Math Examination Spring Instructions These problems should be viewed as essa questions. Before making a calculation, ou should explain in words what our strateg is. Please write our solutions on our own paper. Each of the problems counts for points.. Evaluate the iterated integral π x cos(x) d dx. Solution. Observe that x cos(x) is the -partial derivative of sin(x). Then π x cos(x) d dx = π sin(x) = = dx π = (sin(x) ) dx = cos(x) π = cos(π ) + cos() =.. Describe the solid whose volume is given b the spherical-coordinate triple integral π π (You do not need to evaluate the integral.) ρ sin(φ) dρ dφ dθ. Solution. Notice that ρ sin(φ) dρ dφ dθ is dv, the volume element, expressed in spherical coordinates. Since ρ goes from to, the solid is contained in the shell between a sphere of radius and a sphere of radius. The angle φ goes from to π, its maximum range, so this angle does not restrict the solid. The angle θ goes from to π, half its maximum range, so the solid lies in the half-space where >. Thus the solid is a hemisphere of radius with a concentric hemisphere of radius removed. The problem does not ask for the value of the integral. But that value works out to 4π. March 8, Page of 6 Dr. Boas
Math Examination Spring x. Evaluate the double integral D xe da, where D is the region in the first quadrant bounded b the line =, the line x =, and the parabola = x. Solution. The convenient method is to integrate with respect to first: D xe da = x xe d dx = xe =x dx = ( ) = x e x dx u=x = ( e u ) du = [ e u u ] = [ ] (e ) ( ) = e. 4. Find the volume of the solid bounded b the clinder x + = and the planes z = and x + z =. Solution. One method is to use clindrical coordinates. The variable z goes from to x, or r cos(θ). Since the equation x + = describes a circle of radius, the angle θ is unrestricted, and r goes from to. Here is the iterated integral: π r cos(θ) r dz dr dθ = π = π = π = π. ( r cos(θ))r dr dθ [ r r cos(θ) ] [ cos(θ) ] dθ dθ March 8, Page of 6 Dr. Boas
Math Examination Spring 5. Rewrite the integral in the order dx d dz. f(x,, z) dz dx d as an iterated integral Solution. The solid lies in the first octant (since all three variables start at ). The solid lies under the parabolic clinder z = (with axis along the xaxis) and between the plane x = and the z plane. Also the solid stops at the plane =. Consequentl, the variable x goes from to ; then goes from z to, and z goes from to 4 (since the equation z = sas that z = 4 when = ). Here is the iterated integral: 4 z f(x,, z) dx d dz. 6. Set up an integral for the surface area of the part of the paraboloid z = x + that lies between the plane z = and the plane z = 4. (You do not need to evaluate the integral.) Solution. The surface area element is ( ) ( ) z z + + = + (x) x + () = + 4(x + ). Since z goes from to 4, the surface lies over the region in the x-plane between the circle = x + and the circle 4 = x + (which has radius ). Because of the circular smmetr, it is convenient to write the integral in polar coordinates: π + 4r r dr dθ. The problem does not ask for the value of the integral. But the value works out to π ( 7 7 5 ) 5. 6 March 8, Page of 6 Dr. Boas
Math Examination Spring x 7. Let D be the square with vertices (, ), (, ), (, ), and (, ). Rewrite the double integral D (x+) 5 da as an integral with respect to du dv, where u = x + and v = x. (You do not need to evaluate the integral.) Solution. Compute the Jacobian: (u, v) (x, ) = u x v x u v = =. The area-magnification factor for the change of variables is the absolute value of the Jacobian, so du dv = dx d, or dx d = du dv. The side of the square in the first quadrant of the x-plane is part of the line x + =, which turns into the equation u =. Similarl, the parallel side in the third quadrant turns into the equation u =. The side in the fourth quadrant is part of the line x =, which corresponds to v =, and the parallel side in the second quadrant corresponds to v =. Consequentl, the new integral in the uv-plane is u5 du dv. The problem does not ask for the value of the integral. But the value is b smmetr! 8. Find the work done b the force field F (x, ) = ı + x ȷ on a particle that moves in a straight line from the point (, ) to the point (5, ). Solution. One method is to observe that F (x, ) = (x ). Therefore the work integral F d r equals x (5,) (,) = 5 4 =. Another method is to parametrize the line via x = + 4t = t t. March 8, Page 4 of 6 Dr. Boas
Math Examination Spring Then dx = 4 dt and d = dt, so F d r = dx + x d [ = ( t) (4) + ( + 4t)( t)( ) ] dt ( = t t + ) dt = 4 5 + =. 9. Evaluate the line integral C ds when the parametric equations of C are x = t and = t, where t. Solution. The arc length element ds equals (dx ) ( ) d (t ) + dt = + dt = t dt dt + dt. Therefore C ds = 4t t + dt = 4 ( t + ) = 4 [ ].. What does it mean to sa that a vector field F is a conservative vector field? Solution. A conservative vector field is the same thing as a gradient vector field: there exists a function f such that F = f. March 8, Page 5 of 6 Dr. Boas
Math Examination Spring Optional bonus problem for extra credit Find the volume of an egg whose eggshell has the equation x 8 + 5 + z =. Solution. The problem asks for D dv, where D is the region inside the surface. One method is to change variables via x = u 8 and = v 5 and z = w. Then dx d dz = du dv dw, and the equation of the surface in the new coordinates becomes u + v + w =, a sphere of radius. Since the volume of a ball of radius r equals 4 πr, the volume of a ball of radius equals 4 π. Accordingl, D dv = ( ) 4 π, or 8π. (If ou forgot the formula for the volume of a ball of radius, then ou could compute it in spherical coordinates as which works out to 4π.) π π ρ sin(φ) dρ dφ dθ, March 8, Page 6 of 6 Dr. Boas