prac exam 2B solutions

Thus hz z c and putting everything together we obtain

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Unformatted text preview: k into f , we get: f (x, y, z ) = ex yz + y 2 z + h(z ). Differentiating with respect to z and comparing, we get: fz = ex y + y 2 + h′ (z ) = ex y + y 2 + 1, so h′ (z ) = 1. Thus h(z ) = z + c, and putting everything together we obtain: f (x, y, z ) = ex yz + y 2 z + z + c. z 7. φ= π 6 a) S is part of the sphere x2 + y 2 + z 2 = 4, so its normal vector points 1 ˆ radially outwards, straight away from the origin. So n = 2 x, y, z (the factor 1 is there because | x, y, z | = 2 at all points of S ), and 2 S 2r 1 x, y, z z2 =. 2 2 Using the spherical angles φ, θ to parametrize...
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