MATH 285, HONORS MULTIVARIABLE CALCULUS, FALL 1999

Answers to the Final Exam

Problem 1 Find an equation of the tangent plane to the surface z + 1 = xe^{y} cos z at the point (1, 0, 0).

Answer: We can write the surface as f(x, y, z) = 1, where f(x, y, z) = xe^{y} cos z − z. The tangent plane that we are looking for will be parallel to the level surface of f at the given point. We also know that the gradient is orthogonal to the level surface, and thus the gradient of f will be a normal vector for the desired tangent plane. We calculate ∇f = ⟨e^{y} cos z, xe^{y} cos z, −xe^{y} sin z − 1⟩ = ⟨1, 1, −1⟩. We need an equation for the plane through r_{0} = (1, 0, 0) with normal vector n = ⟨1, 1, −1⟩. Such an equation is r · n = r_{0} · n, or x + y − z = 1.

Problem 2 Calculate ∫∫_{D} x^{2} + y^{2} dA, where D is the unit disk x^{2} + y^{2} ⩽ 1.

Answer: This is best calculated in polar coordinates. ∫∫_{D} x^{2}+y^{2} dA = ∫∫_{D} r^{2}·r drdθ = ∫_{0}^{2π} ∫_{0}^{1} r^{3} drdθ = 2π · \frac{1}{4} = π/2.

Problem 3 The parametric equations

\begin{aligned}x & = & t^{3} − t \\ y & = & t^{4} − t^{2} \end{aligned}

describe a curve which has two loops, as shown in the illustration. Calculate the area enclosed by one of the loops.

Answer: The curve passes through the origin at t = −1, t = 0, and t = 1. Thus, one of the loops is traced out for 0 ⩽ t ⩽ 1 (the other loop is symmetrical). We calculate the area:

∫ x dy = ∫_{0}^{1} xy′ dt
= ∫_{0}^{1}(t^{3} − t)(4t^{3} − 2t)dt
= ∫_{0}^{1} 4t^{6} − 6t^{4} + 2t^{2} dt
= [\frac{4}{7}t^{7} − \frac{6}{5}t^{5} + \frac{2}{3}t^{3}]_{0}^{1}
= \frac{4}{7} − \frac{6}{5} + \frac{2}{3}
= \frac{60−126+70}{105}
= \frac{4}{105}.

Problem 4 Minimize x^{2} + 4y^{2} + 9z^{2} subject to the constraint xyz = \frac{4}{3}, for x, y, z > 0.

Answer: We use he method of Lagrange multipliers. Let f(x, y, z) = x^{2} +4y^{2} +9z^{2} and g(x, y, z) = xyz. Since f and g are differentiable, we know that all local minima occur at points (x, y, z) that satisfy

∇f(x, y, z) = λ · ∇g(x, y, z),

for some λ. This means,

⟨2x, 8y, 18z⟩ = λ · ⟨yz, xz, xy⟩.

So we get λ = 2x/yz = 8y/xz = 18z/xy, and, by multiplying everything by xyz/2, x^{2} = 4y^{2} = 9z^{2}. Thus we have y = x/2 and z = x/3. We also know that the point we are looking for satisfies the constraint xyz = 4/3, thus x^{3}/6 = 4/3, thus x^{3} = 8 or x = 2. It follows that ⟨x, y, z⟩ = ⟨2, 1, 2/3⟩ is the only critical point. This is indeed the global minimum.

Problem 5 Compute the outward flux of F(x, y, z) = (x^{2} + sin y)i + y^{2}j + (x − 2z)yk through the surface of the cube 0 ⩽ x ⩽ 2, 0 ⩽ y ⩽ 2, 0 ⩽ z ⩽ 2.

Answer: By the Divergence Theorem, the flux is equal to \begin{matrix}∫ \\ ∫ \\ ∫ \end{matrix}_{E} div F dV, where E is the cube in question. We calculate div F = 2x + 2y − 2y = 2x. Thus, \begin{matrix}∫ \\ ∫ \\ ∫ \end{matrix}_{E} div F dV = ∫_{0}^{2} ∫_{0}^{2} ∫_{0}^{2} 2x dxdydz = 16.

Problem 6 Let F(x, y, z) = (x + e^{y})i + (xe^{y} + 2ye^{z})j + (y^{2}e^{z} + 4z)k. Evaluate ∫_{C} F · dr, where C is the curve with initial point (0, 0, 2) and terminal point (1, 0, 0) shown in the figure.

Answer: We calculate curl F = 0 and conclude that F is conservative. A potential field is easily found: F = ∇f where f = x^{2}/2 + xe^{y} + y^{2}e^{z} + 2z^{2}. So by the fundamental theorem for line integrals, we have ∫_{C} F · dr = f(1, 0, 0) − f(0, 0, 2) = 1.5 − 8 = −6.5.

Problem 7 Evaluate the integral ∫∫_{S} curl F·dS, where S is the part of the sphere x^{2} +y^{2} +z^{2} = 1 with z ⩾ 0, and F(x, y, z) = (z − y)i + \frac{x}{x^{2}+y^{2}} j + e^{xyz}k. Choose the upward orientation for S.

Answer: By Stokes’ Theorem, ∫∫_{S} curl F · dS = ∫_{C} F · r, where C is the boundary of S. Thus, C is the curve x^{2} + y^{2} = 0 with z = 0, oriented in the positive direction. We parameterize C with 0 ⩽ t ⩽ 2π as

\begin{aligned}x(t) & = & cos t \\ y(t) & = & sin t \\ z(t) & = & 0. \end{aligned}

We calculate r′(t) = ⟨− sin t, cos t, 0⟩ and F(x(t), y(t), z(t)) = (0−sin t)i+ \frac{cos t}{cos^{2} t+sin^{2} t}j+e^{cos t·sin t·0}k = ⟨− sin t, cos t, 1⟩. Now we can calculate the line integral: ∫_{C} F · r = ∫_{0}^{2π} F · r′dt = ∫_{0}^{2π}⟨− sin t, cos t, 1⟩ · ⟨− sin t, cos t, 0⟩dt = ∫_{0}^{2π} 1dt = 2π.

Problem 8 Assume that f and g are scalar fields with continuous second-order partial derivatives. Prove that div(∇f × ∇g) = 0.

Answer:

div(∇f × ∇g) = div(⟨f_{x}, f_{y}, f_{z}⟩ × ⟨g_{x}, g_{y}, g_{z}⟩)
= div(⟨f_{y}g_{z} − f_{z}g_{y}, f_{z}g_{x} − f_{x}g_{z}, f_{x}g_{y} − f_{y}g_{x}⟩)
= \frac{∂}{∂x}(f_{y}g_{z} − f_{z}g_{y}) + \frac{∂}{∂y} (f_{z}g_{x} − f_{x}g_{z}) + \frac{∂}{∂z} (f_{x}g_{y} − f_{y}g_{x})
= f_{yx}g_{z} + f_{y}g_{zx} − f_{zx}g_{y} − f_{z}g_{yx}
+f_{zy}g_{x} + f_{z}g_{xy} − f_{xy}g_{z} − f_{x}g_{zy}
+f_{xz}g_{y} + f_{x}g_{yz} − f_{yz}g_{x} − f_{y}g_{xz}
= 0

Problem 9 Use Green’s Theorem to evaluate the line integral ∫_{C} 4xy−4y dx+2x^{2} dy, where C is the positively oriented curve that consists of the line segment from (−2, 0) to (2, 0) and the top half of the circle x^{2} + y^{2} = 4.

Answer: Let ⟨P, Q⟩ = ⟨4xy − 4y, 2x^{2}⟩. By Green’s theorem, the line integral ∫_{C} 4xy − 4y dx + 2x^{2} dy is equal to

∫∫_{D} (\frac{∂Q}{∂x} − \frac{∂P}{∂y}) dA = ∫∫_{D} 4x − (4x − 4) dA = ∫∫_{D} 4 dA,

where D is the region enclosed by C. The answer is 4 times the area of D, or 8π.