Question Find the surface area of that portion of the sphere x^2 + y^2 + z^2 = a^2 that is above xy-plane and within the cylinder x^2 + y^2 = b^2 , 0 < b < a Solution.. i try to find fx and fy.. how am i going to proceed?
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You'll obtain $\displaystyle A=a\iint_{D}\dfrac{dxdy}{\sqrt{a^2-x^2-y^2}}$ with $\displaystyle D\equiv x^2+y^2\leq b^2$ now, use polar coordinates.
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