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sorry, this is not about flux integration... but surface area! sorry about the title!

Find the surface area of the part of the sphere [tex]x^2+y^2+z^2=36 [/tex] that lies above the cone [tex]z=\sqrt{x^2+y^2}[/tex]

[tex]z=\sqrt{36-x^2-y^2} [/tex]

[tex]A(S)=\int\int_D \sqrt{1+\left( \frac{\partial z}{\partial x} \right) ^2 + \left( \frac{\partial z}{\partial y} \right) ^2 } dA[/tex]

[tex] \frac{\partial z}{\partial x} =\frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2x \right) [/tex]

[tex] \frac{\partial z}{\partial y}=\frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2y \right) [/tex]

[tex]A(S)=\int\int_D \sqrt{1+\left( \frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2x \right) \right) ^2 + \left( \frac{1}{2}\left( 36-x^2-y^2 \right) ^{\frac{3}{2}} \left( -2y \right) \right) ^2 } dA[/tex]

is this correct so far? how would I find the ends of integration for a cone?

I've taken the liberty of changing the title of this thread. Since it wasn't even "flux integration" it was bothering me!

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# Homework Help: Fux integral

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