Solutions to Practice Final Exam

Transcription

1 Math V22. Calculus IV, ection, pring 27 olutions to Practice Final Exam Problem Consider the integral x 2 2x dy dx + 2x dy dx x 2 x (a) ketch the region of integration. olution: ee Figure. y (2, ) y = (, ) y = x 2 y = x (, ) x Figure : {(x, y) y, y x y} (b) Reverse the order of integration and evaluate the integral that you get. olution: = x 2 x= y y 2x dy dx + 2x dy dx = 2xdxdy x 2 x y 6x 2 x=y dy = (6y 2 6y)dy = (2y 3 3y 2 ) = 8 y= y= Problem 2 Consider the transformation of R 2 defined by the equations given by x = u/v, y = v.

2 (a) Find the Jacobian (x,y) of the transformaion. (u,v) olution: (x, y) (u, v) = x x u v = y u y v v u v 2 = v (b) Let R be the region in the first quadrant bounded by the lines y = x, y = 2x and the hyperbolas xy =, xy = 2. ketch the region in the uv-plane corresponding to R. olution: The lines y = x and y = 2x in the xy-plane correspond to v = u/v, v = 2u/v in the uv-plane, respectively. The part in the first quadrant can be rewritten as v = u and v = 2u, respectively. The hyperbolas xy =, xy = 2 in the xy-plane correspond to the lines u =, u = 2 in the uv-plane, respectively. v (2, 2) v = 2u (, u = 2 2) (2, u = 2) v = u (, ) u Figure 2: = {(u, v) R u 2, u v 2u} (c) Evaluate R y da. olution: y da = = R 3u 2 du = u3 v 2 v dudv = u=2 = 7 u= 2 2u u v 3 dvdu = v v= 2u v= u dv

3 Problem 3 Let be the boundary of the solid bounded by the paraboloid z = x 2 + y 2 and the plane z =, with outward orientation. (a) Find the surface area of. Note that the surface consists of a portion of the paraboloid z = x 2 + y 2 and a portion of the plane z =. olution: Let be the part of the paraboloid z = x 2 + y 2 that lies below the plane z =, and let 2 be the disk x 2 + y 2, z =. Then is the union of and 2, and Area() = Area( ) + Area( 2 ) where Area( 2 ) = π since 2 is a disk of radius 2. To find Area( ), consider a vector equation of given by r(x, y) = x, y, g(x, y), (x, y), where g(x, y) = x 2 + y 2 and = {(x, y) R 2 x 2 + y 2 }. We have r x r y = g x, g y, = 2x, 2y, r x r y = x 2 + y 2 + Area( ) = r x r y dxdy = x2 + y 2 + dxdy We use polar coordinates x = r cosθ, y = r sin θ, dxdy = rdrdθ. x2 + y 2 + dxdy = r2 + rdrdθ Let u = r 2 +, du = 8rdr. Then o r2 + rdr = 7 r2 + rdrdθ = u /2du 8 = u3/2 2 u= 7 u= = dθ = π 2 6 (7 7 ) Area() = Area( )+Area( 2 ) = π 6 (7 7 )+π = π 6 (7 7+23) 3

4 (b) Use the ivergence Theorem to calculate the surface integral F d, where F = (x + y 2 z 2 )i + (y + z 2 x 2 )j + (z + x 2 y 2 )k. olution: = E, where E is the solid bounded by the paraboloid z = x 2 + y 2 and the plane z =. By the ivergence Theorem, F d = div FdV where divf = x (x + y2 z 2 ) + y (y + z2 x 2 ) + z (z + x2 y 2 ) = + + = 3. We use cylindrical coordinates x = r cos θ, y = r sin θ, z = z, dv = rdzdrdθ. divfdv = 3rdzdrdθ = 3r( r 2 )drdθ E r 2 = (6r 2 3 r ) dθ = 2dθ = 2π o F d = 2π. r=2 r= Problem Let y i + xj F = x 2 + y. 2 Note that F is defined on {(x, y) R (x, y) (, )}. (a) Evaluate C F dr, where C is the circle x 2 + y 2 =, oriented counterclockwise. olution: A vector equation of C is given by r(t) = cost, sin t, E t 2π C F dr = = F(r(t)) = F(cos t, sin t) = sin t, cos t r (t) = sin t, cost F(r(t)) r (t)dt = dt = 2π cost, sin t sin t, cost dt

5 (b) Compute curl F. olution: i j k curl F = x y z y/(x 2 + y 2 ) x/(x 2 + y 2 ) ( ) ( x ) ( y ) = k + x x 2 + y 2 y x 2 + y ( 2 ) (x 2 + y 2 ) x 2x = k + (x2 + y 2 ) y 2y (x 2 + y 2 ) 2 (x 2 + y 2 ) 2 =. (c) Use Green s Theorem to evaluate C 2 F dr, where C 2 is the circle (x 2) 2 + (y 2) 2 =, oriented counterclockwise. olution: C 2 =, where is the disk (x 2) 2 + (y 2) 2. Note that does not contain the origin (, ), and the components x/(x 2 + y 2 ), y/(x 2 + y 2 ) of F are defined and has continuous partial derivatives on. By the vector form of Green s theorem, F dr = curlf kda = da =. C 2 (d) Is F conservative? olution: F is not conservative because the line integral of F along the simple closed curve C is 2π. Problem 5 Let E be a solid in the first octant bounded by the cone z 2 = x 2 + y 2 and the plane z =. Evaluate E xyz2 dv. olution: We use cylindrical coordinates x = r cosθ, y = r sin θ, z = z, 5

6 dv = dxdydz = rdrdθdz. xyz 2 dv = = = W π/2 π/2 = cos(2θ) 2 π/2 sin(2θ) z= r 3 z 3 drdθ = 6 z=r ( ) sin(2θ) r 6 r7 r= dθ = 7 r= θ=π/2 = 56 θ= r r cosθr sin θz 2 rdzdrdθ π/2 π/2 sin(2θ) (r 3 r 6 )drdθ 6 sin(2θ) dθ 56 Problem 6 Use the ivergence Theorem to evaluate F d, where F = e y2 i + (y + sin(z 2 ))j + (z )k, and is the upper hemisphere x 2 + y 2 + z 2 =, z, oriented upward. Note that the surface does NOT include the bottom of the hemisphere. olution : Consider the solid E = {(x, y, z) x 2 + y 2 + z 2, z }. Its boundary E is the union of and the disk = {(x, y, z) R 3 x 2 + y 2, z = }, where is oriented downward. By the ivergence Theorem F d + F d = div FdV W where divf = ) + x (ey2 y (y + sin(z2 )) + (z ) = 2 z divfdv = 2dV = 2volume(E) = volume(b) = π E E 3 where B is the unit ball x 2 + y 2 + z 2. The downward unit normal of is the constant vector k =,,, so F d = F ( k)d = ( z+)d = d = Area( ) = π F d = divfdv F d = π E 3 π = π 3 6

7 Problem 7 Use tokes Theorem to evaluate F dr, where C F = x 2 yi xy 2 j + z 3 k, and C is the curve of intersection of the plane 3x + 2y + z = 6 and the cylinder x 2 + y 2 =, oriented clockwise when viewed from above. olution: Let be the part of the plane 3x + 2y + z = 6 that lies inside the cylinder x 2 + y 2 =, oriented downward. Then C =. By tokes Theorem, F dr = curlf d where C i j k curlf = x y z x 2 y xy 2 z 3 = ( y2 x 2 )k. A vector equation of is given by r(x, y) = x, y, g(x, y), (x, y) where g(x, y) = 6 3x 2y and = {(x, y) R 2 x 2 + y 2 }. We have curlf(r(x, y)) =,, x 2 y 2 r x r y = g x, g y, = 3, 2, r x r y is upward, so curlf d = curlf(r(x, y)) ( r x r y )dxdy =,, x 2 y 2 3, 2, dxdy = (x 2 + y 2 )dxdy We use polar coordinates x = r cos θ, y = r sin θ, dxdy = rdrdθ. (x 2 + y 2 )dxdy = r 3 drdθ = r r=2 r= dθ = dθ = 8π. 7

8 Problem 8 Use tokes theorem to evaluate curlf d, where F = (sin(y + z) yx 2 y3 )i + x cos(y + z)j + cos(2y)k, 3 and consists of the top and the four sides (but not the bottom) of the cube with vertices (±, ±, ±), oriented outward. olution: Let be the bottom of the cube, oriented by the upward unit normal k, and let C be the boundary of (with the positive orientation). Then = C =. By tokes s theorem, curlf d = F dr = curlf d = curlf kd C i j k curl F = x y z sin(y + z) yx 2 y3 x cos(y + z) cos(2y) 3 curlf k = (x cos(y + z)) x y (sin(y + z) yx2 y3 3 ) = cos(y + z) ( cos(y + z) x 2 y 2) = x 2 + y 2 curlf kd = (x 2 + y 2 )d = = ( x3 3 + xy2 ) x= dy = ( 2 x= 3 + 2y2 )dy = o curlf d = 8 3. Problem 9 Write in the form of a + bi: (a) Find all the fourth roots of. (x 2 + y 2 )dxdy ( 2y 3 + 2y3 3 ) y= y= = 8 3 olution: = (cos π + i sin π), so the fourth roots of are ( / cos( π + 2kπ ) + i sin( π + 2kπ ) ), k =,, 2, 3 8

9 k = : k = : k = 2 : k = 3 : π 2(cos + i sin π ) = + i 3π 2(cos + i sin 3π ) = + i 5π 2(cos + i sin 5π ) = i 7π 2(cos + i sin 7π ) = i The fourth roots of are + i, + i, i, i. (b) Evaluate ( i). olution: i = 2( 2 2 i) = 2(cos( π ) + i sin( π )) ( i) = 2 (cos( π ) + i sin( π ) = 25 ( i) = 32i (c) Find all the possible values of ( 2) i. olution: 2 = 2e πi, so where k is any integer. ln( 2) = ln2 + i(π + 2kπ) ( 2) i = e i ln( 2) = e i ln2 π(2k+) = e π(2k+) (cos(ln 2) + i sin(ln2)) = e π(2k+) cos(ln 2) + e π(2k+) sin(ln 2)i where k is any integer. Problem Let f(z) = e iz. (a) Write f(z) in the form u + iv. olution: f(z) = e iz = e i(x+iy) = e y+ix = e y (cosx + i sin x) = e y cos x + ie y sin x 9

10 (b) Is f(z) analytic? olution: f(z) = u(x, y) + iv(x, y), where u(x, y) = e y cos x, v(x, y) = e y sin x, u x = u e y sin x, y = e y cosx, v x = v e y cosx, y = e y sin x. u, v satisfy the Cauchy-Riemann equations u x = v y, v x = u y, so f(z) is analytic. Problem Let f(z) be an analytic function which only takes real values, i.e., Imf(z) =. how that f(z) is a constant function. (Hint: Use Cauchy- Riemann equations.) olution: Let u = Ref, so that f(x + iy) = u(x, y). It suffices to show that u(x, y) is a constant function. f(z) is analytic, so u and v = satisfy the Cauchy-Riemann equations: u x =, u y =. o u(x, y) is a constant function.