Stokes’ theorem

At a Glance

• Frequency: 10 sub-parts across 9 of 13 years (2013, 2014, 2016, 2018, 2019, 2020, 2021, 2022, 2024)
• Priority tier: T1
• Marks (count): 5 (1), 10 (1), 13 (1), 15 (3), 20 (4)
• Average solve time: ~13 min
• Difficulty mix: medium 4, hard 3, easy 3
• Section: B | Dominant type: computation

Why This Chapter Matters

Stokes’ theorem appears in 9 of the last 12 years with a strong lean toward 15- and 20-mark questions. In every case the method is the same: replace a hard integral on one side of the equation with an easier one on the other side. Six of the ten questions ask for a pure evaluation (line integral → surface integral or vice versa); three ask for verification (compute both sides); one is an identity proof. Getting the orientation right via the right-hand rule is the single most common source of sign errors and lost marks.

Minimum Theory

Stokes’ theorem. Let $S$ be a piecewise-smooth oriented surface with boundary $C=\partial S$ traversed with the orientation induced by the right-hand rule: curl the fingers of the right hand in the direction of $C$ and the thumb points in the direction of $\hat n$. Then for any $C^1$ vector field $\vec F$, $\oint_C\vec F\cdot d\vec r=\iint_S(\nabla\times\vec F)\cdot\hat n\,dS.$

Computing $\nabla\times\vec F$. For $\vec F=(P,Q,R)$: $\nabla\times\vec F=\bigl(R_y-Q_z,\;P_z-R_x,\;Q_x-P_y\bigr)$, where subscripts denote partial derivatives.

Surface element for a graph. For $z=g(x,y)$ with upward orientation, the vector area element is $\hat n\,dS=(-g_x,\,-g_y,\,1)\,dx\,dy.$ The projection of $S$ onto the $xy$-plane gives the domain $D$ for the double integral.

Key facts. (1) Stokes’ theorem is independent of the choice of surface: any surface with boundary $C$ gives the same value. Choose the simplest cap. (2) If $\nabla\times\vec F=\vec0$ everywhere in a simply connected domain, the line integral over any closed curve is zero. (3) Exact differentials: $\vec F\cdot d\vec r=d(f)$ on the boundary implies the integral is zero.

Question Archetypes

Three patterns cover all Stokes’ theorem questions.

stokes-evaluation (6 question(s); 2013, 2014, 2018, 2019, 2020, 2021)

Recognition Cues

• “Evaluate the line integral … by Stokes’ theorem” — convert to a surface integral.
• “Evaluate the surface integral $\iint(\nabla\times\vec F)\cdot\hat n\,dS$” — use Stokes to convert to a line integral on the boundary.
• C is described as the intersection of two surfaces (cylinder + plane, sphere + plane, paraboloid + $xy$-plane).

Solution Template

Line integral → surface integral:

1. Identify $\vec F=(P,Q,R)$ from $P\,dx+Q\,dy+R\,dz$.
2. Compute $\nabla\times\vec F$.
3. Choose a capping surface $S$ (usually the flat portion of a plane enclosed by $C$). Project onto the $xy$-plane for the domain $D$.
4. Determine $\hat n$ and orientation (right-hand rule: CCW boundary → upward $\hat n$).
5. Compute $(\nabla\times\vec F)\cdot\hat n\,dS$ and integrate over $D$ (often in polar coordinates).

Surface integral → line integral (via Stokes):

1. Identify the boundary $C=\partial S$ (the curve where $S$ meets the bounding plane or cylinder).
2. Simplify $\vec F$ restricted to $C$ (often $z=0$ or $z=\text{const}$ kills terms).
3. Evaluate $\oint_C\vec F\cdot d\vec r$ directly (look for exact differentials or trig integrals).

Worked Example(s)

2013 Paper 1, 2013-P1-Q8d (15 marks)

Evaluate $\int_C(-y^3\,dx+x^3\,dy-z^3\,dz)$ where $C$ is the intersection of $x^2+y^2=1$ and $x+y+z=1$, via Stokes.

Curl. $\vec F=(-y^3,x^3,-z^3)$; $\nabla\times\vec F=(0,0,3x^2+3y^2)$.

Surface. Cap $S$ = disk in the plane $x+y+z=1$ projected to the unit disk $D$. Plane normal: $\hat n=(1,1,1)/\sqrt3$. Surface element: $dS=\sqrt3\,dx\,dy$.

Flux. $(\nabla\times\vec F)\cdot\hat n\,dS=(3(x^2+y^2)/\sqrt3)\cdot\sqrt3\,dx\,dy=3(x^2+y^2)\,dx\,dy$.

Polar. $\iint_D3(x^2+y^2)\,dx\,dy=3\int_0^{2\pi}\int_0^1r^3\,dr\,d\theta=3\cdot2\pi\cdot\tfrac14=\dfrac{3\pi}{2}$.

$\boxed{\;\int_C(-y^3\,dx+x^3\,dy-z^3\,dz)=\frac{3\pi}{2}.\;}$

2018 Paper 1, 2018-P1-Q8b (13 marks)

Evaluate $\int_C(-y^3\,dx+x^3\,dy+z^3\,dz)$ for $C$ = cylinder $x^2+y^2=1$ $\cap$ plane $x+y+z=1$ (CCW in $xy$-plane).

Same $\vec F$ as 2013 except $-z^3\to+z^3$. The curl is identical: $\nabla\times\vec F=(0,0,3(x^2+y^2))$. CCW in the $xy$-plane fixes the upward normal, giving vector area element $(1,1,1)\,dx\,dy$. Since only the $\hat k$-component of the curl is non-zero, the answer is the same:

$\boxed{\;\frac{3\pi}{2}.\;}$

2014 Paper 1, 2014-P1-Q6c (20 marks)

Evaluate $\int_\Gamma(y\,dx+z\,dy+x\,dz)$ where $\Gamma$ is $x^2+y^2+z^2-2ax-2ay=0$ $\cap$ $x+y=2a$, starting from $(2a,0,0)$ going below the $z$-plane.

Identify. The sphere has centre $(a,a,0)$, radius $\sqrt2\,a$. The plane $x+y=2a$ passes through the centre: $\Gamma$ is a great circle with radius $\sqrt2\,a$, area $2\pi a^2$.

Curl. $\vec F=(y,z,x)$; $\nabla\times\vec F=(-1,-1,-1)$.

Normal to plane $x+y=2a$: $\hat n=(1,1,0)/\sqrt2$.

Orientation. “Going below the $z$-plane” at $(2a,0,0)$ corresponds to $\hat n=(1,1,0)/\sqrt2$ (right-hand rule verified by parametrisation).

$\oint_\Gamma\vec F\cdot d\vec r=(-1,-1,-1)\cdot\frac{(1,1,0)}{\sqrt2}\cdot2\pi a^2=-\sqrt2\cdot2\pi a^2=\boxed{\;-2\sqrt2\,\pi a^2.\;}$

2019 Paper 1, 2019-P1-Q8c-ii (5 marks)

Evaluate $\oint_C e^x\,dx+2y\,dy-dz$ where $C: x^2+y^2=4$, $z=2$.

Curl. $\vec F=(e^x,2y,-1)$; $\nabla\times\vec F=\vec0$. By Stokes, the integral over any surface bounded by $C$ is zero. Alternatively, $\vec F=\nabla(e^x+y^2-z)$ is conservative.

$\boxed{\;0.\;}$

2020 Paper 1, 2020-P1-Q8b (15 marks)

Evaluate $\iint_S(\nabla\times\vec F)\cdot\hat n\,dS$ for $\vec F=(y,x-2xz,-xy)$, $S$ = upper hemisphere $x^2+y^2+z^2=a^2$, $z\ge0$.

Apply Stokes. Boundary $\partial S$ = circle $x^2+y^2=a^2$, $z=0$. On this circle $z=0$, $dz=0$: $\vec F|_{z=0}=(y,x,-xy)$ and $\vec F\cdot d\vec r=y\,dx+x\,dy=d(xy)$. Since $xy$ is single-valued, $\oint d(xy)=0$.

$\boxed{\;0.\;}$

2021 Paper 1, 2021-P1-Q8c (15 marks)

Evaluate $\iint_S(\nabla\times\vec F)\cdot\hat n\,dS$ for $\vec F=(x^2+y-4,3xy,2xy+z^2)$, $S$ = paraboloid $z=4-(x^2+y^2)$ above $xy$-plane.

Boundary. $\partial S$ = circle $x^2+y^2=4$, $z=0$. Parametrise: $x=2\cos\phi$, $y=2\sin\phi$, $z=0$.

$\vec F\cdot d\vec r=(x^2+y-4)(-2\sin\phi)+(3xy)(2\cos\phi)\,d\phi$.

With $x^2+y^2=4$: $x^2+y-4=-4\sin^2\phi+2\sin\phi$; $3xy=6\sin2\phi$.

$\vec F\cdot d\vec r=(-16\sin^3\phi-4\sin^2\phi+24\sin\phi)\,d\phi$.

Integrate over $[0,2\pi]$: $\int\sin^{2n+1}=0$ (odd powers), $\int\sin^2=\pi$:

$\boxed{\;\iint_S(\nabla\times\vec F)\cdot\hat n\,dS=-4\pi.\;}$

Common Traps

• Orientation is the most common lost mark. State explicitly: upward $\hat n$ ↔ CCW boundary when viewed from above. Reversed orientation flips the sign.
• The $\sqrt3$ factor from $dS$ and the $1/\sqrt3$ from $\hat n$ multiply to $1$ for the plane $x+y+z=1$ — effectively giving the flat area element with no extra factor (2013, 2018).
• When applying Stokes to the hemisphere (2020, 2021): the boundary is the $z=0$ circle, not the hemisphere’s “top”. On $z=0$, all $z$-dependent terms in $\vec F$ may simplify or vanish.
• For the exact-differential test (2019, 2020): if $\vec F\cdot d\vec r=d(f)$ on the boundary curve, the integral is zero without further computation.
• On the great-circle problem (2014): the area is $\pi R^2$ where $R$ is the great-circle radius, and the curl dotted with the plane’s normal is the only factor needed.

stokes-verification (3 question(s); 2020, 2022, 2024)

Recognition Cues

• “Verify Stokes’ theorem for $\vec F$ on $S$” — you must compute both sides independently and confirm they match.
• The surface is typically a parabolic cylinder ($z=1-x^2$) or a triangular plane (first-octant portion of $x+y+z=1$).

Solution Template

1. State Stokes’ theorem (required for full marks on the “state and verify” variant).
2. Compute $\nabla\times\vec F$.
3. LHS — surface integral. Use $\hat n\,dS=(-g_x,-g_y,1)\,dx\,dy$ for an upward-oriented graph $z=g(x,y)$; integrate over the projected rectangle or triangle.
4. RHS — line integral. Identify the boundary $C=\partial S$ from the parameter-region boundary. Determine orientation (CCW as seen from above for upward $\hat n$). Parametrise each edge and integrate.
5. State that both sides agree (and their common value).

Worked Example(s)

2020 Paper 1, 2020-P1-Q7a (20 marks) and 2024 Paper 1, 2024-P1-Q7c (20 marks)

Verify Stokes’ theorem for $\vec F=xy\hat i+yz\hat j+xz\hat k$ on $z=1-x^2$, $0\le x\le1$, $-2\le y\le2$ (upward).

Sources:,

Curl. $\nabla\times\vec F=(-y,-z,-x)$.

Surface integral (LHS). $\hat n\,dS=(2x,0,1)\,dx\,dy$. $(\nabla\times\vec F)\cdot\hat n\,dS=(-2xy-x)\,dx\,dy$. $\int_0^1\!\!\int_{-2}^2(-2xy-x)\,dy\,dx=\int_0^1-4x\,dx=-2.$

Boundary (4 edges, CCW from above):

Total RHS $=-19/15+0-11/15+0=-2=$ LHS ✓.

$\boxed{\;\oint_C\vec F\cdot d\vec r=\iint_S(\nabla\times\vec F)\cdot\hat n\,dS=-2.\;}\quad\text{Stokes' theorem verified.}$

2022 Paper 1, 2022-P1-Q7a (20 marks)

Verify Stokes’ theorem for $\vec F=(x,z^2,y^2)$ on $x+y+z=1$ in the first octant.

Curl. $\nabla\times\vec F=(2y-2z,0,0)$.

Surface integral. $\hat n\,dS=\hat n\sqrt3\,dx\,dy=(1,1,1)\,dx\,dy$ (the $\sqrt3$‘s cancel). $(\nabla\times\vec F)\cdot(1,1,1)\,dx\,dy=(2y-2z)\,dx\,dy$. With $z=1-x-y$: integrand $=(2y-2(1-x-y))=(4y+2x-2)\,dx\,dy$. Inner integral collapses to $(1-x)\cdot0=0$. LHS = 0.

Line integral (3 edges of the triangle):

• Edge 1 $(1,0,0)\to(0,1,0)$: integral $=-1/2$.
• Edge 2 $(0,1,0)\to(0,0,1)$: integral $=0$.
• Edge 3 $(0,0,1)\to(1,0,0)$: integral $=+1/2$.
• RHS $=-1/2+0+1/2=0=$ LHS ✓.

$\boxed{\;\oint_C\vec F\cdot d\vec r=\iint_S(\nabla\times\vec F)\cdot\hat n\,dS=0.\;}\quad\text{Stokes' theorem verified.}$

Common Traps

• Boundary orientation on the parabolic cylinder (2020, 2024): the four edges of the rectangle $[0,1]\times[-2,2]$ map to four boundary segments on the surface; the upward normal means traverse them CCW from above. Edges $C_2$ and $C_4$ have $z=0$ and $z=1$ respectively, so $dz=0$ on those edges — most terms vanish.
• Four pieces, not two: the parabolic cylinder’s boundary has four segments (not two open arcs); missing $C_2$ or $C_4$ is a common error.
• The inner integral collapse in the 2022 triangle problem is not a mistake — $(1-x)\cdot0=0$ for all $x$, so both sides are identically zero.
• Always state the theorem when the question says “state and verify.”

stokes-identity-proof (1 question(s); 2016)

Recognition Cues

• “Prove that $\oint_C f\,d\vec r=\iint_S d\vec S\times\nabla f$.”
• An identity involving $f$ (scalar) and a line or surface integral — not an evaluation.

Solution Template

1. Apply classical Stokes to $\vec G=f\vec c$ where $\vec c$ is an arbitrary constant vector.
2. LHS: $\oint_C(f\vec c)\cdot d\vec r=\vec c\cdot\oint_C f\,d\vec r$.
3. RHS: $\nabla\times(f\vec c)=\nabla f\times\vec c$ (since $\nabla\times\vec c=\vec0$). Use the triple product $(\nabla f\times\vec c)\cdot d\vec S=\vec c\cdot(d\vec S\times\nabla f)$.
4. Since LHS = RHS holds for all constant $\vec c$, equate the vectors.

Worked Example(s)

2016 Paper 1, 2016-P1-Q8b (10 marks)

Prove $\oint_C f\,d\vec r=\iint_S d\vec S\times\nabla f$.

Apply Stokes to $\vec G=f\vec c$ ($\vec c$ constant): $\vec c\cdot\oint_C f\,d\vec r=\oint_C(f\vec c)\cdot d\vec r=\iint_S(\nabla f\times\vec c)\cdot d\vec S=\vec c\cdot\iint_S d\vec S\times\nabla f.$

Since this holds for all $\vec c$: $\boxed{\;\oint_C f\,d\vec r=\iint_S d\vec S\times\nabla f.\;}\qquad\blacksquare$

Common Traps

• The arbitrary constant vector $\vec c$ is the key trick: equality of $\vec c\cdot(\ldots)$ for all $\vec c$ implies equality of the vectors.
• Sign in the triple product: $(\nabla f\times\vec c)\cdot d\vec S=\vec c\cdot(d\vec S\times\nabla f)$ (not $\nabla f\times d\vec S$) — the order on the right matches the stated identity.
• $\nabla\times\vec c=\vec0$ kills the second product-rule term; state this explicitly.

Marks-Aware Writing

5-mark questions (2019): State Stokes (one line), compute curl, note it’s zero (or $d(\text{exact})$), conclude. No surface integral needed.

13–15-mark questions (2013, 2018, 2020-Q8b, 2021): Full Stokes setup: state the theorem, compute curl (show each component), write the vector area element, compute the dot product, integrate in polar or Cartesian. Box the answer.

20-mark questions (2014, 2020-Q7a, 2022, 2024): For evaluations: also state the orientation reasoning explicitly. For verifications: show both sides in labelled sub-sections (LHS / RHS), with each boundary edge computed separately in a table, and conclude with “LHS = RHS = …”. The “state Stokes’ theorem” preamble is required when the question asks for it.

Practice Set