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Sample Midterm I Solutions

# Sample Midterm I Solutions - Solutions to the 1st Practice...

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Unformatted text preview: Solutions to the 1st Practice Midterm 1. Evaluate the following integrals. (a) Z 1 " Z √ 1- x 2 e x 2 + y 2 dy # dx We can’t just proceed by taking the anti-derivative of e y 2 , because it doesn’t have an anti- derivative given by elementary functions. Instead, observe that the domain of integration is the part of the unit disk x 2 + y 2 = 1 which lies in the first quadrant, and that the integrand can be written as e r 2 . This makes a switch to polar coordinates look promising. Doing so, we see that this integral can be written Z π/ 2 Z 1 e r 2 · r dr dθ = Z π/ 2 1 2 e r 2 r =1 r =0 dθ = Z π/ 2 1 2 ( e- 1) dθ = π 4 ( e- 1) . (b) Z 3 Z 9 y 2 y cos ( x 2 ) dx dy Again, we immediately have the problem that cos ( x 2 ) doesn’t have an anti-derivative given by elementary functions. The domain of integration is the region bounded by the parabola x = y 2 and the lines y = 0 and x = 9 in the first quadrant. Thus, switching the order of integration gives Z 9 Z √ x y cos ( x 2 ) dy dx = Z 9 cos ( x 2 ) 1 2 y 2 y = √ x y =0 dx = Z 9 1 2 x cos ( x 2 ) dx = 1 4 sin ( x 2 ) 9 = 1 4 sin 81 . 2. Consider the solid cone given as the region between the surface z = p x 2 + y 2 and the plane z = 1 . Suppose that this solid has constant density ρ . (a) Determine the center of mass of this solid. By symmetry, we know that x = y = 0. In order to find z , we first need the total mass. 1 Working in cylindrical coordinates, we have m = Z 2 π Z 1 Z 1 r ρr dz dr dθ = Z 2 π Z 1 ρr (1- r ) dr dθ = Z 2 π ρ 1 2 r 2- 1 3 r 3 r =1 r =0 dθ = Z 2 π 1 6 ρ dθ = 1 3 πρ. To compute the moment about the xy-plane, we again use cylindrical coordinates: M xy = Z 2 π Z 1 Z 1 r zρr dz dr dθ = Z 2 π Z 1 1 2 ρrz 2 z =1 z = r dr dθ = 1 2 ρ Z 2 π dθ Z 1 ( r- r 3 ) dr = ρπ 1 2 r 2- 1 4 r 4 1 = 1 4 πρ. Thus z = M xy /m = 3 / 4, and the center of mass is (0 , , 3 / 4). (b) Determine the moment of inertia of this solid about the z-axis. Again using cylindrical coordinates (and the fact that x 2 + y 2 = r 2 ), we have I z = Z 2 π Z 1 Z 1 r r 2 ρr dz dr dθ = ρ Z 2 π dθ Z 1 r 3 (1- r ) dr = 2 πρ 1 4 r 4- 1 5 r 5 1 = 1 10 πρ. 3. Compute the surface area of the surface given by the graph of z = xy + 1 over the disk x 2 + y 2 ≤ 4 . In order to find the surface area of the graph, we need to integrate p ( ∂f/∂x ) 2 + ( ∂f/∂y ) 2 + 1 over the given disk. We see that ∂f/∂x = y and ∂f/∂y = x , so the surface area is given by Z Z D p x 2 + y 2 + 1 dA, where D is the disk x 2 + y 2 +1. Given that the domain and the integrand are naturally adapted 2 to polar coordinates, we write this integral in polar coordinates to get Z 2 π Z 2 p r 2 + 1 · r dr dθ = Z 2 π 1 3 ( r 2 + 1 ) 3 / 2 r =2 r =0 dθ = 1 3 5 3 / 2- 1 Z 2 π dθ = 2 3 5 √ 5- 1 π....
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Sample Midterm I Solutions - Solutions to the 1st Practice...

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