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Midterm 1 Review

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Math 5C, Solutions to Midterm 1 Review ProblemsWinter 20071. ComputeZCyz dx + 2x dy − y dz where C is the straight line path from (1, 2, 1) to(−1, 3, 0).Solution. The straight line path between the two points has parametric equations(x, y, z) = (−2t + 1, t + 2, −t + 1). Thus, using the definition of line integrals, theintegral becomesZ10(t + 2)(1 − t)(−2) + 2(1 − 2t) − (t + 2)(−1) dt =Z102t2− t dt= 2t3/3 − t2/2]10= 1/6.2. Find the surface area of the surface S, which is parametrized byφ(u, v) =x(u, v) = u − vy(u, v) = u + vz(u, v) = uvfor all (u, v) with u2+ v2≤ 1.Solution. We use the formula S.A. =R R√EG − F2du dv. Here, E = x2u+y2u+z2u=2 + v2, F = xuxv+ yuyv+ zuzv= 1 − 1 + vu = uv, and G = x2v+ y2v+ z2v= 2 + u2.Thus EG − F2= 4 + 2(u2+ v2), andS.A. =Z Zu2+v2≤1p4 + 2(u2+ v2) du dv=Z2π0Z10√4 + 2r2r dr dθ=14Z2π023(4 + 2r2)3/2]10dθ=π3(63/2− 43/2) =π3(6√6 − 8)3. Let S be the top half of the unit sphere (i.e., S is given by x2+ y2+ z2= 1 and z ≥ 0),oriented by the outer normal. IntegrateZ ZSx dy dz + y dz dx + z2dx dy.Solution. S has parametrization x = cos θ sin φ, y = sin θ sin φ and z = cos φ for0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π/2. Notice that this parametrization induces the innernormal on the sphere, and so we need to multiply the integral by −1 when we write itin terms of the parameters θ and φ.1We now compute the Jacobians we need (these are also the components of the normalvector).dy dz =cos θ sin φ sin θ cos φ0 −sin φdθ dφ = −cos θ sin2φ dθ dφ,dz dx =0 −sin φ−sin θ sin φ cos θ cos φdθ dφ = −sin θ sin2φ dθ dφ,dx dy =−sin θ sin φ cos θ cos φcos θ sin φ sin θ cos φdθ dφ = −sin φ cos φ dθ dφ,ThusZ ZSx dy dz + ··· = −Zπ/20Z2π0−cos2θ sin3φ − sin2θ sin3φ − sin φ cos3φ dθ dφ= 2πZπ/20(sin2φ + cos3φ) sin φ dφ= 2πZπ/20(1 − cos2φ + cos3φ) sin φ dφ= −2π(cos φ −13cos3φ +14cos4φ)]π/20=11π6.4. Let S be the surface given by z = xy2−3x2with upper normal n, over the square withvertices (±1, ±1) in the xy-plane. I f w = (z + 3x2)i + yzj + y2k, calculateZ ZSw · n dσ.Solution.Z ZSw · n dσ =Z1−1Z1−1(−wx∂z∂x− wy∂z∂y+ wz) dx dy=Z1−1Z1−1−(xy2− 3x2+ 3x2)(y2− 6x) − y(xy2− 3x2)(2xy) + y2dx dy=Z1−1Z1−16x2y2− xy4− 2x2y4+ 6x3y2+ y2dx dy=Z1−16y2− 4y4/3 dy =5215.25. Let S be the unit sphere, x2+ y2+ z2= 1, oriented outward, and let F be the vectorfield F(x, y, z) = xy2i − xz2j + x2z k. Use the Divergence Theorem to computeZ ZSF · n dσ.Solution.Z ZSF · n dσ =Z Z ZRdiv(F) dxdydz=Z Z ZRy2+ x2dxdydz=Zπ0Z2π0Z10(ρ2sin2φ)ρ2sin φ dρdθdφ=Zπ0Z2π015sin3φ dθdφ=Zπ02π5(1 − cos2φ) sin φ dφ= −2π5(φ −13cos3φ)]π0= −2π5(π + 1/3 − 0 −−1/3) = −2π25−4π156. Let S be the cone x2= y2+ z2, 0 ≤ x ≤ 2, oriented inward (so the normal vectorspoint toward the x-axis). Use Stokes’ Theorem to calculateZ ZS(∇×F) ·n dσ, whereF(x, y, z) = x2i − zj + (y2− z)k.Solution. Recall ∇×F = curl(F), so by Stokes’ Theorem, the surface integral reducesto the line integralHCFTds where C is the boundary of S. Now S is a cone with vertexat the origin, and so its boundary is a circle lying in the plane x = 2. By the right-hand rule, since the normal should be roughly in the direction of the x -axis, the circleC should be traversed in the counterclockwise direction of the yz-plane. Thus C hasparametric equations (x, y, z) = (2, cos t, sin t) for 0 ≤ t ≤ 2π, andZCFTds =ZCx2dx − z dy + (y2− z) dz=Z2π04(0) − sin t(−sin t) + (cos2t − sin t) cos t dt=Z2π0sin2t + (cos2t − sin t) cos t dt=Z2π012(1 − cos(2t)) + (1 −sin t − sin2t) cos t dt=12(t −12sin(2t)) + sin t −12sin2t −13sin3t]2π0= π.37. Let C be the curve given by x = sin t, y = cos t, z = cos(2t) for 0 ≤ t ≤ 2π. UseStokes’ Theorem to evaluateICxz dx + y2dy + z2dz.Solution. Since z = cos(2t) = cos2t − sin2t = y2− x2, the curve C lies on thesurface z = y2− x2(the surfaces z = 2y2− 1 or z = 1 − 2x2would also work). Sincex = sin t, y = cos t traces out the unit circle, the interior of C lies above the unitdisk in the xy-plane. Thus C is the boundary of the surface S given by the graphof z = y2− x2for x2+ y2≤ 1. Furthermore, since C is traversed in the clockwisedirection, S must be given the lower normal according to the right-hand rule (thus wemultiply by −1). Now, using Stokes’ Theorem, we haveICxz dx + y2dy + z2dz =Z ZScurl(xzi + y2j + z2k) · n dσ= −Z ZSx dz dx= −Z Zx2+y2≤1−x(2y) dx dy=Z1−1Z√1−x2−√1−x22xy dy dx=Z1−12x(1 − x2) dx = x2− x4/2]1−1= 0.8. Show that the integralZ(π/2,0,1)(−1,1,3)z2cos(x + y2) dx + 2yz2cos(x + y2) dy + 2z sin(x + y2) dzis independent of path and evaluate it.Solution. To show that the integral is path independent, it suffices to find a functionF (x, y, z) such that the integrand equals dF . To get F , integrate the dx term withrespect to x to get F =Rz2cos(x + y2) dx = z2sin(x + y2) + C(y, z). If we nowdifferentiate this function with respect to y and z (separately), we get the other twoterms of the integrand when we let C(y, z) = 0. Thus the integral becomesZ(π/2,0,1)(−1,1,3)d(z2sin(x + y2)) = z2sin(x + y2)](π/2,0,1)(−1,1,3)= sin(π/2) − 9 sin 0 = 1.49. Let u be the vector fieldu(x, y, z) =yx2+ y2i −xx2+ y2j + z2kon R3minus the z-axis.(a) Show that curl(u) = 0 on this domain.Solution.curl(u) =i j k∂∂x∂∂y∂∂zyx2+y2−xx2+y2z2= 0i + 0j + (x2− y2(x2+ y2)2−x2− y2(x2+ y2)2)k = 0.(b) Show that u is not the gradient vector field of any function F on this domain.(Hint: Find a closed curve C withRCuTds 6= 0.)Solution. Let C be the unit circle in the xy-plane: x = cos t, y = sin t, z = 0for 0 ≤ t ≤ 2π. ThenICuTds =Z2π0(sin tcos2t + sin2t(−sin t) −cos tcos2t + sin2t(cos t) + 0) dt=Z2π0−(sin2t + cos2t) dt= −2π 6= 0.Since this integral is not zero, we know that the integralRuTds is not path-independent in the given domain, and hence u is not a gradient vector field. (Youcould also prove this more directly by trying to solve for F (x, y, z) with ∇F = u,and showing that no solutions


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