two forms of calculate a volume I have trouble relating two forms of finding the volume of a cylinder. First of all you can imagine integrating the area of a disk along the height. Something like this, $V=\int_{0}^{h} \pi r^2 dx = \pi r^2h$. But also you can think of it as integrating the area of a rectangle ($hr$) along the "angles", so the volume would be $V=\int_{0}^{2\pi} hrd\theta=2\pi hr$. I know the last one is wrong. But I really don't see why. Hope you can help
 A: You can't really just integrate along the "angles" because it doesn't make much sense. The way we're visualizing the first one is going along the height and doing $\pi r^2dx$ as a small volume where $dx$ is a small amount of the height and adding up those small volumes from the bottom to the top. The second one is doing $hrd\theta$ where $hr$ is the area of a rectangle and you're multiplying it by a small angle $d\theta$. Adding up the area of a rectangle multiplied by a small angle doesn't really mean anything.
Also, the units don't make sense. Volumes are always in $length^3$ Angles have no dimensions but $r$ and $h$ both have units of length. Let's say meters. So in the first integral, we have units of $\pi r^2 dx$ being $meters^3$ since pi is dimensionless, $r^2$ is $meters^2$ and $dx$ is $meters$ so $meters^2 * meters = meters^3$, a volume. But the second integral has units of $meters^2$ since $h$ is in $meters$, $r$ is in $meters$ and $d\theta$ is dimensionless, we end up having $meters * meters = meters^2$ which isn't a volume at all. It's the dimensions of an area.
Instead, think about cutting the cylinder into very thin wedges, kinda like how you cut a cake. The volume of this wedge will be the height multiplied by the area of the triangle-like shape on top. Notice that the wedge is a very small portion, so we'll be adding all of the volumes of each wedge with an integral to get the total volume.
$$dV = h*dA$$
now if you want the area of that triangle-like portion, you'll see that its sorta like an isosceles triangle. We can call the angle at the top $d\theta$ since each angle is very small and will be adding up to the full $2\pi$. The height of the triangle is approximately $r$ and the base is $r * d\theta$.
Therefore the area should be $\frac{1}{2}r*rd\theta$ or $\frac{1}{2}r^2d\theta$. This gets closer to being true the thinner the wedge is. Now we can replace $dA$ with $\frac{1}{2}r^2d\theta$. Now we have a formula for $dV$ and we'll be able to put it in an integral. We know for the full cylinder, theta will go from $0$ to $2\pi$, so we'll put that in the integral:
$$V = \int_V dV = \int_A h * dA = \int_0^{2\pi} h * \frac{1}{2}r^2d\theta$$
which can be simplified to:
$$V = \frac{1}{2}\int_0^{2\pi}hr^2d\theta$$
This integral ends up being the volume of the cylinder.
Hope this helped!
A: In your second equation, your units are area units, $m^2$.
You should integrate over the volume of infinitesimal sectors.
$$V=\int_{0}^{2\pi} h \times \frac{r^2 \theta}{2}d\theta=\pi r^2h$$
