MATH 52 FINAL EXAM SOLUTIONS


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1 MAH 5 FINAL EXAM OLUION. (a) ketch the region R of integration in the following double integral. x xe y5 dy dx R = {(x, y) x, x y }. (b) Express the region R as an xsimple region. R = {(x, y) y, x y } (c) Evaluate the integral by changing the order of integration. y [ ] x=y xe y5 dxdy = x e y5 dy = = e x= y4 e y5 dy = [ ] ey5. (a) Let be a solid cone whose base is the disk x + y in the xyplane and whose vertex is the point (,, ). et up, but do not evaluate, a triple integral in cylindrical coordinates which computes the moment of inertia of around the xaxis. In cylindrical coordinates, so = {(r, θ, z) z r, r, θ π}, I x = y + z dv = π r (r sin θ + z )r dz dr dθ (b) Let be the solid ball of radius centered at (,, ). et up, but do not evaluate, a triple integral in spherical coordinates which computes the moment of inertia of around the zaxis. Hint: First consider the change of variables u = x, v = y, w = z. After the change of variables, the sphere takes the form u +v +w =, and the zaxis becomes the line (u, v) = (, ). his means the distance from the zaxis is (u + ) + v, and therefore I z = π π [(ρ cosθ sin φ + ) + (ρ sin θ sin φ) ]ρ sin φ dρ dφ dθ. onsider the curve parametrized by r(t) = (e t, et + e t ) for t.
2 (a) Find the arclength of. ince r (t) = (e t, e t e t ), s = = ds = e t + e t dt = = e e + r (t) dt = [ et e t (b) Find the xcoordinate of the centroid of. 4e t + e 6t e t + e t dt ] x = xds = e t (e t + e t ) dt s s = e 4t + dt = [ ] s s e4t + t = ( s e4 + ) (c) Let be the surface obtained by rotating around the line x =. Find the area of. By Pappus heorem, area() = s d, where d is the distance travelled by the centroid. Now the centroid travels along a circle of radius + x, hence d = π( + x). o area() = πs( + x) = π( e e + + e4 + ) 4. (a) onsider the change of variables u = x + 4 y, v = y. Find the Jacobian x (x, y) (u, v) Hence (u, v) (x, y) = x y x = + y x = + v = ( + 4v ) y x (x, y) (u, v) = + 4v
3 (b) Find the area of the region in the first quadrant enclosed by the ellipses x 4 +y =, x + 4 y = 4 and the lines y = x and y = 4x. he area is therefore A = dv du = 6 + 4v + 4v dv = 8 w=4 v= + w dw = [ arctan(w)]8 4 = (arctan(8) arctan(4)) 5. Let be the solid enclosed by the planes z = x + y, x + y = and the paraboloid z = y. et up, but do not evaluate a triple integral in rectangular coordinates which computes the volume of by (a) regarding as xsimple. If we regard as xsimple we have a cylinder z = y and two graphs x = z y and x = y. Now the two graphs intersect along a line z + y =. his line intersects the parabola z = y in two points: (y, z) = (, ) and (y, z) = (, 4). o for the coordinate ranges are as follows: y, y z y and z y x y. o vol() = y= y y z=y x=z y dxdz dy (b) regarding as zsimple. If we regard as zsimple we have a cylinder x+y = and two graphs z = x+y and z = y. Now the graphs intersect along a parabola x = y y. his parabola intersects a line x + y = in two points: (x, y) = (, ) and (x, y) = (6, ). o for the coordinate ranges are as follows: y, y y x y and y z x + y. o vol() = y= y x=y y x+y z=y dxdz dy 6. Let be the curve parametrized by r(t) = (t, sin t) for t π/. (a) ompute y dx. ince r (t) = (t, cost), we have y dx = π/ [ t sin t dt = ] π/ t cost + sin t =
4 (b) Let R be the region in the plane bounded by the yaxis, the line y =, and the curve. Use Green s heorem to find the area of R. By Green s heorem, y dx = da = area(r) R R Now the boundary of R consists of the curve, followed by the line segment L from (π /4, ) to (, ) and the line segment L from (, ) to (, ). hese segments are parametrized by (π /4( t), ) and (, t), respectively, for t. hus y dx = y dx + y dx + y dx R L L = + which implies that area(r) = π /4. π /4 dt + dt = π /4 7. Let be the portion of the cylinder x + y = x which lies above the xyplane and below the surface z = x. (a) Write down a parametrization of. Be sure to specify the domain. here are several possible solutions. Here are four of them. First parametrize the circle x + y = x by x = + cost, y = sin t for t π. hen the zcoordinate varies from to x = ( + cos t). o r(t, z) = ( + cost, sin t, z) t π, z ( + cost). A variant of the first solution is to let z = u( + cost) with u varying from to : r(t, u) = ( + cos t, sin t, u( + cos t) ) t π, u. Using polar coordinates to describe the circle, we have r = cosθ for π/ θ π/, so x = r cos θ = cos θ and y = r sin θ = cosθ sin θ. he z coordinate again varies from to x = 4 cos 4 θ. o r(θ, z) = ( cos θ, cosθ sin θ, z) π/ θ π/, z 4 cos 4 θ. A variant of the previous parametrization is r(θ, u) = ( cos θ, cosθ sin θ, 4u cos 4 θ) π/ θ π/, u. (b) Find the area of. Using the first parametrization above, we have r t = ( sin t, cost, ) and r z = (,, ) 4
5 so and therefore r t r z = (cost, sin t, ) = area() = π (+cos t) dz dt = π ( + cost) dt = π 8. Let be the surface parametrized by r(u, v) = (v, u, uv) over the domain = [, ] [, ] in the uvplane. (a) Find the area of. ince r u = (, u, v) and r v = (v,, u), we have r u r v = ( u, v, 4uv) = 8u 4 + 8v 4 + 6u v = (u + v ). hus area() = = d = r u r v du dv (u + v ) du dv = 6 (b) Find the xcoordinate of the centroid of. ince x = v on the surface, x = xd = area() 6 v (u + v ) du dv = Let be the portion of the sphere x + y + z = 4 which lies above the plane z =, and let F(x, y, z) = (x y + z, x + y + z, x + y z). ompute the flux F n d, where n denotes the outward unit normal to. Let be the disk in the plane z = centered at (,, ) with radius. he forms the boundary of the solid region consisting of the portion of the ball x + y + z 4 with z. hus by the ivergence heorem F n d = divfdv = dv =. Now F n d = 5 F n d + F n d,
6 where in the last integral n = (,, ). Hence F n d = F n d = x + y + z d = x y + d, since z = on. Now by the symmetry of and the fact that x and y are odd functions, this reduces to d = area() = 9π.. Let be the triangle with vertices A = (,, ), B = (,, ) and = (,, ), parametrized in that order. Let F(x, y, z) = (4z, x, y ). (a) Find the area of the triangle. he vectors from A to B and A to are given by v = (,, ) and w = (,, ), respectively. he area of the triangle is half the area of the parallelogram spanned by these vectors, so area( AB) = v w = ( 4,, 4) = (b) Find the equation of the plane in which the triangle lies. From part (a) we know that (,, ) is a normal vector to the plane, so using the point (,, ) we see that the equation of the plane is x + y + z = 5. (c) ompute curl F. curl F = (y, 8z, 4x). (d) ompute F dr. Let denote the portion of the plane bounded by. hen by tokes heorem, F dr = curlf n d where the direction of n must be the same as v w from the solution of part (a). hat is, n = (,, ). Using this together with the results of parts (a), (b) and (c), we get curlf n d = 4y+8z+8xd = d = area =. Let F(x, y, z) = (xy + z, yz + x, zx + y ). (a) Find a potential for F. f(x, y, z) = x y + z x + y z. 6
7 (b) Let be the curve parametrized by r(t) = (cost, sin t, t) for t π. ompute F dr he curve begins at (,, ) and ends at (,, π), so by the Fundamental heorem of Line Integrals, F dr = f(,, π) f(,, ) = π.. Let be the solid region which lies below the surfaces z = x and z = y and above the xyplane. (a) Find the volume of. As a ysimple region, takes the form {(x, y, z) x, z x, z y z}. Using the symmetry of the region, we have vol() = 4 = 4 = 8 x z [ ( z)/ ] x [ x 4 x4 ] = dy dz dx = 4 dx = 4 x z dz dx ( x ) dx (b) Find the outward flux through the boundary of of the vector field F(x, y, z) = (zx, z y, z z x). By the ivergence heorem, F n d = divfdv = dv = vol() = 4 7
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