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Double integral-Surface area

  1. Apr 27, 2009 #1
    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?

    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Apr 27, 2009 #2


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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.
  4. Apr 27, 2009 #3
    Awesome, I see. Thanks!
  5. Apr 28, 2009 #4


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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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