how do you find the surface area of a cylinder using integrals
with height of 4 and radius of 1?
I really want to find the surface area of just the side of the cylinder, not the top and bottom
calculus
how do you find the surface area of a cylinder using integrals
with height of 4 and radius of 1?
I really want to find the surface area of just the side of the cylinder, not the top and bottom
Best Answer
If you'd really want to go all out, let the cylinder represented in cylindrical coordinates $(r,\theta,z)$ where $r$ is the radius from the $z$-axis, $\theta$ is the azimuthal angle. Now the surface area of a small element of the cylinder will be given by $dA = rd\theta dz$. We seek to integrate around the cylinder $0\leq \theta \leq 2\pi$ and $0\leq z\leq 4$ with a fixed radius $1$. The area of the cylinder is then the integral, \begin{equation} \iint_AdA = \int_{0}^4\int_{0}^{2\pi} d\theta dz = 8\pi \end{equation} as required.
From basic geometry, the surface area is $A = 2\pi\cdot4 = 8\pi$.