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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

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  • $\begingroup$ You want to use calculus to get the same answer you learned in high school geometry (or earlier)? Start with an integral that gives the circumference (arclength of a circle). $\endgroup$ – hardmath Dec 6 '14 at 5:20
  • $\begingroup$ The curved surface area = height of the cylinder x circumference of the base circle $\endgroup$ – user_of_math Dec 6 '14 at 5:21
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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 $A = 2\pi\cdot4 = 8\pi$.

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    $\begingroup$ Thanks, I was thinking of volume when I checked with the simple geometry to get $4\pi$. $\endgroup$ – user197985 Dec 7 '14 at 2:45

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