Double integral-Surface area

  • #1
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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:

http://image.cramster.com/answer-board/image/cramster-equation-20094271716356337644939541787501925.gif

http://image.cramster.com/answer-board/image/cramster-equation-20094271718496337644952983975006480.gif

How would I do this in cylindrical co-ordinates?

Thanks!
 
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Answers and Replies

  • #2
HallsofIvy
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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.
 
  • #3
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Awesome, I see. Thanks!
 
  • #4
HallsofIvy
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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 [itex]18\pi[/itex], doing that to the cylinder from -4 to 4 would give a rectangle with width [itex]18\pi[/itex] and height 8. Its area is [itex]144\pi[/itex].
 

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