Chapter 35 — Exercises

34 problems across scalar line integrals, vector line integrals (work), the Fundamental Theorem for Line Integrals, path independence, Green's theorem (both forms), and applications. Tiered ⭐ (routine) to ⭐⭐⭐⭐ (challenge). Work everything by hand; the boxed formulas of §35.2, §35.3, §35.5, and §35.7 are all you need.

Tier Count Problems
⭐ Routine 8 A1–A4, B1–B2, C1, D1
⭐⭐ Standard 12 A5–A6, B3–B5, C2–C3, D2–D3, E1–E2, F1
⭐⭐⭐ Substantial 10 B6, C4, D4, E3–E4, F2–F3, G1–G2, G3
⭐⭐⭐⭐ Challenge 4 E5, F4, G4, G5

Part A — Scalar Line Integrals

A1 ⭐ Evaluate $\int_C (x+y)\,ds$ where $C$ is the segment from $(0,0)$ to $(1,2)$.

Solution$\mathbf{r}(t)=\langle t,2t\rangle$, $0\le t\le 1$; $\mathbf{r}'=\langle 1,2\rangle$, $|\mathbf{r}'|=\sqrt5$. Integrand $f=t+2t=3t$. So $\int_0^1 3t\sqrt5\,dt = \sqrt5\cdot\tfrac32 = \tfrac{3\sqrt5}{2}$.

A2 ⭐ Evaluate $\int_C x\,ds$ where $C$ is the upper semicircle $x^2+y^2=4$, $y\ge 0$.

Solution$\mathbf{r}(t)=\langle 2\cos t,2\sin t\rangle$, $0\le t\le\pi$; $|\mathbf{r}'|=2$. So $\int_0^\pi 2\cos t\cdot 2\,dt = 4\sin t\big|_0^\pi = 0$. (By symmetry the positive and negative $x$ halves cancel.)

A3 ⭐ Find the length of one turn of the helix $\mathbf{r}(t)=\langle\cos t,\sin t,t\rangle$, $0\le t\le 2\pi$, by computing $\int_C 1\,ds$.

Solution$\mathbf{r}'=\langle -\sin t,\cos t,1\rangle$, $|\mathbf{r}'|=\sqrt{\sin^2 t+\cos^2 t+1}=\sqrt2$. So $\int_0^{2\pi}\sqrt2\,dt = 2\sqrt2\,\pi$.

A4 ⭐ Evaluate $\int_C y\,ds$ where $C$ is the segment from $(0,0)$ to $(3,4)$.

Solution$\mathbf{r}(t)=\langle 3t,4t\rangle$, $|\mathbf{r}'|=\sqrt{9+16}=5$, $y=4t$. So $\int_0^1 4t\cdot5\,dt = 20\cdot\tfrac12 = 10$.

A5 ⭐⭐ A wire follows $y=x^2$ from $x=0$ to $x=1$ with linear density $\rho(x,y)=x$. Find its mass.

Solution$\mathbf{r}(t)=\langle t,t^2\rangle$, $|\mathbf{r}'|=\sqrt{1+4t^2}$, $\rho=t$. $M=\int_0^1 t\sqrt{1+4t^2}\,dt$. Let $u=1+4t^2$, $du=8t\,dt$: $M=\tfrac18\int_1^5 u^{1/2}\,du = \tfrac18\cdot\tfrac23(5^{3/2}-1)=\tfrac{5\sqrt5-1}{12}\approx 0.848$.

A6 ⭐⭐ Evaluate $\int_C xy\,ds$ over the quarter circle $\mathbf{r}(t)=\langle\cos t,\sin t\rangle$, $0\le t\le\pi/2$.

Solution$|\mathbf{r}'|=1$, $xy=\cos t\sin t=\tfrac12\sin 2t$. $\int_0^{\pi/2}\tfrac12\sin 2t\,dt=\tfrac12\big[-\tfrac12\cos 2t\big]_0^{\pi/2}=\tfrac14(1+1)=\tfrac12$.

Part B — Vector Line Integrals (Work)

B1 ⭐ Compute $\int_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F}=\langle x,y\rangle$ along the segment from $(0,0)$ to $(1,1)$.

Solution$\mathbf{r}(t)=\langle t,t\rangle$, $\mathbf{r}'=\langle 1,1\rangle$, $\mathbf{F}=\langle t,t\rangle$. $\mathbf{F}\cdot\mathbf{r}'=2t$. $\int_0^1 2t\,dt=1$.

B2 ⭐ Compute $\int_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F}=\langle 0,x\rangle$ along the segment from $(0,0)$ to $(1,1)$.

Solution$\mathbf{r}(t)=\langle t,t\rangle$, $\mathbf{F}=\langle 0,t\rangle$, $\mathbf{F}\cdot\mathbf{r}'=0\cdot1+t\cdot1=t$. $\int_0^1 t\,dt=\tfrac12$.

B3 ⭐⭐ Compute the circulation $\oint_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F}=\langle y,-x\rangle$ around the unit circle (counterclockwise).

Solution$\mathbf{r}(t)=\langle\cos t,\sin t\rangle$; $\mathbf{F}=\langle\sin t,-\cos t\rangle$, $\mathbf{r}'=\langle-\sin t,\cos t\rangle$. Dot $=-\sin^2 t-\cos^2 t=-1$. $\int_0^{2\pi}(-1)\,dt=-2\pi$.

B4 ⭐⭐ Work done by $\mathbf{F}=\langle x^2,xy\rangle$ along $y=x^2$ from $(0,0)$ to $(1,1)$.

Solution$\mathbf{r}(t)=\langle t,t^2\rangle$, $\mathbf{r}'=\langle 1,2t\rangle$, $\mathbf{F}(\mathbf{r})=\langle t^2,t^3\rangle$. Dot $=t^2+2t^4$. $\int_0^1(t^2+2t^4)\,dt=\tfrac13+\tfrac25=\tfrac{11}{15}$.

B5 ⭐⭐ Compute $\int_C \mathbf{F}\cdot d\mathbf{r}$ for $\mathbf{F}=\langle x,y\rangle$ along the quarter circle from $(1,0)$ to $(0,1)$. Explain the result geometrically.

Solution$\mathbf{r}(t)=\langle\cos t,\sin t\rangle$, $0\le t\le\pi/2$; $\mathbf{F}(\mathbf{r})=\langle\cos t,\sin t\rangle$, $\mathbf{r}'=\langle-\sin t,\cos t\rangle$. Dot $=-\cos t\sin t+\sin t\cos t=0$. Integral $=0$: the radial force is perpendicular to the circular motion, so it does no work.

B6 ⭐⭐⭐ Compute $\int_C P\,dx+Q\,dy$ for $P=y$, $Q=x^2$ along the segment from $(0,0)$ to $(1,1)$, then along $y=x^2$ from $(0,0)$ to $(1,1)$. Are they equal? Is $\mathbf{F}=\langle y,x^2\rangle$ conservative?

SolutionSegment $\mathbf{r}=\langle t,t\rangle$: $\mathbf{F}=\langle t,t^2\rangle$, $\mathbf{r}'=\langle 1,1\rangle$, dot $=t+t^2$, $\int_0^1=\tfrac12+\tfrac13=\tfrac56$. Parabola $\mathbf{r}=\langle t,t^2\rangle$: $\mathbf{F}=\langle t^2,t^2\rangle$, $\mathbf{r}'=\langle 1,2t\rangle$, dot $=t^2+2t^3$, $\int_0^1=\tfrac13+\tfrac12=\tfrac56$. Equal here — but check the curl: $Q_x=2x$, $P_y=1$, unequal except at $x=\tfrac12$. **Not conservative**; these two particular paths coincidentally agree. Equality on *all* paths is what defines conservative (§35.6).

Part C — Fundamental Theorem for Line Integrals

C1 ⭐ Show $\mathbf{F}=\langle 2x,2y\rangle$ is conservative and find a potential $f$.

SolutionCurl test: $Q_x=\partial(2y)/\partial x=0=\partial(2x)/\partial y=P_y$. ✓ Take $f=x^2+y^2$; then $\nabla f=\langle 2x,2y\rangle=\mathbf{F}$.

C2 ⭐⭐ Using the potential from C1, evaluate $\int_C \mathbf{F}\cdot d\mathbf{r}$ from $(0,0)$ to $(3,4)$ along any path.

SolutionBy the Fundamental Theorem for Line Integrals (§35.5), $\int_C\mathbf{F}\cdot d\mathbf{r}=f(3,4)-f(0,0)=(9+16)-0=25$.

C3 ⭐⭐ For $\mathbf{F}=\langle 2xy,x^2+2y\rangle$, evaluate $\int_C \mathbf{F}\cdot d\mathbf{r}$ from $(1,1)$ to $(2,3)$.

SolutionCurl test: $Q_x=2x=P_y$. ✓ Build $f$: $f_x=2xy\Rightarrow f=x^2 y+g(y)$; $f_y=x^2+g'(y)=x^2+2y\Rightarrow g'(y)=2y\Rightarrow g=y^2$. So $f=x^2y+y^2$. Result $=f(2,3)-f(1,1)=(12+9)-(1+1)=19$.

C4 ⭐⭐⭐ Recall Worked Example 4 of §35.3: $\mathbf{F}=\langle y,x\rangle$ from $(0,0)$ to $(1,1)$ gave work $1$ by direct integration. Re-derive it in one step using the Fundamental Theorem, and explain why the parabola never appeared.

Solution$\mathbf{F}=\langle y,x\rangle=\nabla(xy)$ since $\nabla(xy)=\langle y,x\rangle$. So $f=xy$ and $\int_C\mathbf{F}\cdot d\mathbf{r}=f(1,1)-f(0,0)=1-0=1$. Because $\mathbf{F}$ is conservative, the integral depends only on the endpoints; *any* path from $(0,0)$ to $(1,1)$ gives $1$, so the specific parabola is irrelevant. This is exactly the original FTC (Chapter 14) wearing 2D clothing.

Part D — Path Independence and Closed Loops

D1 ⭐ Show $\oint_C \mathbf{F}\cdot d\mathbf{r}=0$ for every closed curve when $\mathbf{F}=\nabla(x^2+y^2)$.

Solution$\mathbf{F}$ is conservative with potential $f=x^2+y^2$. On a closed curve the start and end coincide, so $f(\text{end})-f(\text{start})=0$ (closed-loop corollary, §35.5).

D2 ⭐⭐ Decide whether $\mathbf{F}=\langle e^x\cos y,-e^x\sin y\rangle$ is conservative; if so, find $f$.

Solution$Q_x=\partial(-e^x\sin y)/\partial x=-e^x\sin y$; $P_y=\partial(e^x\cos y)/\partial y=-e^x\sin y$. Equal ✓, conservative. $f_x=e^x\cos y\Rightarrow f=e^x\cos y+g(y)$; $f_y=-e^x\sin y+g'(y)=-e^x\sin y\Rightarrow g'=0$. So $f=e^x\cos y$.

D3 ⭐⭐ For $\mathbf{F}=\langle -y,x\rangle$, compute $\int_C\mathbf{F}\cdot d\mathbf{r}$ along (a) the upper semicircle from $(1,0)$ to $(-1,0)$, and (b) the segment from $(1,0)$ to $(-1,0)$. Conclude.

Solution(a) $\mathbf{r}(t)=\langle\cos t,\sin t\rangle$, $0\le t\le\pi$: $\mathbf{F}=\langle-\sin t,\cos t\rangle$, $\mathbf{r}'=\langle-\sin t,\cos t\rangle$, dot $=\sin^2 t+\cos^2 t=1$, $\int_0^\pi=\pi$. (b) $\mathbf{r}(t)=\langle 1-2t,0\rangle$, $0\le t\le 1$: $\mathbf{F}=\langle 0,1-2t\rangle$, $\mathbf{r}'=\langle-2,0\rangle$, dot $=0$, integral $=0$. Different answers ($\pi\ne 0$) $\Rightarrow$ **not path-independent**, not conservative (curl $=2\ne 0$).

D4 ⭐⭐⭐ The field $\mathbf{F}=\left\langle \dfrac{-y}{x^2+y^2},\dfrac{x}{x^2+y^2}\right\rangle$ passes the curl test everywhere it is defined. Compute $\oint_C\mathbf{F}\cdot d\mathbf{r}$ around the unit circle and explain why it is not $0$.

SolutionOn the unit circle $x^2+y^2=1$, so $\mathbf{F}(\mathbf{r})=\langle-\sin t,\cos t\rangle$ and $\mathbf{r}'=\langle-\sin t,\cos t\rangle$; dot $=1$. $\oint=\int_0^{2\pi}1\,dt=2\pi\ne 0$. The curl test gives conservative *only on simply-connected domains*; here the domain is the punctured plane $\mathbb{R}^2\setminus\{0\}$, and the loop encircles the hole. Curl-free does not imply conservative when a loop cannot shrink to a point (§35.6 Math Major Sidebar).

Part E — Green's Theorem (Circulation Form)

E1 ⭐⭐ Use Green's theorem to evaluate $\oint_C y\,dx+x\,dy$ around the unit circle.

Solution$P=y$, $Q=x$; $Q_x-P_y=1-1=0$. $\iint_D 0\,dA=0$. (The field is conservative, $\nabla(xy)$, so any closed loop gives $0$.)

E2 ⭐⭐ Use Green's theorem to evaluate $\oint_C (x^2-y^2)\,dx+2xy\,dy$ around the unit square $[0,1]^2$ (counterclockwise).

Solution$P=x^2-y^2$, $Q=2xy$; $Q_x-P_y=2y-(-2y)=4y$. $\iint_{[0,1]^2}4y\,dA=\int_0^1\!\!\int_0^1 4y\,dy\,dx=4\cdot\tfrac12=2$.

E3 ⭐⭐⭐ Use Green's theorem on $\oint_C (y^2-x)\,dx+3xy\,dy$, where $C$ bounds the triangle $(0,0),(1,0),(1,1)$ counterclockwise.

Solution$P=y^2-x$, $Q=3xy$; $Q_x-P_y=3y-2y=y$. Triangle: $0\le x\le 1$, $0\le y\le x$. $\iint_D y\,dA=\int_0^1\!\!\int_0^x y\,dy\,dx=\int_0^1\tfrac{x^2}{2}\,dx=\tfrac16$.

E4 ⭐⭐⭐ Verify Green's theorem for $\mathbf{F}=\langle -y,x\rangle$ on the unit disk by computing both sides.

SolutionDouble integral: $Q_x-P_y=1-(-1)=2$, so $\iint_D 2\,dA=2\pi(1)^2=2\pi$. Line integral: $\mathbf{r}(t)=\langle\cos t,\sin t\rangle$, $\mathbf{F}\cdot\mathbf{r}'=\sin^2 t+\cos^2 t=1$, $\oint_0^{2\pi}1\,dt=2\pi$. ✓ Both equal $2\pi$.

E5 ⭐⭐⭐⭐ Evaluate $\oint_C (3y-e^{\sin x})\,dx+(7x+\sqrt{y^4+1})\,dy$ where $C$ is the circle $x^2+y^2=9$ counterclockwise.

SolutionThe integrand looks brutal directly, but Green's theorem only needs the partials: $Q_x=7$, $P_y=3$, so $Q_x-P_y=4$. $\oint_C=\iint_D 4\,dA=4\cdot(\text{area of disk of radius }3)=4\cdot 9\pi=36\pi$. The nasty $e^{\sin x}$ and $\sqrt{y^4+1}$ terms vanish under differentiation — the whole point of choosing Green's theorem.

Part F — Green's Theorem as an Area Formula

F1 ⭐⭐ Use $\text{Area}=\tfrac12\oint_C(x\,dy-y\,dx)$ to find the area of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Solution$x=a\cos t$, $y=b\sin t$; $dx=-a\sin t\,dt$, $dy=b\cos t\,dt$. $x\,dy-y\,dx=ab\cos^2 t+ab\sin^2 t=ab$. Area $=\tfrac12\int_0^{2\pi}ab\,dt=\pi ab$.

F2 ⭐⭐⭐ Use Green's theorem to evaluate $\oint_C x\,dy$ around the unit square $[0,1]^2$, and explain why $\oint_C x\,dy$ equals the enclosed area.

Solution$P=0$, $Q=x$; $Q_x-P_y=1$. $\oint_C x\,dy=\iint_D 1\,dA=\text{area}=1$. In general choosing $P,Q$ with $Q_x-P_y=1$ turns the boundary integral into the area; $P=0,Q=x$ is one such choice, the symmetric one being $P=-y/2$, $Q=x/2$.

F3 ⭐⭐⭐ Find the area enclosed by one arch of the cycloid $x=t-\sin t$, $y=1-\cos t$, $0\le t\le 2\pi$, and the $x$-axis, using a Green's-theorem area integral.

SolutionUse $\text{Area}=-\oint_C y\,dx$ taken counterclockwise; along the $x$-axis return $y=0$ contributes nothing, leaving the arch. Along the arch $dx=(1-\cos t)\,dt$, $y=1-\cos t$, traversed from $t=2\pi$ back to $0$ for CCW orientation, giving Area $=\int_0^{2\pi}(1-\cos t)^2\,dt=\int_0^{2\pi}(1-2\cos t+\cos^2 t)\,dt=2\pi-0+\pi=3\pi$.

F4 ⭐⭐⭐⭐ A simple polygon has vertices $(x_1,y_1),\dots,(x_n,y_n)$ counterclockwise (with $(x_{n+1},y_{n+1})=(x_1,y_1)$). Derive the shoelace formula $\text{Area}=\tfrac12\sum_{i=1}^n (x_i y_{i+1}-x_{i+1}y_i)$ from Green's area integral.

Solution$\text{Area}=\tfrac12\oint_C(x\,dy-y\,dx)$. The boundary is a union of edges; on edge $i$ parametrize linearly $\mathbf{r}(s)=(1-s)\langle x_i,y_i\rangle+s\langle x_{i+1},y_{i+1}\rangle$, $0\le s\le 1$. Then $x\,dy-y\,dx$ integrated over the edge yields $x_iy_{i+1}-x_{i+1}y_i$ (the cross term; the constant pieces cancel between $x\,dy$ and $-y\,dx$). Summing edges gives $\tfrac12\sum(x_iy_{i+1}-x_{i+1}y_i)$. This is the discrete embodiment of the planimeter (§35.7).

Part G — Applications and Theory

G1 ⭐⭐⭐ A mass of $1$ kg is lifted from $(0,0)$ to $(3,4)$ (meters, $y$ vertical) under gravity $\mathbf{F}_{\text{grav}}=\langle 0,-g\rangle$, $g=9.8$. Find the work done by gravity, along any path.

SolutionGravity is conservative with potential energy $U=gy$ (per unit mass), so $\mathbf{F}_{\text{grav}}=-\nabla U=\langle 0,-g\rangle$. Work by gravity $=U_i-U_f=g(0)-g(4)=-9.8\cdot4=-39.2$ J. Negative because gravity opposes the rise; path-independent by the Fundamental Theorem for Line Integrals.

G2 ⭐⭐⭐ A spring exerts $\mathbf{F}=-kx\,\hat{\mathbf{i}}$. Find the work the spring does as its end moves from $x=0$ to $x=a$, and identify the stored elastic potential energy.

Solution$W_{\text{spring}}=\int_0^a(-kx)\,dx=-\tfrac12 k a^2$. The spring does negative work as it is stretched, storing elastic PE $U=\tfrac12 k a^2$ (so $\mathbf{F}=-\nabla U=-kx\,\hat{\mathbf{i}}$). This is the variable-force work integral of Chapter 14, now read as a 1D line integral.

G3 ⭐⭐⭐ For a fluid with velocity $\mathbf{u}=\langle 1+y,x\rangle$, compute the circulation $\oint_C\mathbf{u}\cdot d\mathbf{r}$ around the square $[0,1]^2$ using Green's theorem, and interpret.

Solution$P=1+y$, $Q=x$; 2D curl $=Q_x-P_y=1-1=0$. Circulation $=\iint_D 0\,dA=0$. The flow is irrotational over the square — no net swirl — so a tiny paddle wheel placed anywhere inside would not spin on average.

G4 ⭐⭐⭐⭐ A charged particle moves through a magnetic field, feeling the Lorentz force $\mathbf{F}=q\mathbf{v}\times\mathbf{B}$. Show that the magnetic force does zero work along any trajectory.

Solution$W=\int_C\mathbf{F}\cdot d\mathbf{r}=\int\mathbf{F}\cdot\mathbf{v}\,dt=\int q(\mathbf{v}\times\mathbf{B})\cdot\mathbf{v}\,dt$. The cross product $\mathbf{v}\times\mathbf{B}$ is perpendicular to $\mathbf{v}$, so $(\mathbf{v}\times\mathbf{B})\cdot\mathbf{v}=0$ at every instant, giving $W=0$. Magnetic forces bend trajectories (Chapter 28) but never change kinetic energy.

G5 ⭐⭐⭐⭐ Prove that $\oint_C\mathbf{F}\cdot d\mathbf{r}=0$ for every closed path in a region if and only if $\mathbf{F}=\nabla f$ for some scalar $f$ on that region.

Solution($\Leftarrow$) If $\mathbf{F}=\nabla f$, the Fundamental Theorem for Line Integrals gives $\oint_C\mathbf{F}\cdot d\mathbf{r}=f(A)-f(A)=0$ for any closed loop. ($\Rightarrow$) Suppose all closed-loop integrals vanish; then line integrals are path-independent (any two paths from $A$ to $B$ form a closed loop, so their integrals are equal). Fix a basepoint $\mathbf{x}_0$ and define $f(\mathbf{x})=\int_{\mathbf{x}_0}^{\mathbf{x}}\mathbf{F}\cdot d\mathbf{r}$ along any path. Differentiating along the coordinate directions shows $\nabla f=\mathbf{F}$ (move from $\mathbf{x}$ a small step $h\hat{\mathbf{i}}$: $\partial f/\partial x=\lim_{h\to0}\tfrac1h\int_{\mathbf{x}}^{\mathbf{x}+h\hat{\mathbf{i}}}\mathbf{F}\cdot d\mathbf{r}=P$). Hence $\mathbf{F}$ is conservative. This is the equivalence of conditions 1 and 3 in §35.6.

Reflection

Line integrals give the integral a sense of direction. Scalar line integrals (Part A) weight a function by arc length to find masses and lengths; vector line integrals (Part B) extract the tangential component of a field to find work and circulation. The Fundamental Theorem for Line Integrals (Parts C–D) reveals that conservative fields forget their path and remember only their endpoints — the calculus face of energy conservation. Green's theorem (Parts E–F) trades a boundary loop for an interior double integral, and in its area form lets you measure a region by walking only its edge. Hold onto these patterns: in Chapters 36 and 37 they stretch across surfaces and fill into volumes, and in Chapter 38 they collapse into the single statement $\int_{\partial M}\omega=\int_M d\omega$.