Correct integration bounds and radius variable for voltage given charge density My professor worked out this problem and I'm confused with the use of radius.
The hemisphere has a radius $a$. The charge density if a function where $\rho = Ar\cos^2(\theta)$ using spherical coordinates. You solve for $q$, with $dq = \rho\ dV$. You place this value into voltage from a point charge and solve? There is a $r$ variable in the charge density, the equation for voltage from a point charge and from the spherical coordinates. Then you integrate with $dr$.
I'm confused because the $r$ from the point charge I thought was voltage as a function of distance away from the point charge. While the $r$ in spherical and charge density are as a result of the shape. Yet in the work they interact. I imagine the $r$ in point charge as a value from distance, while the $r$ in spherical and charge density shape the object.
Here is a picture of the solved problem. Thank you

 A: Yes, the voltage a distance $r$ from a point charge $q$ is $$V=\frac{1}{4\pi\epsilon_0}\cdot\frac{q}{r}$$ where we have set $V=0$ at $r\to\infty$.
Therefore, if you want to determine the potential at some location $\mathbf r$ of a charge distribution with volume charge density $\rho=\text dq/\text d\tau'$, we treat each little charge element $\text dq=\rho\,\text d\tau'$ located at postion $\mathbf r'$as a point charge. Then the potential at $\mathbf r$ due to this charge element is $$\text dV=\frac{1}{4\pi\epsilon_0}\cdot\frac{\text dq}{|\mathbf r-\mathbf r'|}=\frac{1}{4\pi\epsilon_0}\cdot\frac{\rho\,\text d\tau'}{|\mathbf r-\mathbf r'|}$$
so that the total potential at position $\mathbf r$ is just the total contribution of each charge element:
$$V=\iiint\frac{1}{4\pi\epsilon_0}\cdot\frac{\rho\,\text d\tau'}{|\mathbf r-\mathbf r'|}$$
You haven't specified, but it looks like the problem asked you to find the potential at the origin, so $\mathbf r=\bf 0$, and
$$V=\iiint\frac{1}{4\pi\epsilon_0}\cdot\frac{\rho\,\text d\tau'}{|-\mathbf r'|}=\iiint\frac{1}{4\pi\epsilon_0}\cdot\frac{\rho\,\text d\tau'}{r'}$$
which is what your work shows.
I think your issue here is that there are technically two positions to worry about in general. 1) The position where you are calculating the voltage at ($\mathbf r$), and 2) the position of each charge element that you are integrating over ($\mathbf r'$). Since you are looking at the voltage at the origin, we can get rid of $\mathbf r$, and then drop the prime on $\mathbf r'$ to arrive at what you have.
A: 
well my friend you seem to be committing a very trivial mistake in perceiving the $'r'$ in the charge density and the $'r'$ in the denominator of the potential as the same! so to avoid the confusion use different notation for the problem as described in the figure above !
