1. Evaluate the line integral

   ó õ C  xdy-ydx

   where C is the curve r( t) = á2t,3t ñ , t in [ 0,1] . Solution: The pullback yields

   ó õ C  xdy-ydx = ó õ 1 0  æ è x  dy dt -y  dx dt ö ø dt = ó õ 1 0  2t( 3) -3t( 2) dt = ó õ 1 0  0dt = 0

2. Test for exactness. If exact, find its potential: F(x,y) = á x²+y²,xy ñ Solution: Since M = x²+y², N = xy, and P = 0, the curl of F is given by

   curl( F) = á 0-0,0-0,2x-x ñ ¹ 0

   Thus, the field is not conservative and thus does not have a potential.
3. Test for exactness. If exact, find its potential: F(x,y) = á sin( x+y) ,sin( x+y) ñ Solution: Since M = N = sin( x+y) and P = 0, the curl of F is given by

   curl( F) = á 0-0,0-0,cos( x+y)-cos( x+y) ñ = 0

   Thus, F is conservative and its potential is given by

   U( x,y) = ó õ sin( x+y) dx = -cos( x+y)+C( y)

   Since U_y = sin( x+y) = N, the function C( y) must satisfy C_y = 0. Thus,

   U( x,y) = -cos( x+y) +G

   for some constant G.
4. Test for exactness. If exact, find its potential: F(x,y,z) = á ye^x,e^x+1,e^z ñ Solution:The curl of F is given by

   curl( F) = á 0-0,0-0,ex-ex ñ = 0

   Thus, the field is conservative and its potential is given by

   U( x,y,z) = ó õ yexdx = yex+C( y,z)

   However, U_y = e^x+C_y, so that

   ex+Cy = ex+1 Cy = 1 C = y+k( z)

   Thus, the potential is given by U( x,y,z) =  ye^x+y+k(z) . To determine k, we notice that

   Uz = k¢( z) = ez        Þ        k( x) = ez+G

   for some constant G. Thus, U( x,y,z) =  ye^x+y+e^z+G.
5. Evaluate the integral below using the fundamental theorem for line integrals

   ó õ ( 1,1,1) ( 0,0,0)   ( x+y+z) (dx+dy+dz)

   Solution: The vector field is F( x,y,z) = á x+y+z, x+y+z, x+y+z ñ , which has a curl of

   curl( F) = á 1-1,1-1,1-1 ñ = 0

   Thus, the potential of F is

   U( x,y,z) = ó õ ( x+y+z) dx =  x2 2 +xy+xz+C( y,z)

   Since U_x = x+C_y, we have

   x+Cy = x+y+z Cy = y+z C =  y2 2 +yz+k( z)

   Thus, the potential at this point is

   U( x,y,z) =  x2 2 +xy+xz+  y2 2 +yz+k(z)

   However, U_z = x+y+k^¢( z) , so that

   x+y+k¢( z) = x+y+z k¢( z) = z k =  z2 2 +G

   Thus, the potential is

   U( x,y,z) =  x2 2 +xy+xz+  y2 2 +yz+  z2 2 +G

   and the line integral is

   ó õ ( 1,1,1) ( 0,0,0)   ( x+y+z) (dx+dy+dz) =  x2 2 +xy+xz+  y2 2 +yz+  z2 2 ê ê 1 0  =  1 2 +1+1+  1 2 +1+  1 2 =  9 2

6. Explain why the integral ò_{( 0,0,0)} ^(1,1,1) xdy+ydx+zdz is independent of path. Then calculate the integral along two different paths from ( 0,0,0) to (1,1,1) . Solution: The vector field is F( x,y,z) = á y,x,z ñ . The curl of F(x,y,z) is

   curl( F) = á 0-0,0-0,1-1 ñ = 0

   so F is conservative. One path from ( 0,0,0) to ( 1,1,1) is given by

   r( t) = á t,t,t ñ ,       t  in  [ 0,1]

   The line integral over this curve is

   ó õ ( 1,1,1) ( 0,0,0)   xdy+ydx+zdz = ó õ 1 0  æ è x  dy dt +y  dx dt +z  dz dt ö ø dt = ó õ 1 0  3tdt =  3t2 2 ê ê 1 0  =  3 2

   Another curve that passes from ( 0,0,0) to (1,1,1) is given by

   r( t) = á t,t2,t3 ñ,        t  in  [ 0,1]

   The line integral over this curve is

   ó õ ( 1,1,1) ( 0,0,0)   xdy+ydx+zdz = ó õ 1 0  æ è x  dy dt +y  dx dt +z  dz dt ö ø dt = ó õ 1 0  t( 2t) +t2( 1) +t3(3t2) dt = ó õ 1 0  3t2+3t5dt = t3+  3t6 6 ê ê 1 0  = 1+  1 2 =  3 2

7. Let R be the unit square. Use Green's theorem to evaluate the line integral

   ¶R  y2dx+x2dy

   Solution: Green's theorem implies that

   ¶R  y2dx+x2dy = ó õ ó õ R  ¶ ¶x ( x2) -  ¶ ¶y ( y2) dA = ó õ ó õ R ( 2x-2y) dA = ó õ 1 0  ó õ 1 0  ( 2x-2y) dydx = 0

8. Let R denote the upper half of the unit disk. Evaluate using Green's theorem:

   ¶R  ( xy) ( dx+dy)

   Solution: Green's theorem implies that

   ¶R  xydx+xydy = ó õ ó õ R  ¶ ¶x (xy) -  ¶ ¶y ( xy) dA = ó õ ó õ R   ( y-x) dA = ó õ p 0  ó õ 1 0  ( rsin( q) -rcos( q) ) rdrdq = ó õ p 0  ó õ 1 0  ( r2sin( q)-r2cos( q) ) drdq = ó õ p 0   r3 3 sin( q) -  r3 3 cos( q) ê ê 1 0  dq =  1 3 ó õ p 0  ( sin( q) -cos(q) ) dq =  1 3 ( -cos( q) -sin( q) ) | 0p =  2 3

9. Evaluate by using Green's theorem to convert to a line integral over the boundary (D is the unit disk): 

   ó õ ó õ D   -x ( x2+y2+1) 3/2 dA

   Solution: Let's let

   Nx =  -x ( x2+y2+1) 3/2 ,       so  that    N = ó õ  -x ( x2+y2+1) 3/2 dx

   Letting u = x²+y²+1 implies that du = -2xdx, and

   N =  -1 2 ó õ  du u3/2 =  1 u1/2 =  1 (x2+y2+1) 1/2

   Thus, Green's theorem says that

   ó õ ó õ D   -x ( x2+y2+1) 3/2 dA = ó (ç) õ ¶D   1 ( x2+y2+1) 1/2 dy = ó õ 2p 0   1 ( x2+y2+1) 1/2  dy dt dt

   However, x²+y²+1 = 2 on the unit circle, so that

   ó õ ó õ D   -x ( x2+y2+1) 3/2 dA =  1 Ö2 ó õ 2p 0   dy dt dt =  1 Ö2 y( t) | 02p =  1 Ö2 ( y( 2p) -y( 0) )

   Since the curve is closed-and thus, the endpoint and beginning point are the same-we must have y( 2p) = y( 0) . Thus,

   ó õ ó õ D   -x ( x2+y2+1) 3/2 dA = 0

10. Find the area enclosed by the curve r( t) = á cos²( t) ,cos( t) sin(t) ñ , t in [ 0,p] , using Green's theorem. Solution: The area is given by

    Area =  1 2 C  xdy-ydx =  1 2 ó õ p 0  æ è x  dy dt -y  dx dt ö ø dt =  1 2 ó õ p 0  ( cos2( t) ( -sin( t) sin( t) +cos( t) cos(t) ) -cos( t) sin( t) ( -2cos( t) sin( t) ) ) dt =  1 2 ó õ p 0  ( -sin2( t) cos2( t) +cos4( t) +2cos2( t)sin2( t) ) dt =  1 2 ó õ p 0  cos2( t) ( cos2(t) +sin2( t) ) dt =  1 2 ó õ p 0  cos2( t) dt =  1 2 ó õ p 0   1 2 +  1 2 cos( 2t) dt =  1 4 t+  1 8 sin( 2t) ê ê p 0  =  p 4

11. Calculate the surface area of the surface S parameterized by r( u,v) = á ucos( v) ,usin( v) ,u² ñ for u in [ 0,1] and v in [ 0,2p] . Solution: To begin with, r_u = á cos(v) ,sin( v) ,2u ñ and r_v = á -usin( v) ,ucos( v),0 ñ . Their cross-product is

    ru×rv = á -2u2cos( v) , -2u2sin( v) , u ñ

    Thus, | | r_u×r_v| |² = 4u⁴cos²( v) +4u⁴sin²( v)+u² = 4u⁴+u², so that

    dS = 4u2+1  dudv = u 4u2+1  dvdu

    The surface area is thus given by

    S = ó õ ó õ S dS = ó õ 1 0  ó õ 2p 0  u 4u2+1  dvdu = 2p ó õ 1 0  u 4u2+1  du

    We let w = 4u²+1, so that dw = 8udu and

    S =  2p 8 ó õ 5 1  w1/2dw =  5pÖ5 6 -  p 6

12. Compute the flux of the vector field F( x,y,z) = á y,x,z ñ through the surface S parameterized by

    r( u,v) = á ucos( v) ,usin( v) ,u2 ñ ,        u  in  [ 0,1],    v  in  [ 0,2p]

    Solution: To begin with, r_u = á cos(v) ,sin( v) ,2u ñ and r_v = á -usin( v) ,ucos( v),0 ñ . Their cross-product is

    ru×rv = á  -2u2cos( v) ,-2u2sin( v) ,u ñ

    Thus, dS = á  -2u²cos( v),-2u²sin( v) ,u ñ dudv and

    Flux = ó õ ó õ S F·dS = ó õ 2p 0  ó õ 1 0  á y,z,x ñ ·ru×rvdudv = ó õ 2p 0  ó õ 1 0  á usin( v),u2,ucos( v) ñ · á \allowbreak-2u2cos( v) ,-2u2sin( v) ,u ñ dudv = ó õ 2p 0  ó õ 1 0  ( -2u3sin( v) cos( v) -2u4sin( v) +u2cos( v)) dudv = ó õ 2p 0  æ è -  1 2 sin( v) cos(v) -  2 5 sin( v) +  1 3 cos( v) ö ø dv = 0

13. Show that if F( x,y,z) = áxy+2z,yz+2x,xz+2y ñ , then curl( F) = á 2-y,2-z,2-x ñ . Then evaluate

    ó õ ó õ S  curl( F) · dS

    when S is the surface of the pyramid with vertices ( 2,0,0) , ( 2,2,0) , ( 0,2,0) , ( 0,0,0) , and ( 1,1,2) that is not contained in the xy-plane. Solution: Stoke's theorem implies that

    ó õ ó õ S curl( F) ·dS = ¶S  F·dr =  ó õ ó õ B  curl( F) · dS

    where ¶S is the square with vertices ( 2,0,0) , ( 2,2,0) , ( 0,2,0) , ( 0,0,0) and B is base [ 0,2] ×[ 0,2] in the xy-plane. In the base B, the unit normal is the unit vector k, so that

    ó õ ó õ S curl( F) ·dS = ó õ ó õ B á2-y,2-z,2-x ñ ·k  dydx = ó õ 2 0  ó õ 2 0  ( 2-x) dydx = 4

14. Use Stoke's theorem for differential forms to calculate

    ó õ ó õ ¶S xy  dy^dz-z2  dy^dx

    when S is the solid cube [ 0,1] ×[ 0,1]×[ 0,1] . Solution: Applying the d operator and using Stoke's theorem implies that

    ó õ ó õ ¶S xy  dy^dz-z  dy^dx = ó õ ó õ S d( xy  dy^dz-z  dy^dx) = ó õ ó õ S d( xy) ^dy^dz-d(z2) ^dy^dx = ó õ ó õ S ( ydx+xdy) ^dy^dz-2z  dz^dy^dx = ó õ ó õ S ydx^dy^dz+xdy^dy^dz+2z  dy^dz^dx = ó õ ó õ S ydx^dy^dz-2z  dy^dx^dz = ó õ ó õ S  ( y+2z) dx^dy^dz = ó õ 1 0  ó õ 1 0  ó õ 1 0  ( y+2z) dxdydz =  3 2

15. Compute the flux of the vector field F( x,y,z) = á x,y,z ñ through the surface of a sphere S with radius R centered at the origin. Then show that the divergence theorem produces the same result. Solution: The unit sphere is parameterized by

    r( f,q) = á Rsin( f) cos( q) ,Rsin( f) sin(q) ,Rcos( f) ñ ,    f  in  [ 0,p] ,    q  in  [ 0,2p]

    Thus, r_f = á Rcos( f) cos(q) ,Rcos( f) sin( q) ,-Rsin( f) ñ and r_q = á-Rsin( f) sin( q) ,Rsin( f) cos( q) ,0 ñ and

    rf×rq = á R2sin2( f) cos( q) ,R2sin2(f) sin( f) ,R2cos( f) sin( f) ñ = Rsin( f) á Rsin( f) cos( q) ,Rsin( f) sin( f),Rcos( f) ñ = á x,y,z ñ Rsin( f)

    Thus, the flux is given by

    Flux = ó õ ó õ ¶S F·dS = ó õ ó õ ¶S á x,y,z ñ · á x,y,z ñ  Rsin( f) dfdq = ó õ ó õ ¶S ( x2+y2+z2)  Rsin(f) dfdq

    
    However, if ( x,y,z) is on the unit sphere, then x²+y²+z² = R² and

    Flux = ó õ ó õ ¶S R3sin( f) dfdq = R3 ó õ 2p 0  ó õ p 0  sin( f) dfdq = 4pR3

    The divergence theorem, on the other hand, implies that

    ó õ ó õ ¶S F·dS = ó õ ó õ ó õ S  div( F) dV = ó õ ó õ ó õ S  ¶ ¶x x+  ¶ ¶y y+  ¶ ¶z zdV = ó õ ó õ ó õ S 3dV = 3( Volume  of  sphere) = 3  4pR3 3 = 4pR3