Calculus II, Spring 2016 (http://www.math.nthu.edu.tw/ wangwc/)

Solutions to selected problems in HW for Week 15

1. Section.16.3: Problem 26. Check the component test conditions:

∂_{y} (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} ) = −\frac{yz}{(x^{2} + y^{2} + z^{2})^{3/2}} = ∂_{z} (\frac{y}{\sqrt{x^{2} + y^{2} + z^{2}}} )

∂_{x} (\frac{y}{\sqrt{x^{2} + y^{2} + z^{2}}} ) = −\frac{xy}{(x^{2} + y^{2} + z^{2})^{3/2}} = ∂_{y} (\frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}} )

∂_{x} (\frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} ) = −\frac{xz}{(x^{2} + y^{2} + z^{2})^{3/2}} = ∂_{z} (\frac{x}{\sqrt{x^{2} + y^{2} + z^{2}}} )

Since the natural domain D = {(x, y, z) ∈ ℝ^{3} : (x, y, z) ≠ (0, 0, 0)} is simply connected, we can conclude that the line integral is independt of path.

2. Section.16.4: Problem 10. Let F = Mi + Nj and R the region enclosed by the curve C. Since M and N have continuous first partial derivatives everywhere, we can apply Green’s Theorem.

∮_{C} F · T ds = ∫∫_{R}(N_{x} − M_{y}) dxdy ∫∫_{R}(2 − 3) dxdy = (−1)Area(R) = −π\sqrt{2}

∮_{C} F · n ds = ∫∫_{R}(M_{x} + N_{y}) dxdy = ∫∫_{R}(1 − 1) dxdy = 0

3. Section.16.4: Problem 38. Let F = Mi + Nj and C an arbitrary piecewise smooth simple closed curve enclosing a region R. Since M and N have continuous first partial derivatives everywhere, we can apply Green’s Theorem.

∮_{C} F · T ds = ∫∫_{R} 1 − (\frac{1}{4}x^{2} + y^{2}) dxdy

which admits the maximal value as R = \{(x, y) : 1 − \frac{1}{4}x^{2} + y^{2} ≥ 0\}, that is, as C is the curve \frac{1}{4}x^{2} + y^{2} = 1.

4. Section.16.4: Problem 39. a. First compute ∇f = (\frac{2x}{x^{2}+y^{2}} , \frac{2y}{x^{2}+y^{2}} ). Since the components of ∇f are not defined

at (0, 0), we cannot apply Green’s Theorem on the region enclosed by any circle C : x^{2} + y^{2} = a^{2}, a > 0. Thus we compute the line integral directly. Let r(t) = (a cos t, a sin t), 0 ≤ t ≤ 2π. (counterclockwise)

∮_{C} ∇f · n ds = ∫_{0}^{2π} (\frac{2 cos t}{a}, \frac{2 sin t}{a}) · (a cos t, a sin t) dt = 4π.

b. Let R be the region enclosed by K. If (0, 0) lies ouside K, then the components of ∇f have continuous derivaives at (0, 0), and thus we can apply Green’s Theorem.

∮_{K} ∇f · n ds = ∫∫_{R} ∂_{x} (\frac{2x}{x^{2} + y^{2}} ) + ∂_{y} (\frac{2y}{x^{2} + y^{2}} ) dxdy

= ∫∫_{R} \frac{2(y^{2} − x^{2})}{(x^{2} + y^{2})^{2}} + \frac{2(x^{2} − y^{2})}{(x^{2} + y^{2})^{2}} dxdy = 0

If (0, 0) lies inside K, then we choose a circle curve C : x^{2} + y^{2} = a^{2} with a > 0 small enough so that C totally lies inside K (no intersection). Then the new region R enclosed by these two curves K and C doesn’t contain (0, 0), and thus we can apply Green’s Theorem.

∮_{K} ∇f · n ds ⟲ + ∮_{C} ∇f · n ds ⟳
= ∮_{K} (\frac{2x}{x^{2} + y^{2}} dy − \frac{2y}{x^{2} + y^{2}} dx) ⟲ + ∮_{C} (\frac{2x}{x^{2} + y^{2}} dy − \frac{2y}{x^{2} + y^{2}} dx) ⟳
= ∫∫_{R} ∂_{x} (\frac{2x}{x^{2} + y^{2}} ) + ∂_{y} (\frac{2y}{x^{2} + y^{2}} ) = 0
⇒ ∮_{K} ∇f · n ds + (−4π) = 0
⇒ ∮_{K} ∇f · n ds = 4π

5. Homework assignment: Problem 2. Let D = {(x, y, z) ∈ ℝ^{3} : x^{2} + y^{2} ≠ 0}.

(a) Check directly.

(b)

∂_{z}f = 0 ⇒ f(x, y, z) = g(x, y)

⇒ ∂_{x}f = ∂_{x}g = \frac{x}{\sqrt{x^{2} + y^{2}}}

⇒ g(x, y) = \sqrt{x^{2} + y^{2}} + h(y)

⇒ ∂_{y}f = \frac{y}{\sqrt{x^{2} + y^{2}}} + h′(y) = \frac{y}{\sqrt{x^{2} + y^{2}}}

⇒ h(y) = C

⇒ f(x, y, z) = \sqrt{x^{2} + y^{2}} + C

Thus F is conservative on D.

(c) Let r(t) = (cos t)i + (sin t)j, 0 ≤ t ≤ 2π. Then

∫_{0}^{2π} G(r(t)) · r′(t) ds = ∫_{0}^{2π}(− sin t, cos t) · (− sin t, cos t) dt = 2π ≠ 0.

Thus G is not conservative on D.

(d) We can determine whether H is conservative on D just by checking whether

(∗) ∮_{{(x,y,z_{0}): x^{2}+y^{2}=d^{2}}} H · T ds = 0

for any z_{0} ∈ ℝ and d ≥ 0. To see this, we need to verify that ∮_{K} H · T ds = 0 for any loop K which surrounds the z-axis from our assumption (∗). For other loop K which doesn’t surround the z-axis, we can apply Stokes’ Theorem on the piecewise smooth oriented surface enclosed by K to conclude that

∮_{K} H · T ds = ∫∫_{S} ∇ × H · n = 0

where the last equality comes from the component test conditions.

Now let K be a loop which surrounds the z-axis. Let C = {(x, y, z_{0}) : x^{2} + y^{2} = d^{2}} be chosen well so that C and K form the boundary of a “two-sided” piecewise smooth oriented surface S. Then we can apply Stokes’ Theorem on S:

∮_{K} H · T ds + ∮_{C} H · T ds = ∫∫_{S} ∇ × H · n = 0

where the last equality comes from the component test conditions. By our assumption (∗), ∮_{C} H · T ds = 0. Thus,

∮_{K} H · T ds = 0.

6. Homework assignment: Problem 3. (1) D = {(x, y, z) ∈ ℝ^{3} : (x, y, z) ≠ (0, 0, 0)} = ℝ^{3} \ {(0, 0, 0)}.

(2) Check directly.

(3) Yes.

(4) Yes. Since D is simply connected (by (3)) and F satisfies the component test conditions on D (by (2)), F is conservative on D. In fact, one after integration, it is easy to find the F = ∇f where f(x, y, z) = \sqrt{x^{2} + y^{2} + z^{2}}.