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Math Help - Surface integral

  1. #1
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    Surface integral

    Evaluate the surface integral:

    \iint_S (1+x^2+y^2)dS where S is part of the surface of the paraboloid z=1-x^2-y^2 in the second octant.

    ---------
    First of all, I hope the second octant is where x is negative but y and z are positive? (well, or 0).

    Second,

    Let f(x,y,z)=z+x^2+y^2-1 then f=0 and \nabla f = <2x,2y,1> which is normal to the surface S.
    EDIT: so |\hat{n}\cdot\hat{k}|=\frac{1}{\sqrt{4x^2+4y^2+1}}

    Then in polar coords, in the 2nd octant, we have:
    0\le r \le 1 since the shadow D on the xy-plane is a unit circle and,
    \pi/2 \le \theta \le \pi in the 2nd oct.

    So the integral is:
    \int_{\pi/2}^\pi \int_0^1 (1+r^2)\sqrt{4r^2+1} r dr d\theta. Right so far?

    Now the problem is, I don't see any simple way to integrate this... I don't think u-substitution will help me here, but could be wrong.
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  2. #2
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    Quote Originally Posted by scorpion007 View Post
    Evaluate the surface integral:

    \iint_S (1+x^2+y^2)dS where S is part of the surface of the paraboloid z=1-x^2-y^2 in the second octant.

    ---------
    First of all, I hope the second octant is where x is negative but y and z are positive? (well, or 0).

    Second,

    Let f(x,y,z)=z+x^2+y^2-1 then f=0 and \nabla f = <2x,2y,1> which is normal to the surface S.
    EDIT: so |\hat{n}\cdot\hat{k}|=\frac{1}{\sqrt{4x^2+4y^2+1}}

    Then in polar coords, in the 2nd octant, we have:
    0\le r \le 1 since the shadow D on the xy-plane is a unit circle and,
    \pi/2 \le \theta \le \pi in the 2nd oct.

    So the integral is:
    \int_{\pi/2}^\pi \int_0^1 (1+r^2)\sqrt{4r^2+1} r dr d\theta. Right so far?

    Now the problem is, I don't see any simple way to integrate this... I don't think u-substitution will help me here, but could be wrong.
    Because you have "r dr", the first thing I would do is let u= 4r^2+ 1 to simplify that square root. du= 8r dr or (1/8)d= rdr. \sqrt{4r^2+ 1}= u^{1/2}. Since 4r^2+ 1= u, 4r^2= u- 1 and r^2= u/4- 1/4, r^2+ 1= u/4+ 3/4.
    \int_0^1 (1+r^2)\sqrt{4r^2+ 1} rdr= \int_1^5(\frac{u}{4}+ \frac{3}{4})u^{1/2}du= \int_1^5(\frac{1}{4}u^{3/2}+ \frac{3}{4}u^{1/2})du
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  3. #3
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    ah, thank you. I think I can solve it now
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