1. ompute oln: (8x + 36xy)ds = Math 235 - Review for Exam 3 (8x + 36xy)ds, where c(t) = (t, t 2, t 3 ) on the interval t 1. 1 (8t + 36t 3 ) 1 + 4t 2 + 9t 4 dt = 2 3 (1 + 4t2 + 9t 4 ) 3 2 1 = 2 3 ((14) 3 2 1). 2. If a wire with linear density ρ(x, y) lies along a plane curve, its moment of inertia about the x-axis and y-axis are defined by I x := y 2 ρ(x, y)ds, I y := x 2 ρ(x, y)ds. Find the moments of inertia of the right half of the circle x 2 + y 2 = 16 if ρ 1. olution: The curve is parameterized by (t) = (4 cos t, 4 sin t), π/2 t π/2. We compute I x as The other integral is computed in a similar way. π/2 π/2 π/2 I x = (4 sin t) 2 (t) dt = 16 sin 2 t 4 dt = 64 sin 2 t dt = 32π. π/2 π/2 π/2 3. If g is continuously differentiable and α, β are constants, compute αg(x 2 + y 2 )dx + βg(x 2 + y 2 )dy, where is the circle of radius 2 (about the origin) rotating counterclockwise. olution: Parameterize the circle as (t) = (2 cos t, 2 sin t), t 2π. We then get αg(x 2 + y 2 )dx + βg(x 2 + y 2 )dy 2π = (αg(4) d dt 2 cos t + βg(4) d 2 sin t)dt dt 2π = ( 2αg(4) sin t + 2βg(4) cos t)dt = You can also use Green s theorem to get the same answer. 4. ompute F d, where F (x, y, z) = (3x 2 y 2 z, 2x 3 yz, x 3 y 2 ) and is a curve from (3, 2, 1) to (1, 2, 3) (hint: there is an easy way to do this problem). oln: If you recognize that F = f for f(x, y, z) = x 3 y 2 z, then F d = f(1, 2, 3) f(3, 2, 1) = 12 18 = 96. 5. ompute the line integral of F (x, y) = (2x sin y, x 2 cos y 3y 2 ) along the straight line from ( 1, ) to (5, 1). olution: Evaluating the line integral directly can be cumbersome. To make things easier, check to see if F is conservative, that is to say F = φ for some scalar function φ. We need to find (if possible) a function φ such that φ = (φ x, φ y) = F = (2x sin y, x 2 cos y 3y 2 ). omparing the first components, we get the equation φ x = 2x sin y which means that φ = x 2 sin y + h(y). Using this and comparing the second components, we get φ y = x 2 cos y 3y 2 = x 2 cos y + h (y) and so h(y) = y 3. Thus φ = x 2 sin y y 3 (plus a constant which is not really important here). By the fundamental theorem of calculus (for line integrals) we get (5,1) (5,1) F d = φ d = φ(5, 1) φ( 1, ) = 25 sin 1 1. ( 1,) ( 1,) 1
2 6. (a) If is a straight line segment from (a 1, b 1 ) to (a 2, b 2 ), compute (in terms of the a s and b s) the quantity ydx + xdy. (b) If is now a positively oriented closed curve enclosing a domain, what (in words) does the quantity ydx + xdy represent. (c) Use (a) and (b) to compute the area of the pentagon with vertices (, ), (2, 1), (1, 3), (, 2) and ( 1, 1). olution: (a) The line from (a 1, b 1 ) to (a 2, b 2 ) is parameterized by (t) = (a 1, b 1 ) + t(a 2 a 1, b 2 b 1 ) = ( a 1 + t(a 2 a 1 ), b 1 + t(b 2 b 1 ) ), t 1. The quantity ydx + xdy can be computed as { } [b 1 + t(b 2 b 1 )](a 2 a 1 ) + [a 1 + t(a 2 a 1 )](b 2 b 1 ) dt, which after simplifying equals Use the above to compute the line integral { a 1 b 2 b 1 a 2 } dt. 1 ydx + xdy = (a 1 b 2 b 1 a 2 )dt = a 1 b 2 b 1 a 2. A cautionary note: In the past, students have attempted to use Green s theorem to compute the above line integral. This is not valid since using Green s theorem requires that line integral be taken over a closed loop (with the proper orientation). A line segment does not quality as a closed loop. (b) By Green s theorem (which is valid since we have a closed positively oriented loop) ( d ydx + xdy = dx x d dy ( y) ) da x,y, where is the area enclosed by the curve. A computation shows this integral is equal to 2 da = 2 (area enclosed by ). 7. Let F be the vector field (x, y, z) F (x, y, z) = 4 r 6, where r = x 2 + y 2 + z 2. ompute the work it takes to move along the straight line from (1, 1, 1) to (2, 2, 2). olution: The work is equal to F d, where is the line from (1, 1, 1) to (2, 2, 2). Evaluating this line integral directly can be complicated. To make things easier, we will use the fact that F is a conservative vector field, that is F = φ for some scalar function φ. To find φ note that the equation φ = F says that 4x φ x = (x 2 + y 2 + z 2 ) 3 and so φ = (x 2 + y 2 + z 2 ) 2. Thus, by the fundamental theorem of calculus for line integrals, F d = φ d = φ(2, 2, 2) φ(1, 1, 1) = 1 12 2 1 3 2. 8. Let (t) = (cos t, sin t, t), t 2π and F (x, y, z) = (x 2 y 3 + y, x 3 y 2 + x, z). alculate the line integral F d. olution: alculating the integral directly will be a prohibitive calculation. o try to show that F = φ and then use the fundamental theorem of calculus. To find φ we note that φ = (φ x, φ y, φ z) = (x 2 y 3 + y, x 3 y 2 + x, z). Equating the first coordinate entries, we get φ x = x 2 y 3 + y which means that φ = x 3 y 3 /3 + xy + h(y, z). Equating second coordinate entries we get φ y = x 3 y 2 + x + h y = x 3 y 2 + x which means that h y =. Thus, so far, φ = x 3 y 3 /3 + xy + h(z). Equating third coordinate entries, we get φ z = h (z) = z and so h(z) = z 2 /2 and so φ = x 3 y 3 /3 + xy + z 2 /2. To evaluate the line integral, we get (using the fundamental theorem of calculus for line integrals) F d = φ d = φ((2π)) φ(()) = φ(1,, 2π) φ(1,, ) = 2π 2.
3 9. Let r = x 2 + y 2 + z 2 and F (x, y, z) = er (x, y, z). ompute r (2,, ) to (,, 1) F d, where is the straight line from x olution: The vector field F is conservative with potential function φ(x, y, z) = e 2 +y 2 +z 2, i.e., φ = F. Thus (,,1) φ d = φ(, 1, ) φ(2,, ) = e e 2. (2,,) 1. ompute (x 2 y 3 + y, x 3 y 2 + x, z) d, where is the top portion of the circle of radius 2 in the xy plane, centered at the origin, going counterclockwise. olution: From a previous problem, (x 2 y 3 + y, x 3 y 2 + x, z) = φ, where φ = x 3 y 3 /3 + xy + z 2 /2. Thus ( 2,,) φ d = φ( 2,, ) φ(2,, ) =. (2,,) 11. Let F be the vector field F (x, y) = (2xe x2 sin y, e x2 cos y). Find the work it takes to move along a straight line from (, ) to (1, π/2). olution: The work is equal to the line integral F d, where is the line from (, ) to (1, π/2). To calculate this line integral, we will find a scalar function φ such that F = φ. Looking at the first coordinate entries of the equation φ = (φ x, φ y) = (2xe x2 sin y, e x2 cos y) we see that φ x = 2xe x2 sin y which means φ = e x2 sin y + h(y). Equating the second coordinate entries yields φ y = e x2 cos y + h (y) = e x2 cos y which means that h (y) =. Thus we can take φ = e x2 sin y. Using the fundamental theorem of calculus for line integrals, we get (1,π/2) F d = φ d = φ(1, φ/2) φ(, ) = e. (,) 12. Let F (x, y, z) = 1 r 3 (x, y, z), where r = x 2 + y 2 + z 2. Is F conservative? Why? olution: To show that F is conservative, we need to show there is a scalar function φ such that φ = F. To this end, x φ = (φ x, φ y, φ z) = ( (x 2 + y 2 + z 2 ) 3/2, y (x 2 + y 2 + z 2 ) 3/2, z (x 2 + y 2 + z 2 ). ) 3/2 Equating the first coordinate entries in the above equation gives us φ x = x(x 2 + y 2 + z 2 ) 3/2 which says φ = (x 2 + y 2 + z 2 ) 1/2. You can check this potential function φ works for the other entries. Thus F is conservative with potential function φ = (x 2 + y 2 + z 2 ) 1/2. 13. ompute F d where F (x, y, z) = (yz, xz, xy) and (t) = (2 cos t, 3 sin t, 4), t 2π.
4 olution: One can show that F = φ for some potential function φ. Indeed φ = (φ x, φ y, φ z) = (yz, xz, xy) and so equating the first components, we get φ x = yz. This means that φ = xyz + h(y, z). Proceeding as in the previous problems, one can check that φ can be taken to be φ = xzy. Thus, by the fundamental theorem of calculus for line integrals, φ d = φ((2π)) φ(()) = since (2π) = (). 14. Let be any path joining any point on the sphere x 2 + y 2 + z 2 = a 2 to any point on the sphere x 2 + y 2 + z 2 = b 2. If F = 5 v 3 v, where v = (x, y, z), what is F d? olution: Notice that F (x, y, z) = (5x(x 2 + y 2 + z 2 ) 3/2, 5y(x 2 + y 2 + z 2 ) 3/2, 5z(x 2 + y 2 + z 2 ) 3/2 ) and a computation shows that if φ(x, y, z) = (x 2 + y 2 + z 2 ) 5/2, then φ = F. Thus if A = (a 1, a 2, a 3 ) lies on the sphere of radius a (i.e., a 2 1 + a2 2 + a2 3 = a2 ) and B = (b 1, b 2, b 3 ) lies on the sphere of radius b (i.e., b 2 1 + b2 2 + b2 3 = b2 ), then by the fundamental theorem of calculus for line integrals, the line integral is φ(b) φ(a) = b 5 a 5. 15. Parameterize the following surfaces: (a) z = x 2 + y 2, 1 z 4. (b) x 2 + y 2 + z 2 = 1, 1/ 2 z 1/ 2. (c) x 2 + z 2 = 4, y 2. olution: (a) X(u, v) = (u, v, u 2 + v 2 ), 1 u 2 + v 2 16. (b) X(θ, φ) = (sin φ cos θ, sin φ sin θ, cos φ), θ 2π, π/4 φ 3π/4. (c) X(θ, t) = (2 cos θ, t, 2 sin θ), θ 2π, t 2. 16. ompute the surface area of the solid bounded by x 2 + y 2 < z < 2. olution: The surface is a cone parameterized by X : The surface are is Here The surface integral becomes X(r, θ) = (r cos θ, r sin θ, r), = {(r, θ) : r 2, θ 2π}. 1 d. d = X r X θ da r,θ = ( r cos θ, r sin θ, r) da r,θ = 2 r da r,θ. 1 d = 2π 2 2 r dar,θ = 2 r dr dθ = 4π 2. 17. ompute the surface area of the surface bounded by z = x 2 + y 2, z < 4. olution: The surface is parameterized by the function X : given by X(u, v) = (u, v, u 2 + v 2 ), = {(u, v) : u 2 + v 2 < 4}. urface area is given by the surface integral of the constant function 1, that is 1 d which can be computed as 1 X u X v da u,v. A computation shows that X u X v = ( 2u, 2v, 1) = 1 + 4u 2 + 4v 2
5 and so the surface area is then 1 + 4u 2 + 4v 2 da u,v which, after switching to polar coordinates, becomes 2π 2 r 1 + 4r 2 dr dθ = π 6 (17 17 1) 18. onsider the surface parameterized by X(u, v) = (u cos v, u sin v, u 2 ), = {(u, v) : u 1, v 2π} (a) raw this surface (b) ompute its surface area. olution: (a) If x = u cos v, y = u sin v, z = u 2, notice that z = x 2 + y 2 which is a paraboloid opening upwards with height 1. (b) The surface are is 1 d = X u X v da u,v. A computation yields Thus X u X v = ( 2u 2 cos v, 2u 2 sin v, u) = u 1 + 4u 2. 2π 1 X u X v da u,v = u 1 + 4u 2 du dv = π 6 (5 5 1). 19. If z = f(x, y), where (x, y). Write down a double integral, involving the function f, which represents the surface area of the graph of f(x, y). olution: Parameterize the surface by X(u, v) = (u, v, f(u, v)), (u, v). The surface area is then X u X u da u,v. Note that and so Put this altogether to get the surface area is then X u X v = (1,, f u) (, 1, f v) = ( f u, f v, 1) X u X v = 1 + fu 2 + f v 2. 1 + f 2 u + f 2 v dau,v. 2. Evaluate the surface integral z 2 d where is the portion of the cone z = x 2 + y 2 for which 1 x 2 + y 2 4. olution: First parameterize the surface by X :, Note that The surface integral becomes X(r, θ) = (r cos θ, r sin θ, r), = {(r, θ) : 1 r 2, θ 2π}. d = X r X θ da r,θ = ( r cos θ, r sin θ, r) da r,θ = 2 r da r,θ. z 2 d = r 2 2π 2 3 15 2rdA rθ = 2r drdθ = 2π 1 2 21. Evaluate (x + y + z)d across the rectangle with vertices (1, 1, 1), (2, 3, 4), ( 1, 2, 1), and (, 4, 4). oln: Parametrize the surface by (1, 1, 1) + u(1, 2, 3) + v( 2, 1, ), u 1, v 1. In this case, the length of T u T v is which is 7. Thus, (x + y + z)d = 1 1 7 (3 + 6u v)dudv = 11 7. 2 i j k 1 2 3 2 1,
6 22. Evaluate z = 3. (x 2 + y 2 )d, where is the surface of the cone z 2 = 3(x 2 + y 2 ) bounded by z = and olution: This cone is parameterized by X(u, t) = (u cos t, u sin t, 3u), t 2π, u 3/ 3. Note that X u X t = ( 3u cos t, 3u sin t, u) and so X u X t = 2u. Thus 2π 3/ 3 (x 2 + y 2 )d = 2u 3 dudt = 9π. 23. ompute the following surface integral xd, where is part of the plane x + y + z = 1 in the first octant {(x, y, z) : x, y, z }. olution: The surface can be parameterized by X : by X(u, v) = (u, v, 1 u v), = {(u, v) : v 1 u, u 1}. A computation shows that d = X u X v da u,v = (1, 1, 1) da u,v = 3dA u,v and so xd = u 3 dv du = 1 1 u 3 u dv du = 3 6. 24. Let the velocity field of a fluid be described by F = x i +y j (measured in meters per second). ompute how many cubic meters of fluid per second are crossing the surface of z = 4 x 2 y 2, z, in the direction of increasing z. 2π 2 2π oln: (x, y, ) (2x, 2y, 1)dxdy = 2r 2 rdrdθ = 8dθ = 16π. 25. Use Green s Theorem to show that the area contained by an ellipse x2 a + y2 2 b = 1 is πab. 2 oln: Parametrize the ellipse by x = a cos (θ) and y = b sin (θ). Green s Theorem states that Area = dxdy = 1 xdy ydx = 1 2π (a cos (θ)b cos (θ) (b sin (θ))( a sin (θ)))dθ = πab. 2 2 26. ompute (x, y) d, where is the triangle with vertices (1, ), (, 1), ( 1, ) in the clockwise direction. olution: By Green s theorem (were denotes the interior of the above triangle), (x, y) d = ( d dx y d x)da =. dy 27. Evaluate the line integral F d where F (x, y) = ( 3y 3 + x 3, 3x 3 y 3 ) and is the circle of raduis 1 centered at the origin going counterclockwise.
7 olution: By Green s theorem ( 3y 3 + x 3, 3x 3 y 3 { d ) (dx, dy) = dx (3x3 y 3 ) d dy ( 3y3 + x 3 ) } da. x 2 +y 2 <1 onverting to polar coordinates yields 2π 1 r 2 r dr dθ = 9π 2. 28. ompute F d where F (x, y) = (ye xy + e x, xe xy + e y ) and (t) is the path along the boundary of R = [, 1] [, 1] going counterclockwise. olution: By Green s theorem, (ye xy + e x, xe xy + e y ) (dx, dy) = R { d dx (xexy + e y ) d dy (yexy + e x ) } da =. 29. uppose that f xx + f yy =. ompute 2π Hint: Try to write the above as a line integral. ( fx (cos t, sin t) cos t + f y (cos t, sin t) sin t ) dt. olution: Notice that the above integral can be written as 2π (f x(cos t, sin t) d dt sin t fy(cos t, sin t) d cos t)dt. dt Letting (t) = (cos t, sin t), t 2π, we note that the above is equal to f xdy f ydx, x 2 +y 2 =1 where the integration is counterclockwise. By Green s theorem, this integral is equal to (f xx + f yy)da which is equal to zero, by the hypothesis of the problem. 3. ompute the flux integral bounded by z = 4 x 2 y 2 and the xy-plane. x 2 +y 2 1 F n d, where F = (3x, xz, z 2 ) and is the surface of the solid olution: You can compute the surface integral directly but it might take a while since the surface is in two pieces (the top part of the upsidedown paraboloid and the xy-plane). However, using the divergence theorem (Gauss theorem), the computation is much easier. By the divergence theorem we get F n d = F dv, where is the solid enclosed by (which in this case is a upside-down paraboloid which intersects the xy plane in a circle of radius 2 and peaks along the z-axis at z = 4). A computation shows that F = 3 + 2z and so F dv = (3 + 2z)dV, which after switching to cylindrical coordinates becomes 2π 2 4 r 2 (3 + 2z)r dz dr dθ = 136π 3.
8 31. Evaluate the surface integral F nd, bd(w ) where F = (1, 1, z(x 2 + y 2 )) and W is the solid cylinder x 2 + y 2 1, z 1. olution: By Gauss theorem, bd(w ) which, after switching to polar coordinates, becomes F nd = W F dv = W 2π 1 1 r 3 dz dr dθ = π 2. (x 2 + y 2 )dv 32. If n is the outward unit normal for a smooth closed surface which encloses a solid, what is div n dv? olution: In this problem you need to work the divergence theorem backwards. Note, by the divergence theorem, that n nd = ndv, where is the surface which bounds. However n n = n 2 = 1, since n is the outward unit normal (and hence has norm equal to 1). Thus ndv = surface area of. 33. If v is a tangent vector to a smooth closed surface which bounds a solid, what is div v dv? olution: Again, use the divergence theorem backwards to get vdv = v nd = since v is tangent to and n is the normal to. 34. Evaluate F n d, where F (x, y, z) = (x, y, z) and is the boundary of the solid region enclosed by z = 1 x 2 y 2 and z =. olution: Use Gauss (divergence) theorem to evaluate the integral. Note that F = 3 and so F nd = F dv = 3dV, z 1 x 2 y 2 z 1 x 2 y 2 which, after switching to cylindrical coordinates, becomes 2π 1 1 r 2 3 r dz dr dθ = 3π 2. 35. Let v = (x, y, z) and F = v / v 3. (a) ompute F n d directly. x 2 +y 2 +z 2 =1 (b) ompute ( F )dv directly. (c) omment. x 2 +y 2 +z 2 1
9 olution: (a) ince we are dealing with the unit sphere, then n = v / v = v. Also notice that F = v (on the sphere). Thus F n d = 2 v d = 1d x 2 +y 2 +z 2 =1 x 2 +y 2 +z 2 x 2 +y 2 +z 2 =1 which is equal to the surface area of the unit sphere, i.e., 4π. (b) A routine computation shows that F = and so ( F )dv =. x 2 +y 2 +z 2 1 (c) Thought it seems like something is wring with Gauss theorem (which says (a) and (b) should be the same), nothing is wrong here since the vector field F is not differentiable at the origin. Thus Gauss theorem cannot be applied. 36. Let be the unit sphere centered at the origin (,, ) and F (x, y, z) = (sin(xyz), sin(x 2 y 2 z 2 ), 1). ompute where v = F v d, 1 (x, y, z). x2 + y 2 + z2 olution: The actual vector field is not important here. The important thing to notice is that since we are dealing with the unit sphere, the vector v is the outward unit normal to (check this!). By Gauss theorem (since we are working with a closed surface) F v d = ( F )dv = x 2 +y 2 +z 2 <1 since ( F ) = for any vector field F. 37. A function g(x, y, z) is said to be harmonic if g xx + g yy + g zz =. If g is harmonic and is the boundary surface of a region R, with outer normal n, compute g n d. olution: By Gauss theorem, since we are assuming g is harmonic. g n d = R ( g)dv = R (g xx + g yy + g zz)dv = 38. Verify the ivergence Theorem for the region bounded by the cone z = x 2 + y 2 and the plane z = 2 and F (x, y, z) = (x, 3xz, y). olution: oing this directly: There are two parts to the surface; 1 : the plane z = 2, and 2 : the cone z = x 2 + y 2. For the 1, the parameterization is X(u, v) = (u, v, 2), u 2 + v 2 4. The outward unit normal is clearly n = (,, 1) and d = X u X v da = da. Thus 1 = (u, u 2 +v 2 4 = u 2 +v 2 4 F nd 6u, v) (,, 1)dA vda. witching to polar coordinates gives 1 F nd =. For the surface 2, the parameterization is X(u, v) = (u, v, u 2 + v 2 ), u 2 + v 2 4. Note that u X u X v = ( u 2 + v, v 2 u 2 + v, 1). 2 Unfortunately, this is neither unit not outward (it points up - not down!). Notice that X u X v = 2 and so the outward unit normal is n = 1 u ( 2 u 2 + v, v 2 u 2 + v, 1). 2
1 A computation shows that u 2 F nd = ( + 3uv v)da, u 2 2 u 2 +v 2 + v2 4 which, after switching to polar coordinates, is equal to π8/3. By using Gauss theorem, we get F nd = F dv = 1 2 2π 2 2 1dV = rdzdrdθ = 8 r 3 π. 39. Is there a vector field G such that G = (xy 2, yz 2, zx 2 )? olution: No. Note that ( G). We know from elementary facts that the the divergence of the curl is equal to zero. 4. For the vector field F = (3z, 4x, 2y) and the surface z = 4 x 2 y 2, z, evaluate in two ways: (a) directly and (b) using tokes theorem. ( F ) n d olution: The surface is parameterized by X : 2 A computation shows that and moreover the outward unit normal n is given by Thus X(u, v) = (u, v, 4 u 2 v 2 ), = {(u, v) : u 2 + v 2 4}. X u X v = (2u, 2v, 1) = 1 + 4u 2 + 4v 2 n = which, after switching to polar coordinates, is equal to 16π. tokes theorem says Xu Xv X u X = (2u, 2v, 1) v 1 + 4u 2 + 4v. 2 ( F ) nd = (2, 3, 4) (2u, 2v, 1)dA u,v, ( F ) nd = F d, where here (t) = (2 cos t, 2 sin t, ), t 2π is the circle which bounds the surface. The line integral can be computed as 2π 2π 2π F d = F ((t)) (t)dt = (, 8 cos t, 4 sin t) ( 2 sin t, 2 cos t, )dt = 16 cos 2 tdt = 16π, which, thankfully, matches the answer we got before. 41. Let be a region in the plane which is bounded by a smooth positively oriented curve. Imagining as a surface and letting F (x, y, z) = (P (x, y), Q(x, y), ), use tokes theorem to derive Green s theorem. olution: tokes theorem says that ( F ) nd = F d. In this special case where is a surface consisting of a region in the plane, n = (,, 1) and d = da (regular area measure in the plane). A computation shows that F = (P, Q, ) = Q x P y. Putting this altogether we get which is Green s theorem. F d = ( F ) nd = (Q x P y)da 42. Evaluate F n d, where F (x, y, z) = (x 2 e yz, y 2 e xz, z 2 e xy ) and is the hemisphere x 2 + y 2 + z 2 = 4, z >. olution: Without the use of tokes theorem, the direct calculation of his integral would be prohibitive. By tokes theorem, F n d = F d, where (t) = (2 cos t, 2 sin t, ) is the boundary (with the proper orientation) of the hemisphere. The line integral can be computed directly as 2π 2π F d = F ((t)) (t)dt = (4 cos 2 t, 4 sin 2 t, ) ( 2 sin t, 2 cos t, )dt, which after a computation (using substitution) equals zero.
11 43. onsider the surface, x 2 + y 2 + 3z 2 = 1 and any differentiable vector field F. ompute the following integral F n d, where n is a outward unit normal to in two distinct ways. olution: The first way is to use Gauss theorem. F n d = R ( F )dv = since ( F ) = (the divergence of the curl is zero). For the second way, cut the surface in half (say along the xy plane) into two pieces + (the top half) and (the bottom half) and let denote the curve in the plane which lies at their intersection. Use tokes theorem on + to get that + F n d = F d, where circulates counterclockwise (important!). Use tokes theorem on to get that F n d = F d, where circulates clockwise (important!). Notice that the last line integral is the negative of the other (since they circulate in opposite directions) and so their sum is zero. 44. Verify tokes theorem for F = (z 2, x 2, y 2 ) and the surface is the part of the plane x + y + z = 1 lying in the first octant. olution: oing this integral directly: First parameterize the triangle as X(u, v) = (u, v, 1 u v), v 1 u, u 1. From the equation of the plane x + y + z = 1 we know that the vector (1, 1, 1) is normal to this triangle and is points out. normal is thus n = 1 (1, 1, 1). The d = X u X 3 v = 3dA and so Using tokes theorem: We need to compute 1 1 u 1 ( F ) nd = (2v, 2(1 u v), 2u) (1, 1, 1) 3dvdu = 1. 3 1 2 3 F d, where 1, 2, 3 are the sides of the triangle. They are parameterized by Note that and it is easy to check that and so the total line integral is equal to 1 (as it should be). 1 (t) = (1 t, t, ), 2 (t) = (, 1 t, t), 3 (t) = (t,, 1 t), t 1. F d = z 2 dx + x 2 dy + y 2 dz 1 2 3 1 2 3 1 z 2 dx + x 2 dy + y 2 dz = (1 t) 2 dt = 1 i 3 The outward unit 45. Let F = (x 2 + y, yz, x z 2 ) and be the triangle defined by 2x + y + 2z = 2, x, y, z. ompute ( F ) n d. olution: By tokes theorem, the above integral is equal to F d = (x 2 + y)dx + yzdy + (x z 2 )dz, where is the triangle with vertices (1,, ), (, 2, ), (,, 1) - traversed in that order. The three boundary curves are 1 (t) = (1 t, 2t, ), 2 (t) = (, 2 2t, t), 3 (t) = (t,, 1 t), t 1. One can show that 1 (x 2 + y)dx + yzdy + (x z 2 )dz = ((1 t) 2 + 2t)( 1)dt = 4 1 3 1 (x 2 + y)dx + yzdy + (x z 2 )dz = ((2 2t)t( 2) t 2 )dt = 1 2 1 (x 2 + y)dx + yzdy + (x z 2 )dz = (t 2 + t (1 t) 2 ( 1))dt = 7 3 6 The integral is the sum of the integrals above which is 7/6.
12 46. If F (x, y, z) = A(x, y, z) i + B(x, y, z) j + (x, y, z) k such that there is some point P such that ( F )(P ) (,, ). oes there exist a scalar function φ such that F = φ? olution: No! If F = φ, then ( φ) = (,, ). 47. tate tokes Theorem, and explain why both integrals in tokes Theorem will be if the function F = f for some f (there is a different reason for the two integrals). oln: tokes Theorem states that under suitable conditions on the function and the surface, (curlf ) d = F d. If F = f for some f, then curlf = f =, so (curlf ) d =. In the line integral, if F = f for some f, then F d = f(c(b)) f(c(a)). ince is a simple closed curve, c(b) = c(a) and hence F d =.