1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $$ z_0=\frac{hr}a. $$

$\endgroup$
1
$\begingroup$

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 $c=\frac ha r$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.