Limits of a triple integral when finding the volume of a cone I was trying to evaluate the volume of a cone with height $h$ and width $2a$. My limits were as follows (which are wrong):
$$0 \le r \le a $$
$$0 \le \theta \le 2\pi$$
$$0 \le z \le h$$
And the main flaw is that the lower bound for $z$ should be $\frac{h}{a}r$ instead of $0$. Now, this makes since to me since my logic would be scaling a circle up in $z$ forming a cylinder. Here, we need to have the radius of the circle being scaled up to decrease with height. So, why does  $\frac{h}{a}r$ satisfy this?
 A: You need to think of the "side" of the cone (when you see it as a triangle) as a line that is $0$ when $r=0$ and $a$ when $r=h$. For any $r$, the side of the cone is at a height $z_0$; and the square triangle with sides $h$ and $a$ will be similar to the square triangle with sides $z_0$ and $r$. So 
$$
\frac ha=\frac{z_0}r, 
$$
and so 
$$
z_0=\frac{hr}a. 
$$
A: Draw the cone point down, with its vertex at the origin.
The points of the cone all have $0\le\theta\le 2\pi$.
Now that you've made that decision, you don't need to think about $\theta$ anymore.  So from now on, just look at a general cross-section of the cone in which $\theta$ is constant.  No matter what value of $\theta$ you choose, this cross-section is a right triangle in the $rz$-plane with height $h$ and base $a$.
Here's a picture: I've shaded the right triangle that I'm talking about.  (Sorry: I don't have enough reputation to embed images.)
Image: triangular cross-section
The points of this triangular  cross-section all have $0\le r\le a$.
Now you don't have to think about $r$ anymore, so from now on, just look at a cross-section of the triangle in which $r$ is constant.  No matter what value of $r$ you choose, this is a vertical segment, as shown in the image above.
The points of that vertical segment all have $c\le z\le h$, and we need to find $c$.  There are similar triangles in the figure, which you can use to show that $\frac cr=\frac ha$, and therefore $c=\frac ha r$.
