Practice Exam 5

Vector Fields & Line Integrals

Questions10

Time90 min

Points100

SketchRequired for Q1, 6

1Reading vector fields8 pts

Sketch each field and compute its 2D curl \(Q_x - P_y\): (a) \(\mathbf{F} = \langle 0, x\rangle\); (b) \(\mathbf{F} = \langle x, -y\rangle\); (c) \(\mathbf{F} = \dfrac{\langle -y, x\rangle}{\sqrt{x^2+y^2}}\).

Solution

(a) Vertical arrows, up for \(x > 0\), down for \(x < 0\), longer farther from the \(y\)-axis — a shear. \(Q_x - P_y = 1 - 0 = 1\). A paddle wheel spins counterclockwise everywhere.
(b) Arrows flow in along the \(y\)-axis and out along the \(x\)-axis — hyperbolic (saddle) flow. \(Q_x - P_y = 0 - 0 = 0\). It is \(\nabla\!\left(\tfrac12(x^2 - y^2)\right)\): conservative.
(c) Counterclockwise circles, every arrow of length 1. With \(r = \sqrt{x^2+y^2}\): \(Q_x = \dfrac{r - x^2/r}{r^2} = \dfrac{y^2}{r^3}\), \(P_y = -\dfrac{r - y^2/r}{r^2} = -\dfrac{x^2}{r^3}\), so \(Q_x - P_y = \dfrac{x^2+y^2}{r^3} = \dfrac1r\). Curl decreases with distance even though the speed does not.

2Scalar line integral10 pts

A wire follows the helix \(\mathbf{r}(t) = \langle\cos t, \sin t, t\rangle\), \(0\le t\le2\pi\), with linear density \(\rho(x,y,z) = z\). Find its mass.

Solution

\[ |\mathbf{r}'(t)| = \sqrt{\sin^2t + \cos^2t + 1} = \sqrt2,\qquad m = \int_C z\,ds = \int_0^{2\pi} t\cdot\sqrt2\,dt = \sqrt2\left[\frac{t^2}{2}\right]_0^{2\pi} = 2\sqrt2\,\pi^2\approx 27.9 \]

A scalar line integral needs \(ds = |\mathbf{r}'|\,dt\) and does not care about direction. The wire's total length is \(2\sqrt2\pi\approx8.9\), and its average density is \(\pi\) (the average of \(z = t\) over \([0, 2\pi]\)), so \(m = 8.9\times3.14\approx27.9\). ✓

3Vector line integral, direct10 pts

Evaluate \(\int_C y^2\,dx + x\,dy\) along the parabola \(y = x^2\) from \((0,0)\) to \((2,4)\). Is the field conservative?

Solution

Parametrise \(x = t\), \(y = t^2\), \(0\le t\le2\); \(dx = dt\), \(dy = 2t\,dt\).

\[ \int_0^2\left[(t^2)^2\cdot1 + t\cdot2t\right]dt = \int_0^2 (t^4 + 2t^2)\,dt = \frac{32}{5} + \frac{16}{3} = \frac{176}{15}\approx 11.73 \]

Conservative? \(P = y^2\), \(Q = x\): \(P_y = 2y\), \(Q_x = 1\). Not equal, so no. Along the straight segment \(\langle t, 2t\rangle\) instead: \(\int_0^2 (4t^2 + 2t)\,dt = \tfrac{32}{3} + 4\approx 14.67\) — a different answer, as path dependence demands.

4Voltage as a line integral · EE10 pts

An electric field is \(\mathbf{E} = \langle y, x, 0\rangle\) V/m. Find the potential difference \(V_B - V_A = -\int_A^B\mathbf{E}\cdot d\mathbf{r}\) from \(A = (0,0,0)\) to \(B = (2,3,0)\), and explain why the path did not need to be specified.

Solution

\(P_y = 1 = Q_x\) and the field is defined everywhere, so \(\mathbf{E}\) is conservative: \(\mathbf{E} = \nabla f\) with \(f = xy\) (check: \(f_x = y\), \(f_y = x\) ✓). By the fundamental theorem for line integrals,

\[ \int_A^B\mathbf{E}\cdot d\mathbf{r} = f(B) - f(A) = (2)(3) - 0 = 6\qquad\Longrightarrow\qquad V_B - V_A = -6\ \text{V} \]

The potential function is \(V = -xy\) (so that \(\mathbf{E} = -\nabla V\)). The path was irrelevant precisely because a potential exists — that is what makes “the voltage between two points” a meaningful phrase. Moving in the direction of \(\mathbf{E}\) lowers the potential, as it should.

5Finding a potential in 3D10 pts

Show that \(\mathbf{F} = \langle 2xy + z^3,\ x^2 + 2y,\ 3xz^2 + 1\rangle\) is conservative, find a potential \(f\), and evaluate \(\int_C\mathbf{F}\cdot d\mathbf{r}\) along any path from \((0,0,0)\) to \((1,2,1)\).

Solution

Curl test: \(R_y - Q_z = 0 - 0 = 0\); \(P_z - R_x = 3z^2 - 3z^2 = 0\); \(Q_x - P_y = 2x - 2x = 0\). ✓ Conservative on all of \(\mathbb{R}^3\).

\[ f = \int(2xy + z^3)\,dx = x^2y + xz^3 + g(y,z) \] \[ f_y = x^2 + g_y \overset{!}{=} x^2 + 2y\ \Rightarrow\ g = y^2 + h(z);\qquad f_z = 3xz^2 + h'(z) \overset{!}{=} 3xz^2 + 1\ \Rightarrow\ h = z \] \[ f(x,y,z) = x^2y + xz^3 + y^2 + z \] \[ \int_C\mathbf{F}\cdot d\mathbf{r} = f(1,2,1) - f(0,0,0) = 2 + 1 + 4 + 1 = 8 \]

6Green's theorem10 pts

Evaluate \(\displaystyle\oint_C \left(3y - e^{\sin x}\right)dx + \left(7x + \sqrt{y^4 + 1}\right)dy\), where \(C\) is the circle \(x^2 + y^2 = 9\) traversed counterclockwise.

Solution

Direct evaluation is impossible. Green's theorem with \(P = 3y - e^{\sin x}\), \(Q = 7x + \sqrt{y^4+1}\):

\[ Q_x - P_y = 7 - 3 = 4,\qquad \oint_C = \iint_D 4\,dA = 4\cdot\pi(3)^2 = 36\pi \]

The ugly terms depend on only one variable each, so they vanish under \(Q_x - P_y\) — they are conservative pieces that contribute nothing around a closed loop. The orientation is counterclockwise as required; clockwise would give \(-36\pi\).

7Area by a boundary integral10 pts

Use Green's theorem to find the area enclosed by the astroid \(x = \cos^3t\), \(y = \sin^3t\), \(0\le t\le2\pi\).

Solution

\[ A = \frac12\oint_C x\,dy - y\,dx = \frac12\int_0^{2\pi}\Big[\cos^3t\cdot3\sin^2t\cos t - \sin^3t\cdot(-3\cos^2t\sin t)\Big]dt \] \[ = \frac32\int_0^{2\pi}\sin^2t\cos^2t\,(\cos^2t + \sin^2t)\,dt = \frac32\int_0^{2\pi}\frac{\sin^22t}{4}\,dt = \frac38\cdot\pi = \frac{3\pi}{8} \]

using \(\sin t\cos t = \tfrac12\sin2t\) and \(\int_0^{2\pi}\sin^22t\,dt = \pi\). Sanity: the astroid fits inside the unit circle (area \(\pi\)) and is much pinched — \(3\pi/8\approx1.18\), about 37% of the circle. ✓ A Cartesian area integral for this curve would be brutal.

8Curl, divergence and a potential12 pts

For \(\mathbf{F} = \langle e^x\sin y,\ e^x\cos y,\ z^2\rangle\): compute \(\nabla\cdot\mathbf{F}\) and \(\nabla\times\mathbf{F}\). What does the curl tell you? Find a potential if one exists.

Solution

\[ \nabla\cdot\mathbf{F} = e^x\sin y - e^x\sin y + 2z = 2z \] \[ \nabla\times\mathbf{F} = \Big\langle \partial_y(z^2) - \partial_z(e^x\cos y),\ \partial_z(e^x\sin y) - \partial_x(z^2),\ \partial_x(e^x\cos y) - \partial_y(e^x\sin y)\Big\rangle = \langle 0,\ 0,\ e^x\cos y - e^x\cos y\rangle = \mathbf{0} \]

Zero curl on all of \(\mathbb{R}^3\) (simply connected) ⇒ conservative. Nonzero divergence is irrelevant to that question — a field can have sources and still have a potential (the Coulomb field does).

\[ f = \int e^x\sin y\,dx = e^x\sin y + g(y,z);\quad f_y = e^x\cos y + g_y = e^x\cos y\Rightarrow g = h(z);\quad f_z = h' = z^2\Rightarrow h = \tfrac{z^3}{3} \] \[ f(x,y,z) = e^x\sin y + \frac{z^3}{3} \]

9Ampère's law · EE10 pts

The magnetic field of a long wire on the \(z\)-axis carrying current \(I\) is \(\mathbf{B} = \dfrac{\mu_0 I}{2\pi}\dfrac{\langle -y, x\rangle}{x^2+y^2}\) (in the plane). (a) Show \(\nabla\cdot\mathbf{B} = 0\). (b) Find \(\oint\mathbf{B}\cdot d\mathbf{r}\) around the square \(|x|\le1\), \(|y|\le1\), counterclockwise, without parametrising the square. (c) What is the circulation around a circle that does not enclose the origin?

Solution

(a) With \(k = \mu_0I/2\pi\) and \(r^2 = x^2+y^2\):

\[ \partial_x\!\left(\frac{-ky}{r^2}\right) = \frac{2kxy}{r^4},\qquad \partial_y\!\left(\frac{kx}{r^2}\right) = -\frac{2kxy}{r^4},\qquad \nabla\cdot\mathbf{B} = 0 \]

(b) From the Unit 5 examples, \(Q_x - P_y = 0\) wherever \(\mathbf{B}\) is defined. Apply Green's theorem to the region between the square and a small circle of radius \(\varepsilon\) around the origin (both counterclockwise, inner boundary reversed): \(\oint_{\text{square}} - \oint_{\text{circle}} = \iint 0\,dA = 0\). So the square's circulation equals the circle's, which is \(\mu_0 I\) (computed in the lesson). \(\oint_{\text{square}}\mathbf{B}\cdot d\mathbf{r} = \mu_0 I\) — Ampère's law, independent of the loop's shape.

(c) A loop not enclosing the wire bounds a region where \(Q_x - P_y = 0\) throughout, so by Green's theorem the circulation is zero. No enclosed current, no circulation.

10Flux form of Green's theorem10 pts

Find the outward flux \(\oint_C\mathbf{F}\cdot\mathbf{n}\,ds\) of \(\mathbf{F} = \langle x^3, y^3\rangle\) across the circle \(x^2+y^2 = a^2\).

Solution

\[ \oint_C\mathbf{F}\cdot\mathbf{n}\,ds = \iint_D\nabla\cdot\mathbf{F}\,dA = \iint_D 3(x^2 + y^2)\,dA = 3\int_0^{2\pi}\!\!\int_0^a r^2\cdot r\,dr\,d\theta = 3\cdot2\pi\cdot\frac{a^4}{4} = \frac{3\pi a^4}{2} \]

Direct check: on the circle the outward unit normal is \(\langle x, y\rangle/a\), so \(\mathbf{F}\cdot\mathbf{n} = (x^4 + y^4)/a\). With \(x = a\cos t\), \(y = a\sin t\), \(ds = a\,dt\): \(\int_0^{2\pi}a^4(\cos^4t + \sin^4t)\,dt = a^4\cdot2\pi\cdot\tfrac34 = \tfrac{3\pi a^4}{2}\). ✓ (Each of \(\cos^4\) and \(\sin^4\) averages to \(3/8\).) This is the two-dimensional Gauss's law: the divergence plays the role of charge density.

Scoring

Score	Where you are
90–100	Solid. On to surface integrals.
75–89	Right theorems, wrong bookkeeping — usually \(dy = 2t\,dt\) becoming \(dy = dt\), or a lost orientation sign.
60–74	Parametrising when a potential exists, or hunting for a potential when the curl is nonzero. Run the conservative test first on every problem.
< 60	Re-read Sections 2–3 with the line-integral lab open and redo Q3–Q5.

Where the points usually go

Using \(dt\) for \(dy\) (Q3); the sign in \(V = -\int\mathbf{E}\cdot d\mathbf{r}\) (Q4); an \(x\) surviving in \(g'(y)\) and being ignored (Q5); orientation of the loop in Green's theorem (Q6); and thinking nonzero divergence rules out a potential (Q8).