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Math Help - Integration

  1. #1
    s7b
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    Integration

    Please Help, I'm not sure how to set up this integration;

    Integrate the given function over the giver surface: h(x,y,z)=yz, over the part of the sphere x^2+y^2+z^2=4 that lies above the cone z=sqrt(x2+y2)
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  2. #2
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    Quote Originally Posted by s7b View Post
    Please Help, I'm not sure how to set up this integration;

    Integrate the given function over the giver surface: h(x,y,z)=yz, over the part of the sphere x^2+y^2+z^2=4 that lies above the cone z=sqrt(x2+y2)
    the surface is S: \ z=\sqrt{4 - x^2 - y^2} and to find the region of integration, D, we have to intersect S and the cone, which gives us x^2 + y^2 = 2. thus D=\{(x,y): \ x^2 + y^2 \leq 2 \}.

    now we have dS= \sqrt{z_x^2 + z_y^2 + 1} \ dx dy=\frac{2}{\sqrt{4-x^2-y^2}} \ dx dy=\frac{2}{z} \ dx dy. therefore the surface integral is: \int \int_D h(x,y,z) \ dS=2 \int \int_{x^2 + y^2 \leq 2} y \ dx dy=0.
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  3. #3
    s7b
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    can you convert into polar coordinates once you have the integrals of y dxdy to find the limits easier??
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    Quote Originally Posted by s7b View Post

    can you convert into polar coordinates once you have the integrals of y dxdy to find the limits easier??
    sure you can or just use the symmetry and the fact that the integrand is an odd function.
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