Given the volume shown below, I would like to determine the shaded area (area of the circle) which varies linearly as a function of the height $l$ of the object between radius $a$ and radius $b$. I can obtain the expression using basic geometry but I wondered if I could do this using a surface integral?
The area of a circle is $\pi r^2$ and using the formula for a straight line I obtain the expression for the area of the circle as $$\pi(a+\frac{b-a}{l}z)^2$$
I want to get this expression using calculus. I note that the unit vector normal to the surface is in the $\hat{Z}$ direction and therefore I pick $dS = rdrd\phi$
The surface integral for the area is $$A = \int_{s} ds = \int_{a}^{b} rdr \int_{0}^{2\pi} d\phi $$
The result of which is
$$A = \pi(b^2 – a^2)$$
This however doesn't take into account the height of the volume. What did I do incorrectly?
Best Answer
Your limits for $r$ are $a$ to $b$. Instead they should be $0$ to $(a+(b-a)z/l)$.