy$-plane. The question from what I remember stated to set up the integral in spherical coordinates with $x^2+y^2+z^2=1$ and $y>0$ of the function:$$\int\int\int yDV$$ And as Professor McClure talked about in class:$$x^2+y^2+z^2=\rho^2$$$$x=\rho\cos(\theta)\sin(\phi)$$$$y=\rho\sin(\theta)\sin(\phi)$$$$z=\rho\cos(\phi)$$Setting $x^2+y^2+z^2=\rho^2$ we get $\rho^2=1$, which is $\rho=\sqrt{1}$ or $\rho=1$. Knowing that $\rho=1$ and that there is no inner restriction $\rho$ has to go from $0$ to $1$. And seeing as $\phi$ is only relevant to any restrictions on the $z$-axis, $\phi$ has to go from $0$ to $\pi$. Then for $\theta$ we can note that it stated that $y>0$, meaning that $\sin(\theta)$ should never be negative (At least this is how I interpreted it, I could be wrong). So referring back to the ancient unit circle we can see that $\sin(\theta)$ is positive from $0$ to $\pi$ and we get that we need to integrate with respect to $\theta$ from $0$ to $\pi$. Also to note that the Cartesian $y$ is equal to $\rho\sin(\theta)\sin(\phi)$, having an end result of:$$\int_0^{\pi}\int_0^{\pi}\int_0^1 \rho\sin(\theta)\sin(\phi)\rho^2\sin(\phi)\delta \rho \delta \phi \delta \theta$$ Or at least this is how I understand it to be.