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Surface Area of 3d graph

  1. Jul 7, 2008 #1
    Find the area of the sphere x^2 + y^2 + (z-2)^2 = 4 that lies inside paraboloid z = x^2 + y^2

    2. Relevant equations

    3. The attempt at a solution

    when i take the equation of the spere and replace x^2 + y^2 with z i get: z(z-3) = 0
    so they intersect at the plane z = 3.

    were supposed to use the double integration rule. im not sure what the parameters would be. when i convert to polar, i know feta is from 0 to 2pie. i tried making r from 0 to 3 but i got the wrong answer. idk whether i integraded the wrong equation or used the wrong parameters or both
  2. jcsd
  3. Jul 7, 2008 #2


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    How are we supposed to know what you did wrong if you won't tell us what you did. What did you integrate? What did you get?
  4. Jul 7, 2008 #3
    i dont know how to type the integration symbols

    the double integral was ||Rx X Ry||
    where x = x, y = y, z = x^2 + y^2

    so basically (skipping some steps) the integral becomes: (1 + 4x^2 + 4y^2)^.5
    which becomes: r(1 + 4r^2)^.5 when put in polar form
    and like i said, i tried 0<feta<2pi and 0<r<3

    i got like pi/6 * some quantity. but its supposed to be 4pi
  5. Jul 7, 2008 #4


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    It looks like you are finding the area of the paraboloid that lies in the sphere. The question asks for the area of the sphere that lies inside the paraboloid. It's the upper cap of the sphere.
  6. Jul 7, 2008 #5


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    You might also notice z=r^2 on the paraboloid. The limits for z are 0->3. The limits for r are not 0->3.
  7. Jul 7, 2008 #6


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    Hi jaredmt! :smile:

    (have a theta: θ and a pi: π and a squared: ² and an integral: ∫)
    That's right! :smile:

    So you're now trying to find the area of a cap of a sphere of radius 2 from "height" 2 to "height" 1 (and you're told to use ∫∫).
    Hint: you can use either x and y parameters, or latitude and longitude parameters.

    They both work (well, why wouldn't they? :rolleyes:), so you may as well try both of them! :smile:

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