Line integrals

At a Glance

• Frequency: 8 sub-parts across 8 of 14 years (2015, 2017, 2018, 2019, 2020, 2021, 2022, 2025)
• Priority tier: T1
• Marks (count): 12 (1), 13 (1), 15 (5), 8 (1)
• Average solve time: ~10 min
• Difficulty mix: easy 4, medium 4
• Section: B | Dominant type: computation

Why This Chapter Matters

Line integrals appear in 8 of the last 14 years, mostly as 15-mark questions. Green’s theorem — the 2D version of Stokes — converts a closed loop integral into a double integral over the enclosed region and back. Five of the eight questions use Green’s theorem directly (three evaluation, two verification). The remaining three cover the vortex field, path-dependence, and a direct circulation. Every question has a well-defined method; the only judgement call is whether to apply Green’s or parametrise directly.

Minimum Theory

Line integral. For a vector field $\vec F=(P,Q)$ and curve $C$: $\int_C\vec F\cdot d\vec r=\int_C P\,dx+Q\,dy.$ Parametrise $C$: substitute $x(t),y(t)$, $dx=x'(t)dt$, $dy=y'(t)dt$, integrate over $t$.

Green’s theorem. For a closed, counterclockwise-oriented curve $C$ bounding a region $R$: $\oint_C P\,dx+Q\,dy=\iint_R\!\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA.$ The double integral is the flux of the “curl” ($\hat k$ component of $\nabla\times\vec F$) through the region.

Conservative fields. $\vec F$ is conservative iff $\partial Q/\partial x=\partial P/\partial y$ everywhere on a simply connected domain. Then $\oint_C\vec F\cdot d\vec r=0$ for every closed $C$, and the integral between two points is path-independent.

The vortex field. $\vec F=(-y,x)/(x^2+y^2)$ satisfies $\partial Q/\partial x=\partial P/\partial y$ everywhere except the origin. For a closed $C$: integral $=0$ if $C$ doesn’t enclose origin; $=2\pi n$ if $C$ winds $n$ times around origin. This is a topological (winding number) result.

Question Archetypes

Five patterns cover all line integral questions.

greens-theorem (3 question(s); 2015, 2017, 2018)

Recognition Cues

• “Using Green’s theorem, evaluate $\oint_C\vec F\cdot d\vec r$” or “evaluate $\iint_R(\nabla\times\vec F)\cdot\hat k\,dA$.”
• The closed curve bounds a simple region (rectangle, parabola-line, two parabolas).

Solution Template

1. Identify $P$ and $Q$; compute $Q_x-P_y$.
2. Describe the region $R$ (find intersection points; state which curve is upper/lower).
3. Set up and evaluate $\iint_R(Q_x-P_y)\,dA$.
4. Check orientation: CCW gives the positive sign in Green’s theorem.

Worked Example(s)

2015 Paper 1, 2015-P1-Q8c (12 marks)

Evaluate $\oint_C e^{-x}(\sin y\,dx+\cos y\,dy)$ around rectangle with vertices $(0,0),(\pi,0),(\pi,\pi/2),(0,\pi/2)$.

$P=e^{-x}\sin y$, $Q=e^{-x}\cos y$. $Q_x-P_y=-e^{-x}\cos y-e^{-x}\cos y=-2e^{-x}\cos y$.

$I=-2\!\underbrace{\int_0^\pi e^{-x}dx}_{1-e^{-\pi}}\cdot\underbrace{\int_0^{\pi/2}\cos y\,dy}_{1}=-2(1-e^{-\pi})=\boxed{\;2(e^{-\pi}-1).\;}$

2017 Paper 1, 2017-P1-Q8c-ii (8 marks)

Green’s theorem for $\vec F=(x^2+y^2,x^2-y^2)$ around $R:\{1\le y\le2-x^2\}$, CCW.

$Q_x-P_y=2x-2y$. Region: $x\in[-1,1]$ (from $2-x^2=1$), $y$ from $1$ to $2-x^2$.

The $2x$ term is odd in $x$ and integrates to zero over $[-1,1]$: $\iint_R(2x-2y)\,dA=2\int_0^1(-3+4x^2-x^4)\,dx=2\!\left(-3+\tfrac43-\tfrac15\right)=\boxed{\;-\frac{56}{15}.\;}$

2018 Paper 1, 2018-P1-Q8c (13 marks)

Integrate $(\nabla\times\vec F)\cdot\hat k$ for $\vec F=(xy^2,y+x)$ over first-quadrant region $y=x^2$ to $y=x$, using Green’s.

$(\nabla\times\vec F)\cdot\hat k=Q_x-P_y=1-2xy$. Region: $x\in[0,1]$, $y$ from $x^2$ to $x$.

$\iint_R(1-2xy)\,dA=\int_0^1\int_{x^2}^x(1-2xy)\,dy\,dx=\int_0^1(x-x^2-x^3+x^5)\,dx=\tfrac12-\tfrac13-\tfrac14+\tfrac16=\boxed{\;\frac{1}{12}.\;}$

Common Traps

• Orientation. CCW is the standard (positive) sign. Verify the vertex listing before concluding CCW.
• Sign of $Q_x-P_y$: both $\partial_x Q$ and $\partial_y P$ may have the same sign. Here $Q_x=-e^{-x}\cos y$ and $P_y=e^{-x}\cos y$ — they don’t cancel; they add.
• For the 2018 curl problem: Green’s theorem says $\iint(Q_x-P_y)=\oint_{\partial R}\vec F\cdot d\vec r$; the double integral IS what’s asked, so no boundary integral needed.
• In the parabola-line region (2017): odd-function terms in $x$ vanish over $[-1,1]$ by symmetry — state this explicitly.

greens-verification (2 question(s); 2022, 2025)

Recognition Cues

• “Verify Green’s theorem” — must compute both the line integral and the double integral independently and confirm they agree.
• Region is a triangle or the area between two parabolas/curves.

Solution Template

1. Double integral side. Compute $Q_x-P_y$; integrate over $R$.
2. Line integral side. Orient the boundary CCW. Parametrise each segment/arc; evaluate each contribution; sum.
3. State equality.

Worked Example(s)

2022 Paper 1, 2022-P1-Q6c (15 marks)

Verify Green’s for $(3x^2-8y^2)dx+(4y-6xy)dy$ over triangle $x=0$, $y=0$, $x+y=1$.

Double integral. $Q_x-P_y=-6y+16y=10y$: $\iint_R 10y\,dA=5\int_0^1(1-x)^2\,dx=5/3$.

Line integral (CCW: $(0,0)\to(1,0)\to(0,1)\to(0,0)$):

• Side 1 ($y=0$): $\int_0^1 3x^2\,dx=1$.
• Side 2 ($x=1-t,y=t$): $\int_0^1(-3+4t+11t^2)\,dt=8/3$.
• Side 3 ($x=0$): $\int_1^0 4y\,dy=-2$.

Sum $=1+8/3-2=5/3$. Both sides equal $5/3$. ✓

$\boxed{\;\oint_C=\iint_R=\frac53.\;}$

2025 Paper 1, 2025-P1-Q7b (15 marks)

Verify Green’s for $(xy+y^2)dx+x^2\,dy$ over region bounded by $y=x$ and $y=x^2$.

Double integral. $Q_x-P_y=2x-(x+2y)=x-2y$: $\iint_R(x-2y)\,dA=\int_0^1(x^4-x^3)\,dx=-1/20$.

Line integral (CCW: parabola $C_1$ then line $C_2$):

• $C_1$ ($y=x^2$): $\int_0^1(3x^3+x^4)\,dx=3/4+1/5=19/20$.
• $C_2$ ($y=x$, reversed): $\int_1^0 3x^2\,dx=-1$.

Sum $=19/20-1=-1/20$. Both sides $=-1/20$. ✓

Common Traps

• CCW orientation on the triangle: $(0,0)\to(1,0)\to(0,1)$ — going right then diagonally back is CCW ✓.
• Parametrising Side 2 of the triangle: $x=1-t$, $dx=-dt$; the minus sign propagates into the integrand.
• For the 2025 two-curve region: the parabola $C_1$ is the lower boundary (traversed left-to-right), the line $C_2$ is the upper boundary (traversed right-to-left for CCW orientation).
• Always state “Green’s theorem verified” and quote the common value.

circulation (1 question; 2019)

Recognition Cues

• “Find the circulation of $\vec F$ round the curve $C$” — a closed loop, direct integration required (not Green’s, or Green’s as a check).
• Two arcs meeting at two points; best to parametrise each arc separately.

Worked Example(s)

2019 Paper 1, 2019-P1-Q6b (15 marks)

$\vec F=(2x+y^2,3y-4x)$; circulation round $C$: $y=x^2$ from $(0,0)$ to $(1,1)$, then $x=y^2$ back to $(0,0)$.

Arc $C_1$ ($y=x^2$, $x:0\to1$, $dy=2x\,dx$): $\int_{C_1}=\int_0^1\big[(2x+x^4)+2x(3x^2-4x)\big]dx=\int_0^1(2x+x^4+6x^3-8x^2)\,dx=\frac{1}{30}.$

Arc $C_2$ ($x=y^2$, $y:1\to0$, $dx=2y\,dy$): $\int_{C_2}=\int_1^0\big[2y(2y^2+y^2)+(3y-4y^2)\big]dy=\int_1^0(6y^3+3y-4y^2)\,dy=-\frac53.$

$\oint=\frac{1}{30}-\frac53=\boxed{\;-\frac{49}{30}.\;}$

Cross-check via Green’s: $Q_x-P_y=-4-2y$; $\iint_R(-4-2y)\,dA=-49/30$ ✓.

Common Traps

• Arc $C_2$: the natural parameter is $y$ (since $x=y^2$), running from $1$ to $0$ (reversed). Keep the reversed limits; don’t re-orient.
• The closed loop here is clockwise (out along $y=x^2$, back along $x=y^2$). For the Green’s cross-check, the area integral also gives $-49/30$ (matching; no extra sign needed since the parametric calculation already incorporates orientation).
• $Q_x-P_y=-4-2y$ (not $-4-2x$): $P=2x+y^2$, $\partial P/\partial y=2y$.

contour-singularity (1 question; 2021)

Recognition Cues

• $\vec F=(-y,x)/(x^2+y^2)$ or $\vec F=(-y,x)/r^2$ — the “vortex” field.
• Closed curve, possibly enclosing the origin.
• The answer depends only on whether the origin is enclosed.

Worked Example(s)

2021 Paper 1, 2021-P1-Q6c (15 marks)

Evaluate $\oint_C(-y\hat\imath+x\hat\jmath)/(x^2+y^2)\cdot d\vec r$ for arbitrary closed $C$.

$P=-y/(x^2+y^2)$, $Q=x/(x^2+y^2)$. Check: $Q_x=P_y=(y^2-x^2)/(x^2+y^2)^2$ everywhere except origin.

If $C$ does not enclose origin: Green’s theorem applies; $Q_x-P_y=0$, so $\oint=0$.

If $C$ encloses origin: Deform $C$ to a small circle $C_\varepsilon$ (both give the same integral since $Q_x=P_y$ between them). On $C_\varepsilon$: $x=\varepsilon\cos\theta$, $y=\varepsilon\sin\theta$; $P\,dx+Q\,dy=d\theta$. $\oint_{C_\varepsilon}=\int_0^{2\pi}d\theta=2\pi.$

$\boxed{\;\oint_C=\begin{cases}0&C\text{ does not enclose origin,}\\2\pi&C\text{ encloses origin (CCW).}\end{cases}\;}$

Common Traps

• The field is locally curl-free (closed 1-form) but not globally conservative — the potential $\theta=\arctan(y/x)$ is multi-valued. This is the key distinction from ordinary conservative fields.
• The deformation argument: since $Q_x-P_y=0$ between $C$ and $C_\varepsilon$, the integrals over them are equal (by Green’s applied to the annular region).
• On the small circle the integrand simplifies to $d\theta$ — the $\varepsilon^2$ factors cancel exactly.

path-dependence (1 question; 2020)

Recognition Cues

• “Compute $\int_C\vec F\cdot d\vec r$ along [multiple paths]; is the result the same? Explain.”
• If the results differ: state $\nabla\times\vec F\ne\vec0$, hence not conservative.

Worked Example(s)

2020 Paper 1, 2020-P1-Q6b (15 marks)

$\vec A=(3x^2+6y,-14yz,20xz^2)$. Compute $\int\vec A\cdot d\vec r$ from $(0,0,0)$ to $(1,1,1)$ along three paths.

Path (i) $x=t,y=t^2,z=t^3$: $\int_0^1(9t^2-28t^6+60t^9)\,dt=3-4+6=5$.

Path (ii) Three straight edges: segment 1 gives $1$; segment 2 gives $0$; segment 3 gives $20/3$. Total: $1+0+20/3=23/3$.

Path (iii) Straight line $x=y=z=t$: $\int_0^1(3t^2+6t-14t^2+20t^3)\,dt=5-11/3+3=13/3$.

Results: $5$, $23/3$, $13/3$ — all different.

Why: $\nabla\times\vec A=14y\hat\imath-20z^2\hat\jmath-6\hat k\ne\vec0$. Since the field is not conservative, the integral is path-dependent.

$\boxed{\;\text{Path-dependent (curl}\ne\vec0\text{).}}$

Common Traps

• On polygonal path (ii): only the moving coordinate has a nonzero differential on each segment; zero out the other two.
• State the explanation in terms of $\nabla\times\vec A\ne\vec0$ — it’s required for full marks.
• Arithmetic: keep fractions exact ($20/3$, $23/3$, $13/3$).

Marks-Aware Writing

8-mark questions (2017): Green’s theorem in three lines — compute $Q_x-P_y$, set up the area integral, evaluate. No need for a preamble beyond “by Green’s theorem.”

12-13-mark questions (2015, 2018): State Green’s theorem (one line); compute $Q_x-P_y$; describe the region; evaluate the double integral. Show all steps including the separability or inner integral.

15-mark questions (2019, 2020, 2021, 2022, 2025): For Green’s evaluation: full working as above. For verification (2022, 2025): state the theorem, compute both sides in clearly labelled subsections, conclude. For circulation (2019): parametrize each arc, show the integrand explicitly, sum. For path-dependence: compute all three paths, then compute the curl and state it’s nonzero.

Practice Set