Question 1: Applications & Change of Variables

Find the mass of a lamina bounded by y = 5%, a: = 1, and y = 0 with the density function a(m,y) = a: + 4 :1 Let E be the solid below the plane 2 = 2 and above the cone 2 = 3:2 -l 3?. Find the mass of E if the density p(m,y, z) = 2. Enter your answer as a reduced fraction. (Note that the factor of 11' is already there!) Find the mass of the solid bounded below by the circular cone 2 = V 3:2 + y2 and above by the hemisphere 2: = V 6.252 m2 y2 if the density {3(3), y, z) = 3:2 + y2 + 2:2. (Note that the factor of1r is already there!) Find the center of mass of a lamina as pictured below, with inner radius 2 and outer radius 6, with a: density function a(a:,y) = m 55%| M \"ii || Find the center of mass of the solid 5' bounded by the paraboloid z = 2.732 + 2y2 and the plane z = 8. Assume the density is constant. Hint: Think about the shape before calculating - you can greatly reduce the number of calculations to be done. Let E be the solid below 2: = 8 m2 y2 and above the square [2, 2] X [2, 2] Given the solid has a constant density of 6, nd the moment of inertia of E about the zaxis. :1 Electric charge is distributed over the triangular region B shown below so that the charge density at (58,:9) is 0156,31) = 2:33;, measured in coulumbs per square meter (C/mz). Find the total charge on D Find the Jacobian of the transformation x = 4u + 2v, y = u2 + v2x - 2y Find dA, where R is the parallelogram enclosed by the lines -5x + 6y 2x - 2y = 0, 2x - 2y =9, - 5x + 6y = 1, - 5x + 6y =7Find (2: + 4y) d.A where R is the parallelogram with vertices (0,0), (-1,5), (3,2), and (2, 7). Use the transformation x = -u + 3v, y = 5u + 2v