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  1. #1
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    Surface Integral Qu

    Hi, I am new to this forum and would be grateful if anyone could help me in this surface integral question.

    Find the area of the surface of the cone z^2=3(x^2+y^2) cut out by the paraboloid z=x^2+y^2.

    The answer is 6pi but I'm not sure how exactly to get this.
    Here is my working

    We are trying to find (int z) dS

    z=sqrt(3x^2+3y^2). So dz/dx=(3x)/(sqrt(3x^2+3y^2) and dz/dy=(3y)/sqrt(3x^2+3y^2). Using the formula dS= sqrt((dz/dx)^2+(dz/dy)^2+1), we are left with

    Double integral sqrt(3x^2+3y^2)*2 dxdy. In polar coordinates this gives

    Double integral sqrt(3r^2)*2r drdtheta. The radius is sqrt3, so the first integral is from 0 to sqrt3.

    The problem is that I thought the second integral is from 0 to 2pi. But this gives a final answer of 12pi. So the second integral must be from 0 to pi.

    Can anyone explain why this would be the case, or have I made a mistake somewhere else?
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  2. #2
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    Quote Originally Posted by Jtown View Post
    Hi, I am new to this forum and would be grateful if anyone could help me in this surface integral question.

    Find the area of the surface of the cone z^2=3(x^2+y^2) cut out by the paraboloid z=x^2+y^2.

    The answer is 6pi but I'm not sure how exactly to get this.
    Here is my working

    We are trying to find (int z) dS

    z=sqrt(3x^2+3y^2). So dz/dx=(3x)/(sqrt(3x^2+3y^2) and dz/dy=(3y)/sqrt(3x^2+3y^2). Using the formula dS= sqrt((dz/dx)^2+(dz/dy)^2+1), we are left with

    Double integral sqrt(3x^2+3y^2)*2 dxdy. In polar coordinates this gives

    Double integral sqrt(3r^2)*2r drdtheta. The radius is sqrt3, so the first integral is from 0 to sqrt3.

    The problem is that I thought the second integral is from 0 to 2pi. But this gives a final answer of 12pi. So the second integral must be from 0 to pi.

    Can anyone explain why this would be the case, or have I made a mistake somewhere else?
    The surface area is given by \int \int dS, NOT \int \int z \, dS. Therefore:

    Surface area = 2 \int \int_{R_{xy}} \, dx \, dx

    where the region R_{xy} of integration in the xy-plane is bounded by the circle x^2 + y^2 = 3.
    Last edited by mr fantastic; July 28th 2008 at 10:44 PM.
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