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Math Help - Surface Integral Help

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Surface Integral Help

    It's been 2 years since I've looked at this stuff, and I'm trying to make sense of all of this. How would I evaluate:

    \iint_\sigma \bigg(x^2+y^2\bigg)z \,dS

    Where \sigma is the portion of the cone z=\sqrt{x^2+y^2} between the planes z=1 and z=2.

    My biggest problem is setting it up properly.

    Any help would be appreciated.

    --Chris
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Chris L T521 View Post
    It's been 2 years since I've looked at this stuff, and I'm trying to make sense of all of this. How would I evaluate:

    \iint_\sigma \bigg(x^2+y^2\bigg)z \,dS

    Where \sigma is the portion of the cone z=\sqrt{x^2+y^2} between the planes z=1 and z=2.

    My biggest problem is setting it up properly.

    Any help would be appreciated.

    --Chris
    Convert to cylindrical polars, then the area element of the cone in question is \sqrt{2}r d\theta dr and the integral becomes:

    I=\sqrt{2} \int_{r=1}^2 \int_{\theta=0}^{2 \pi} r^4 ~d\theta ~dr

    RonL
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