Find the surface area of the part of the sphere x^2+y^2+z^2=36 that lies above t

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Explanation / Answer

Here, we only need the upper hemisphere: z = v(36 - x^2 - y^2). Note that the cone and sphere intersect at z^2 + z^2 = 36 ==> z = v18 ==> The region of integration is x^2 + y^2 + 18 = 36 ==> x^2 + y^2 = 18. So via Cartesian Coordinates, the surface area equals ?? v[1 + (z_x)^2 + (z_y)^2] dA = ?? v[1 + (-x/v(36 - x^2 - y^2))^2 + (-y/v(36 - x^2 - y^2))^2] dA = ?? v[1 + (x^2 + y^2)/(36 - x^2 - y^2)] dA = ?? v[36/(36 - x^2 - y^2)] dA = ?? 6 dA/v(36 - x^2 - y^2). Converting to polar coordinates yields ?(? = 0 to 2p) ?(r = 0 to v18) 6r dr d?/v(36 - r^2) = 2p ?(r = 0 to v18) 6r(36 - r^2)^(-1/2) dr = 2p * -3 * 2v(36 - r^2) {for r = 0 to v18} = 12p (6 - 3v2) = 36p (2 - v2).

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