# Double integral-Surface area

## Main Question or Discussion Point

For a cylinder oriented along the z axis, of radius 3:

x2 + y2 = 9
-4 <= z <= 4

y(x, z) = ±√(9 - x2)

G(x, y, z) = ±√(9 - x2) - y(x,z) = 0

I am trying to set up a double integral to calculate surface area along the sides of the cylinder.

In rectangular coordinates:

How would I do this in cylindrical co-ordinates?

Thanks!

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HallsofIvy
Homework Helper
In cylindrical coordinates x2+ y2= 9 becomes just r2= 9 or r= 3.

The lateral surface of the cylinder is given by r= <x, y, z>= <3 cos(theta), r sin(theta), z>. rtheta= <-3sin(theta), 3cos(theta), 0>, and rz= <0, 0, 1>. The cross product of those is <3cos(theta), 3sin(theta), 0>, which has length 3, so the differential of surface area is 3 dtheta dz. Integrate that for theta from 0 to 2pi and z from -4 to 4.

Awesome, I see. Thanks!

HallsofIvy
A cylinder is, by the way, a "developable surface"- you could cut it along a vertical line and flatten it out. Since a circle of radus 3 have circumference $18\pi$, doing that to the cylinder from -4 to 4 would give a rectangle with width $18\pi$ and height 8. Its area is $144\pi$.