MA 242 Test 3 Solutions ZZ 1 10 points If f is a constant function f x y k and R a b c d compute kdA R Which theorem was used in this problem Solution Z Z dZ b kdxdy k c a c d Z b dy dx k y dc x ba k d c b a a We had to use Fubini s Theorem to compute this integral Z 1 1 Z 2 15 points Evaluate the integral y 0 p x3 1dxdy by reversing the order of integration Solution If we look at the integral in its given Type II form we see that D x y 0 y 1 y x 1 This region can also be described in Type I form In Type I form D x y 0 x 1 0 y x2 Z 1 Z Then 0 1 p x3 Z 1 Z x2 1dxdy y 0 First compute the inner integral 0 R x2 0 p x3 1dydx y x2 x3 1dy y x3 1 y 0 x2 x3 1 Next compute the outer integral via u substitution choose u x3 1 then du 3x2 dx Z 1 p 2 2 3 2 1 x2 x3 1dx x3 1 3 2 x 1 x 0 2 9 9 0 3 15 points Find the moment of inertia about the x axis for the lamina that occupies the region D x y 0 y ex 0 x 1 and has the density function x y y Solution ZZ The moment of inertia about the x axis is Ix y 2 x y dA D Z ex Compute the inner integral 1 0 4 y 3 dy 0 Next compute the outer integral Z 1 4 Z 0 1 y ex 0 e4x 4 4 1 e4x dx 1 4x 1 1 4 e 0 e 1 16 16 Z 0 ex y 3 dydx 4 10 points Set up the triple integral in rectangular coordinates that is needed to find the volume of the solid enclosed by the cylinder x2 y 2 9 and the planes y z 5 and z 1 Set up E as a type I solid region and set up D as a type I plane region Do not compute this integral Solution ZZZ First note that V E dV E The solid region E is of type I and D its projection onto the xy plane is a type I region D is circle of radius 3 centered at the origin So E x y z 5 y z 1 9 x2 y 9 x2 3 x 3 Z 3 Then V 3 Z 9 x2 5 y Z dzdydx 9 x2 1 ZZZ 5 10 points Set up the triple integral in cylindrical coordinates that would be used to evaluate x2 dV E where E is the solid that lies within the cylinder x2 y 2 1 above the plane z 0 and below the cone z 2 4x2 4y 2 Do not compute this integral Solution In cylindrical coordinates E is bounded by the cylinder r 1 the plane z 0 and the cone z 2r So E r z 0 2 0 r 1 0 z 2r ZZZ So 2 Z 2 Z 1 Z x dV E 0 0 2r r2 cos2 rdzdrd 0 6 10 points Find the Jacobian of the transformation x sin y cos Solution x y x y x y 7 10 points On the back of the test Solution a Type 1 b Type 2 c Type 1 d Type 1 sin cos cos sin sin2 cos2