This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 antiderivative 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 antiderivative 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 xyplane, 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 zaxis. 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....
View
Full Document
 Fall '07
 NEEL
 Calculus, Derivative, Integrals, Polar coordinate system, Jacobian, coordinates

Click to edit the document details