Practice exam for Midterm 3 in Math 241 Important note Several of the problems ask you to completely setup but not evaluate a certain integral This means that all the limits of integration are specified and the integrand is in terms of the final variables For example if S is a surface in R3 then an acceptable answer for setting up Z 2 Z 1 v RR 2 u2 v sin u du dv S x y dA would be something like 0 0 1 Let R be the ZZ region shown at right Evaluate y dA 7 points y 1 1 R x 2y 2 1 0 y x R x 2 Consider the solid region E in the positive octant cut off by x y z 1 Completely setup but do not evaluate a triple integral which computes the volume of E 6 points 3 Let R be the region shown at right y v T 1 1 1 1 S R u 1 1 x a Find a transformation T R2 R2 taking S 0 1 0 1 to R 4 points Z b Use your change of coordinates to evaluate x y 2 dA via an integral over S 6 points R Emergency backup transformation If you can t do a pretend you got the answer T u v uv v and do part b anyway 4 Let C be the oriented curve in R2Zshown at right For the vector field F x y x 3 x 2 y use Green s theorem to evaluate F dr 6 points C C 2 1 x 5 Let E be the portion of the positive octant which is inside the unit sphere Use spherical coordiZZZ nates to completely setup but not evaluate the integral x z dV 6 points E 6 Let S be the surface in R3 parameterized by r u v v cos u v sin u u for 0 u and 1 v 1 a Circle the correct picture of S 2 points ZZ y dA b Completely setup but do not evaluate the integral 6 points S c Find the equation for the tangent plane to S at the point 0 0 2 in R3 3 points 7 For each surface S below give a parameterization r D S Be sure to explicitly specify the domain D a The portion of the cylinder x 2 z2 4 where 3 y 3 4 points z x y b The ellipsoid x2 z2 y2 1 4 9 3 points z 0 0 3 0 1 0 2 0 0 x y
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