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Math Help - Surface area of a sphere

  1. #1
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    Surface area of a sphere

    Find the area of the portion of the sphere of radius 2 (centered at the origin) that is in the cone z > \sqrt {x^2+y^2}

    What I have so far is that:

    A(S)= \int  \int \int \sqrt {1+ \frac {x^2}{x^2+y^2}+\frac {y^2}{x^2+y^2} } dA

    =  \int  \int \int \sqrt {2} dA
    = \sqrt {2} \int ^2 _0 \int ^{2 \pi } _0 \int ^ { \pi } _0 p^2 sin \phi d \phi d \theta dp
    Then I have  \frac {16 sqrt {2} \pi }{3} But is wrong, what is wrong?

    Thank you.
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  2. #2
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    A(S)= \int  \int \int \sqrt {1+ \frac {x^2}{x^2+y^2}+\frac {y^2}{x^2+y^2} } dA
    It is not true I believe. It should be a double integral since you have dA.
    Take it from here, I think all should go well.
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  3. #3
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    maybe you could try to find the cone generating angle \theta_0 and then calculate the integral with  \theta=0..\theta_0.

    no need include anything else in the integral besides dA
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
    Find the area of the portion of the sphere of radius 2 (centered at the origin) that is in the cone z > \sqrt {x^2+y^2}

    What I have so far is that:

    A(S)= \int \int \int \sqrt {1+ \frac {x^2}{x^2+y^2}+\frac {y^2}{x^2+y^2} } dA

    =  \int \int \int \sqrt {2} dA
    = \sqrt {2} \int ^2 _0 \int ^{2 \pi } _0 \int ^ { \pi } _0 p^2 sin \phi d \phi d \theta dp
    Then I have  \frac {16 sqrt {2} \pi }{3} But is wrong, what is wrong?

    Thank you.
    Quote Originally Posted by arbolis View Post
    It is not true I believe. It should be a double integral since you have dA.
    Take it from here, I think all should go well.
    In addition, I think you should be using

    z = \sqrt{4-x^2-y^2} in your surface area formula

     <br />
\iint_R \sqrt{1 + z_x^2 + z_y^2}\,dA.<br />
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  5. #5
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    As Danny says, you should write the equation of the sphere as z = \sqrt{4-x^2-y^2}. To find the region over which to integrate, you need to know where the sphere meets the cone, namely where \sqrt{x^2+y^2} = \sqrt{4-x^2-y^2} or in other words x^2+y^2=2. So the integral will be over the region x^2+y^2\leqslant2, and the surface area is given by the integral \iint_{x^2+y^2\leqslant2}\sqrt{1+z_x^2+z_y^2}dxdy = \iint_{x^2+y^2\leqslant2}\frac2{\sqrt{4-x^2-y^2}}dxdy . This cries out for conversion to polar coordinates, and it then becomes \int_0^{2\pi}\!\!\int_0^{\sqrt2}\!\!\frac2{\sqrt{4-r^2}}\,rdrd\theta. From there on, it's easy.
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  6. #6
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    What about a shpere with radius 5 in the same question?

    Quote Originally Posted by Opalg View Post
    As Danny says, you should write the equation of the sphere as z = \sqrt{4-x^2-y^2}. To find the region over which to integrate, you need to know where the sphere meets the cone, namely where \sqrt{x^2+y^2} = \sqrt{4-x^2-y^2} or in other words x^2+y^2=2. So the integral will be over the region x^2+y^2\leqslant2, and the surface area is given by the integral \iint_{x^2+y^2\leqslant2}\sqrt{1+z_x^2+z_y^2}dxdy = \iint_{x^2+y^2\leqslant2}\frac2{\sqrt{4-x^2-y^2}}dxdy . This cries out for conversion to polar coordinates, and it then becomes \int_0^{2\pi}\!\!\int_0^{\sqrt2}\!\!\frac2{\sqrt{4-r^2}}\,rdrd\theta. From there on, it's easy.
    What about if the sphere were radius 5? I seem to be goofing it up just a little somewhere when converting to polar/spherical coordinates. Also, it seems as though the Jacobian should be (p^2)sin(phi) instead of just r because this should be changed to polar and not cylindrical coordinates... right? Any feedback, insight or explanation of this integral problem would be fantastic as I am seriously stuck and puzzled by this problem. Thanks.
    Last edited by alan collins; August 2nd 2009 at 11:09 AM.
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  7. #7
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    I think it would be easier to do the all thing in spherical coordinates.


    Notice that if the height of the cone is z (a generic point z), and the radius of the base is \sqrt{x^2+y^2}, so the generating angle for the cone, \theta_0, is given by tan \theta_0=\frac{\sqrt{x^2+y^2}}{z}

    as z=\sqrt{x^2+y^2}, tan \theta_0=1 and \theta_0=\frac{\pi}{4}

    this makes the integral for the area

     \int_0^{2\pi}\int_0^{\frac{\pi}{4}}R^2sin\theta d\theta d\phi

    it seems much more simpler to me, and no need to change the coordinates in between. the R being 2 or 5 makes no difference you just replace it in the integral as a constant.
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