F = 0. x ψ = y + z (1) y ψ = x + z (2) z ψ = x + y (3)


 Mervin Booth
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1 MATH 255 FINAL NAME: Instructions: You must include all the steps in your derivations/answers. Reduce answers as much as possible, but use exact arithmetic. Write neatly, please, and show all steps. Scientists and engineers uphold very high standards of ethics: the work you submit in this exam must be yours. Be prepared to explain your answers in person. Also, please document your take home final: keep all of your calculations (work not included in your final exam submission until you get a final grade in your class).. (25 pts) Consider the vector field F = y + z, x + z, x + y. Determine whether the vector field is conservative and if so, find an associated scalar potential ψ(x, y, z). F =. Hence, conservative and thus F = ψ(x, y, z). We write x ψ = y + z () y ψ = x + z (2) z ψ = x + y (3) Integrate () with respect to x: ψ = (y + z)x + h(y, z). Differentiate this expression and match to (2): ψ y = x + y h = x + z. Hence y h(y, z) = z, which we integrate with respect to y: h(y, z) = yz + f(z). Hence ψ + (y + z)x + yz + f(z). We differentiate this expression with respect to z and match to (3): z ψ = x + y + f (z) = x + y, hence f (z) = which we integrate in z to get f(z) = c, a constant. Hence ψ = (y + z)x + yz + c.
2 2. (5 pts) Assume g is a scalar function and f a vector function. Use the identity [gf] = fg + g f to compute the divergence of F = r. Here r = xî + yĵ + zˆk, and r 3 r = x 2 + y 2 + z 2, the magnitude of r. r = 3. and r 3 = 3r 5 r. Hence F = rr 3 + r 3 r. Hence F = 3r 3 3r 5 r r = 3r 3 3r 5 r 2 =. 2
3 3. ( pts) Find a vector normal to the surface z = x 2 /4 y 2 /6. It does not have to be a unit vector. let g = z x 2 /4 y 2 /6 =. A level set. Now, Note that this is not a unit vector. N = g. g = ˆk + 2 xî( x2 /4 y 2 /6) /2 + 8 yĵ( x2 /4 y 2 /6) /2 This can be written as If g = z + x 2 /4 y 2 /6 =, then N = ˆk + y [xî + 4z 4ĵ]. N = ˆk y [xî + 4z 4ĵ]. 3
4 4. (5 pts) Find the circulation C F dr, where C is the perimeter of the triangle given by the plane 2x + y + z = 2, in the first octant. The field F = 3xzĵ. Answer: you can do it as a line integral or as a surface integral: Here we use Stokes s Theorem: C F dr = S Fˆndσ. S is the triangle. As a surface integral: We compute curlf = 3( xˆk + zˆk). n = g/ g, where g = 2x + y + z 2 =, or ˆn = 6 (2î + ĵ + ˆk). The differential surface: Hence the integral to be solved is S Fˆndσ = dσ = dxdy ˆn ˆk = 6dxdy. 2 2x As a line integral: F dr = 3xzdy. Hence C F dr = y 2 zdy + 2 3( 4x y + 2)dydx =. 3x(2 y)dy + 3xzdy =, since in the first integral z =, in the second one x =, and in the third z = and the limits of integration are. 4
5 5. ( pts) Find the surface area 2πah, of the side of a cylinder of radius a and height h using either the projection method or the Jacobiantransformation method Answer: parametrize, Then r(u, v) = a cos uî + a sin uĵ + vˆk. r r =< a sin u, a cos u, > u v =<,, >. Hence r u r =< a cos u, a sin u, >. v Finally, < a cos u, a sin u, > = a, u 2π, and v h. Then side ds = h 2π adudv = 2πah. 5
6 6. (25 pts) Let F = P (x, y, z)î + Q(x, y, z)ĵ + R(x, y, z)ˆk be a vector field. Derive the conditions on F that guarantee that C F dr is path independent. Here, r = xî + yĵ + zˆk, the path starts at P and ends at P, two points in 3D space. Write out explicitly the line integral in scalar form. Match the exact differential dψ = ψ dx integral. dx + ψ dy ψ dy + dz dz to the integrand of the line Find conditions on F that associate it to the different components of the exact differential. Write a vector identity that encapsulates the conditions on F found above. Write the answer to the integral in terms of ψ. Answer: C F dr = C P dx+qdy +Rdz. Now if F =, then P dx+qdy +Rdz = dψ, where dψ = x ψdx + y ψdy + z ψdz. Hence C F dr = P P dψ = ψ(p ) ψ(p ). 6
7 7. (3 pts) Let ρ(x, y, z) = z 3 be the density of a solid. Compute its total mass V ρdv, where V is bounded by z =, z = 4+sin(2x)+cos(2y), and π x π and π y π. Hint: exploit the Divergence Theorem. Answer: we use V FdV = S F ds. We let F = 4 z4ˆk, so that F = z 3. So we then compute the surface integral S F ds, where S is the surface of the box. The integral S 4 z4ˆk ds = flat bottom surface 4 z4ˆk ds+ sides of box 4 z4ˆk ds+ wavy top 4 z4ˆk ds. Since z = on the flat bottom surface, the first integral on the right hand side is zero. Since the dot product of F and the normal to the sides of the box is zero, no contribution from that integral. Hence Hence, S wavy top 4 z4ˆk ds = wavy top 4 z4ˆk ds. 4 z4ˆk ds = square 4 z4ˆk ˆn dxdy ˆn ˆk. on the wavy top, F = 4 [4 + sin(2x) + cos(2y)]4ˆk. Hence square π π 4 [4 + sin(2x) + cos(2y)]4 dxdy = π π 4 [4 + sin(2x) + cos(2y)]4 dxdy. The integrand contributes just a few nonzero terms. To find these, first expand a = (4 + cos(2y)), then (a + exp(i2x) exp( i2x) 2i ) 4 will have only the terms a 4 + 3a 2 + 3/8 nonsinusoidal in x. Then we expand (4 + exp(2y)/2 + exp( 2y)/2) 4 and (4 + exp(2y)/2 + exp( 2y)/2) 2 and retain only the nonsinusoidal terms in y. We thus obtain π π π π 4 [4 + sin(2x) + cos(2y)]4 dxdy = π π π π dxdy = 47π2 /4. 7
8 8. (3 pts) Consider a rectangular region D of the x y plane that excludes the origin. Find p such that the circulation on the perimeter of the region D is zero, for where r 2 = x 2 + y 2. F = y3 r p î xy2 r p ĵ, Answer: Compute the curl and set it to zero. We obtain p = 4. That is, if p = 4. curlf = ˆk[ y 2 r p ( r 2 px 2 r 2 py 2 + 4] = ˆk[ y 2 r p ( r 2 r 2 p + 4) =, 8
9 9. (4 pts) Let R be a region in a plane that has a unit normal ˆn = a, b, c and boundary C. Let F = bz, cx, ay. (a) Show that F = ˆn. (b) Show that the area of R is given by C F dr. (c) Consider a curve C given by r = 5 sin t, 3 cos t, 2 sin t, for t 2π. Prove that C lies in a plane by showing that r dr dt is constant for all t. (d) Use part (b) to find the area of the region enclosed by C in part (c). Hint: find the unit normal consistent with the orientation of C. Answer: This can be done as a surface or line integral thanks to Stokes theorem. In part (a) all you need to do is to compute F = ˆn. In part (b), F dr = ( F) ds = ˆn ds = ds = R. C R We compute v = 5 cos t, 3 sin t, 2 cos t, and then the cross product r v = 78 2,, which is a constant vector, for all t. To find a unit normal vector we compute r(π/2) r() = 5,, 2, 3,, and r(3π/2) r() = 5, 3, 2. Taking the cross product and normalizing n = 3 2,, 5. Hence F = 3, 5x, 2y. Thus, F v = 3(52 sin 2 t cos 2 t). Finally, 2π dtf v = 3 2 2π R R dt( ) = 3 3 π. 9
10 . Optional, extra credit. Worth up to 5 pts. Caculate the surface area of a hemisphere x 2 + y 2 + z 2 = 9. Answer: one can work this out using spherical or cylindrical polar coordinates. Recall that the surface of a sphere is 4πr 2, where r is its radius. Hence, the surface of this hemisphere is 8π. Using spherical coordinates, the surface S of a hemisphere is S = π/2 dφ 2π 9 sin φ = 8π π/2 dφ sin φ = 8π. Using cylindrical coordinates and the projection technique we find that the normal to the hemisphere is ˆr = r/ r. Hence ˆk ˆn = z/3, z >. So the surface area can be found by projecting on the x y plane which makes a shadow of a circle of radius 3. Thus S = R dxdy 2π 3 z/3 = 3 dθ rdr 3 = 6π 9 r 2 rdr 9 r 2. then a change of variable u = 9 r 2, so that du/2 = rdr, leads to the expected result after the integration 3π 9 u /2 du = 6πu /2 9 = 8π.. Optional, extra credit. Worth up to 5 pts. CHAPTER4 REVIEW, number 45. Answer: F = κ T, where T = exp( x 2 y 2 z 2 ). Let ρ 2 = x 2 + y 2 + z 2. Hence, F = 2κ exp( ρ 2 )r. We use the divergence theorem to find the total flux across the unit sphere, centered at the origin: F ds = FdV, S we use the volume integral. We need the divf = 2(3 2ρ 2 ) exp( ρ 2 ) unit sphere Fρ2 sin ψdρdφdψ = 2κ Integrating in θ and ψ, 6κπ V π 2π dψ sin ψ dθ dρρ 2 F. ρ 2 (3 2ρ 2 ) exp( ρ 2 )dρ = 6κπ ( 2ρ 3 exp( ρ 2 ) ρ= = 32πκ.
This makes sense. t 2 1 + 1/t 2 dt = 1. t t 2 + 1dt = 2 du = 1 3 u3/2 u=5
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