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Math 5C Solutions to Midterm 1 Review Problems Winter 2007 Z yz dx 2x dy y dz where C is the straight line path from 1 2 1 to 1 Compute C 1 3 0 Solution The straight line path between the two points has parametric equations x y z 2t 1 t 2 t 1 Thus using the definition of line integrals the integral becomes Z 1 Z 1 2t2 t dt t 2 1 t 2 2 1 2t t 2 1 dt 0 3 0 2t 3 t2 2 10 1 6 2 Find the surface area of the surface S which is parametrized by x u v u v y u v u v u v z u v uv for all u v with u2 v 2 1 RR Solution We use the formula S A EG F 2 du dv Here E x2u yu2 zu2 2 2 v F xu xv yu yv zu zv 1 1 vu uv and G x2v yv2 zv2 2 u2 Thus EG F 2 4 2 u2 v 2 and Z Z p S A 4 2 u2 v 2 du dv Z u2 v 2 1 2 Z 1 4 2r2 r dr d 0 Z 0 1 2 2 4 2r2 3 2 10 d 4 0 3 3 2 6 43 2 6 6 8 3 3 3 Let S be the top half of the unit sphere i e S is given by x2 y 2 z 2 1 and z 0 oriented by the outer normal Integrate Z Z x dy dz y dz dx z 2 dx dy S Solution S has parametrization x cos sin y sin sin and z cos for 0 2 and 0 2 Notice that this parametrization induces the inner normal on the sphere and so we need to multiply the integral by 1 when we write it in terms of the parameters and 1 We now compute the Jacobians we need these are also the components of the normal vector dy dz cos sin sin cos d d cos sin2 d d 0 sin dz dx 0 sin d d sin sin2 d d sin sin cos cos dx dy sin sin cos cos d d sin cos d d cos sin sin cos Thus Z Z 2 Z Z x dy dz S 0 Z 2 cos2 sin3 sin2 sin3 sin cos3 d d 0 2 sin2 cos3 sin d 2 0 Z 2 2 1 cos2 cos3 sin d 0 2 cos 1 11 1 2 cos3 cos4 0 3 4 6 4 Let S be the surface given by z xy 2 3x2 with upper normal n over the square with vertices 1 1 in the xy plane If w z 3x2 i yzj y 2 k calculate Z Z w n d S Solution Z Z Z 1Z 1 z z wx w n d wy wz dx dy x y S