Note: Each problem is worth 14 points except numbers 5 and 6 which are 15 points. = 3 2
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1 Math Prelim II Solutions Spring Note: Each problem is worth points except numbers 5 and 6 which are 5 points. x. Compute x da where is the region in the second quadrant between the + y circles x + y and x + y. Solution. In polar coordinates the region is given by the inequalities r and π/ θ π. The integral then becomes π π/ r cos θ π r r dr dθ r dr cos θ dθ r π/ π π/ θ sin θ + + cos θ dθ π π π π/ 8. Let be the region in the first quadrant bounded by the curves y x α and y x β for < β < α. Compute xy da in two ways, integrating in the order dx dy and in the order dy dx, and verify that the answer is the same in both cases. Solution. The graphs of x α and x β intersect at the points, and,. Between these two points the graph of x β is above the graph of x α. Integrating in the order dy dx we then have x β x β xy dy dx x α xy dx x α x β+ x α+ dx x β+ β + xα+ α + If we change to the order dx dy then the integral becomes β + α + α β α + β + y /α y /β xy dx dy x y y/α dy y /α+ y /β+ dy y /β y /α+ /α + y/β+ /β + So we get the same answer in both cases. α α + /α + /β + α β α + β + β β +
2 Math Prelim II Solutions Spring. Consider an integral x y a Sketch the region of integration. fx, y, z dz dy dx. Solution. We can determine the region from the limits of integration. For the dz integration we see that z goes from z the xy-plane to z y a plane sloping down to the right. For the dy and dx integrations we are looking at the shadow of the region in the xy plane, a triangle in the first quadrant cut off by the line y x. The -dimensional region of integration lies above this triangle and below the plane z y, so it has the shape shown in the figure, sort of a pyramid lying on its side. b Set up the integral in the order dx dy dz but do not attempt to evaluate the integral since we have not specified what the function fx, y, z is. Solution. For the order dx dy dz we are projecting to the shadow in the yz-plane, which is the triangle in the first quadrant of the yz-plane cut off by the line z y, or y z/. For the initial dx integration we have x going from x the yz-plane to x y a vertical plane. The integral is then z/ y fx, y, z dx dydz.. Compute the volume of the region bounded by the xy-plane and the two surfaces z x and z y these are parabolic cylinders. Hint: break the region up into several symmetric pieces of equal volume. Solution. The region is shown in the figure on the left below. It is tent-shaped, and the projection of the region to the xy-plane is the square x and y. The region is symmetric under reflection across the yz-plane and the xz-plane. It is also symmetric under reflection across the vertical planes x ±y. We can compute the volume
3 Math Prelim II Solutions Spring by computing the volume of the piece shown in the right half of the figure and multiplying this by 8. For this piece the curved upper surface is part of the surface z x, so the volume of this piece is x x dz dy dx x z x dy dx x y x dx x Therefore the volume of the region is equal to 8. x dy dx x x dx x x There are other ways to compute the volume, using different orders for dx, dy, and dz. For example: z z dx dy dz or y z 5. Find the center of gravity of the constant-density solid cone consisting of all points x, y, z satisfying the inequalities x + y z a, where a is an arbitrary constant. Solution. We use cylindrical coordinates. The inequalities x + y z a then become r z a. The volume of the cone is given by the integral π a a π dv r dz dr dθ r a π rz ar a r r dr dθ a π a dθ dx dz dy π ar r dr dθ a 6 dθ πa Using the rotational symmetry of the cone we can see that the center of gravity lies on the z-axis so its x and y coordinates are. In order to compute the z coordinate we need the integral z dv π a a r r z dz dr dθ π a π a r rz a r Finally, the z coordinate of the center of mass is z z dv πa / dv πa / a π a dr dθ r a dθ a r r dr dθ π a dθ πa
4 Math Prelim II Solutions Spring 6. Find the volume of the region that lies inside the sphere x +y +z z and outside the sphere x + y + z. Solution. The sphere x + y + z has radius with center at the origin. The equation for the other sphere can be rewritten as x + y + z so this sphere has radius and center at,,. In cross section the two spheres look like the circles in the figure at the right. The region we want is the shaded region. We use spherical coordinates to compute the volume. In spherical coordinates the lower sphere has equation ρ and the upper sphere has equation ρ z ρ cosϕ, or ρ cos ϕ. These two spheres intersect in the circle where the plane z intersects the two spheres, so ϕ π/ on this circle. Therefore the region can be described in spherical coordinates as ρ cosϕ ϕ π/ θ π The volume of the region is given by the integral π π/ cos ϕ ρ sin ϕ dρ dϕ dθ π π/ ρ π π dθ sin ϕ cos ϕ dϕ dθ π/ 8 cos ϕ sin ϕ sin ϕ dϕ cos ϕ + π/ cos ϕ π 7 Here we used the substitution u cos ϕ to compute the dϕ integral. 7. Let be the region in R defined by the inequalities y x y and x + y. Compute dx dy by using a change of variables to convert into a square. Hint: xy Rewrite the inequalities y x y as x/y. Solution. We want to simplify the region defined by the inequalities x + y and x/y, and the easiest way to do this is to choose new variables u x/y and v x+y so that the inequalities become u and v, which define a square. To carry out the change of variables from x and y to u and v we will need to express x and y in terms of u, v, i.e., we need to solve for x and y in the system x/y u x + y v.
5 Math Prelim II Solutions Spring Solving the first equation for x gives x uy and substituting this in the second equation gives uy + y v, which can be solved for y to get y v u+. Then the earlier equation x uy gives x uv uv. Thus we have x u+ u+ and y v u+. Next we need to compute the Jacobian determinant: x, y u, v det x u y u x v y v det The change of variables formula gives v u u+ u+ v u+ u+ v + uv u + v u +. xy da x, y xu, vyu, v u, v da v uv v u + da uv da, u+ u+ where is the region in the uv-plane corresponding to, so is the square u, v. Thus we get dv du uv u du v dv lnu lnv lnln ln
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