Gauss divergence theorem

At a Glance

• Frequency: 9 sub-parts across 9 of 13 years (2013, 2017, 2018, 2019, 2021, 2022, 2023, 2024, 2025)
• Priority tier: T1
• Marks (count): 12 (1), 15 (5), 20 (2), 9 (1)
• Average solve time: ~11 min
• Difficulty mix: medium 6, easy 2, hard 1
• Section: B | Dominant type: computation

Why This Chapter Matters

The Gauss divergence theorem appears in 9 of the last 13 years, almost always as a 15-mark Section B question. It transforms a hard surface integral into an easy volume integral — and UPSC designs every question so the volume side is straightforward (constant or separable divergence, standard region). The flux (evaluation) variant is one of the fastest high-mark problems in the paper: compute a divergence, multiply by a volume, and you’re done. The verification variant asks you to do both sides, but the method is completely mechanical once you’ve seen the pattern.

Minimum Theory

Gauss’s divergence theorem. Let $V$ be a closed bounded region in $\mathbb{R}^3$ with piecewise-smooth boundary surface $S$ and outward unit normal $\hat n$. If $\vec F$ is continuously differentiable on $V$, then $\iiint_V(\nabla\cdot\vec F)\,dV=\iint_S\vec F\cdot\hat n\,dS.$

Divergence in coordinates. For $\vec F=(F_1,F_2,F_3)$: $\nabla\cdot\vec F=\frac{\partial F_1}{\partial x}+\frac{\partial F_2}{\partial y}+\frac{\partial F_3}{\partial z}.$ Compute each term independently; terms with no dependence on the differentiation variable vanish.

Closed vs open surfaces. The divergence theorem requires a closed surface (no boundary curve). If $S$ is open (e.g., only the hemisphere, without the disk), close it by adding the missing piece, apply the theorem to the closed surface, then subtract the integral over the added piece.

Standard volume elements. Cylinder: $dV=r\,dr\,d\theta\,dz$; half-disc cross-section $r\in[0,R]$. Sphere/half-ball: $dV=r^2\sin\phi\,dr\,d\phi\,d\theta$.

Question Archetypes

Two patterns cover all divergence theorem questions.

divergence-flux (4 question(s); 2013, 2017, 2018, 2024)

Recognition Cues

• “Using Gauss’s divergence theorem, evaluate the surface integral $\iint_S\vec F\cdot\hat n\,dS$.”
• The surface is a complete closed surface (sphere, full cylinder, ellipsoid) or can be completed trivially.
• The divergence is simple (constant, or separable) — this is the hint that you should use the theorem.

Solution Template

1. Compute $\nabla\cdot\vec F$ — check each partial derivative separately.
2. Identify the closed region $V$ bounded by $S$.
3. Evaluate $\iiint_V\nabla\cdot\vec F\,dV$. Separate the integral if the divergence is constant or factors. Use cylindrical/spherical coordinates as appropriate.
4. State the result.

Worked Example(s)

2024 Paper 1, 2024-P1-Q8b (15 marks)

Evaluate $\iint_S(y^2\hat i+xz^3\hat j+(z-1)^2\hat k)\cdot\hat n\,dS$ over the cylinder $x^2+y^2=16$, $z\in[1,5]$.

Divergence. $\partial_x(y^2)+\partial_y(xz^3)+\partial_z((z-1)^2)=0+0+2(z-1)=2(z-1)$.

Volume integral. In cylindrical: $r\in[0,4]$, $\theta\in[0,2\pi]$, $z\in[1,5]$: $\iiint_V 2(z-1)\,dV=\underbrace{2\pi}_{\int d\theta}\cdot\underbrace{8}_{\int_0^4 r\,dr}\cdot\underbrace{16}_{\int_1^5 2(z-1)\,dz}=256\pi.$

$\boxed{\;\iint_S\vec F\cdot\hat n\,dS=256\pi.\;}$

2018 Paper 1, 2018-P1-Q6d (12 marks)

Evaluate $\iint_S[(x+z)\,dy\,dz+(y+z)\,dz\,dx+(x+y)\,dx\,dy]$ over the sphere $x^2+y^2+z^2=a^2$.

Identify. $\vec F=(x+z,y+z,x+y)$.

Divergence. $1+1+0=2$. (The third component $x+y$ has no $z$-dependence, so its $\partial_z=0$.)

Volume integral. Constant divergence over the ball of radius $a$: $2\cdot\frac{4}{3}\pi a^3=\boxed{\;\frac{8}{3}\pi a^3.\;}$

2017 Paper 1, 2017-P1-Q8c-i (9 marks)

Evaluate the flux of $\vec F=3xy^2\hat i+(yx^2-y^3)\hat j+3zx^2\hat k$ over $y^2+z^2\le4$, $-3\le x\le3$.

Divergence. $3y^2+(x^2-3y^2)+3x^2=4x^2$ (the $\pm3y^2$ cancel).

Volume. Cylinder axis along $x$; disc $y^2+z^2\le4$ has area $4\pi$: $\int_{-3}^{3}4x^2\cdot4\pi\,dx=16\pi\cdot18=\boxed{\;288\pi.\;}$

2013 Paper 1, 2013-P1-Q8c (15 marks)

Evaluate $\iint_S(a^2x^2+b^2y^2+c^2z^2)^{-1/2}\,dS$ on the ellipsoid $ax^2+by^2+cz^2=1$.

Key insight. The outward unit normal on the ellipsoid $F=ax^2+by^2+cz^2-1=0$ is $\hat n=(ax,by,cz)/\sqrt{a^2x^2+b^2y^2+c^2z^2}$. Therefore $\vec r\cdot\hat n=(ax^2+by^2+cz^2)/\sqrt{a^2x^2+b^2y^2+c^2z^2}=1/\sqrt{a^2x^2+b^2y^2+c^2z^2}$ (using the surface equation). So the integrand equals $\vec r\cdot\hat n$ for $\vec F=(x,y,z)$.

Divergence theorem. $\nabla\cdot\vec r=3$; volume of ellipsoid $ax^2+by^2+cz^2\le1$ is $\tfrac{4\pi}{3\sqrt{abc}}$: $I=3\cdot\frac{4\pi}{3\sqrt{abc}}=\boxed{\;\frac{4\pi}{\sqrt{abc}}.\;}$

Common Traps

• The divergence of $\vec F=(x+z,y+z,x+y)$ is $\mathbf{2}$, not 3. The third component has no $z$-dependence: $\partial_z(x+y)=0$.
• The 2017 cylinder’s axis is the $x$-axis (not the $z$-axis); the disc cross-section is in the $yz$-plane with radius $2$, area $4\pi$.
• In the 2013 ellipsoid problem, the normal involves $(ax,by,cz)$, not $(x,y,z)$; the surface equation is used to simplify the numerator.
• $\int_{-3}^3x^2\,dx=18$ (over a symmetric interval of width 6); a common error is writing $9$.
• For the 2024 cylinder: $\int_1^5 2(z-1)\,dz=[(z-1)^2]_1^5=16$, not $8$.

divergence-verification (5 question(s); 2019, 2021, 2022, 2023, 2025)

Recognition Cues

• “State and verify Gauss’s divergence theorem for $\vec F$ over the region $\ldots$”
• You must compute both the volume integral and the surface integral and confirm they are equal.
• The region is a standard solid: cylinder, box, first-octant wedge, half-ball.

Solution Template

1. State the theorem (required when the question says “state and verify”).
2. Volume side. Compute $\nabla\cdot\vec F$; integrate over $V$ in the most convenient coordinates; use symmetry to kill odd-function integrals.
3. Surface side. Identify all faces of $S$; for each face write the outward $\hat n$ and compute $\vec F\cdot\hat n$; integrate. Set up a table if many faces.
4. State equality and conclude.

For open surfaces (hemisphere): close with the flat disk, apply the theorem to the closed surface, subtract the disk’s contribution.

Worked Example(s)

2019 Paper 1, 2019-P1-Q8c-i (15 marks)

State and verify Gauss’s theorem for $\vec F=4x\hat i-2y^2\hat j+z^2\hat k$ over $x^2+y^2\le4$, $0\le z\le3$.

Volume. $\nabla\cdot\vec F=4-4y+2z$. In cylindrical: the $-4y=-4r\sin\theta$ term integrates to zero over $\theta\in[0,2\pi]$: $\iiint_V(4+2z)\,dV=\int_0^2r\,dr\cdot\int_0^{2\pi}d\theta\cdot\int_0^3(4+2z)\,dz=2\cdot2\pi\cdot21=84\pi.$

Surface (3 faces):

• Top ($z=3$, $\hat n=\hat k$): $\vec F\cdot\hat k=z^2=9$; area $4\pi$: $\Rightarrow 36\pi$.
• Bottom ($z=0$, $\hat n=-\hat k$): $-z^2=0$: $\Rightarrow 0$.
• Lateral ($x^2+y^2=4$, $\hat n=(\cos\theta,\sin\theta,0)$, $dS=2\,d\theta\,dz$): $\vec F\cdot\hat n=8\cos^2\theta-8\sin^3\theta$. Integrate: $\int_0^{2\pi}8\cos^2\theta\,d\theta=8\pi$, $\int\sin^3=0$; then $\int_0^3 dz=3$: $\Rightarrow 2\cdot8\pi\cdot3=48\pi$.

$\text{Total flux}=36\pi+0+48\pi=84\pi=\iiint_V\nabla\cdot\vec F\,dV.\quad\text{✓}$

$\boxed{\;\text{Both sides}=84\pi.\;}$

2022 Paper 1, 2022-P1-Q8c (15 marks)

Verify divergence theorem for $\vec F=x\hat i-y\hat j+(z^2-1)\hat k$ over $x^2+y^2\le4$, $z\in[0,1]$.

Volume. $\nabla\cdot\vec F=1+(-1)+2z=2z$: $\iiint_V 2z\,dV=2\!\int_0^1z\,dz\cdot4\pi=4\pi.$

Surface (3 faces):

• Bottom ($z=0$, $\hat n=-\hat k$): $-(z^2-1)|_{z=0}=1$; area $4\pi$: $\Rightarrow4\pi$.
• Top ($z=1$, $\hat n=\hat k$): $(z^2-1)|_{z=1}=0$: $\Rightarrow0$.
• Lateral: $\vec F\cdot\hat n=(x^2-y^2)/2=2\cos2\theta$; $\int_0^{2\pi}\cos2\theta\,d\theta=0$: $\Rightarrow0$.

$\text{Total flux}=4\pi+0+0=4\pi=\iiint_V\nabla\cdot\vec F\,dV.\quad\text{✓}$

$\boxed{\;\text{Both sides}=4\pi.\;}$

2021 Paper 1, 2021-P1-Q7a (20 marks)

Verify divergence theorem for $\vec F=2x^2y\hat i-y^2\hat j+4xz^2\hat k$ over first-octant region $y^2+z^2\le9$, $x\in[0,2]$.

Volume. $\nabla\cdot\vec F=4xy-2y+8xz$. In cylindrical ($y=r\cos\phi$, $z=r\sin\phi$, $r\le3$, $\phi\in[0,\pi/2]$): $\iiint_V\nabla\cdot\vec F\,dV=9\int_0^2[(4x-2)+8x]\,dx=9\int_0^2(12x-2)\,dx=9(24-4)=180.$

Surface (5 faces):

• $x=2$ ($\hat n=\hat i$): $8y$; $\iint 8y\,dA=8\cdot1\cdot9=72$.
• $x=0,z=0,y=0$ faces: each contributes 0 (normal component of $\vec F$ vanishes).
• Curved cylinder ($y^2+z^2=9$, $\hat n=(0,y,z)/3$): after parametrisation with $\int_0^{\pi/2}\cos^3=\int_0^{\pi/2}\sin^3=2/3$: $\Rightarrow108$.

$\text{Total}=72+0+0+0+108=180=\iiint_V\nabla\cdot\vec F\,dV.\quad\text{✓}$

$\boxed{\;\text{Both sides}=180.\;}$

2023 Paper 1, 2023-P1-Q6c (20 marks)

Evaluate $\iint_S(3y^2z^2,4z^2x^2,z^2y^2)\cdot\hat n\,dS$ on the upper hemisphere $x^2+y^2+z^2=1/4$, $z\ge0$; hence verify the divergence theorem.

Close the surface. $S$ is the hemisphere alone (open). Add disk $D$ at $z=0$: $\hat n=-\hat k$ on $D$. Apply the theorem to $S\cup D$:

Volume. $\nabla\cdot\vec F=2zy^2$. In spherical with $r\le1/2$: $\iiint_V 2zy^2\,dV=2\cdot\underbrace{\frac{1}{384}}_{r^5\,dr}\cdot\underbrace{\frac14}_{\cos\psi\sin^3\psi}\cdot\underbrace{\pi}_{\sin^2\phi}=\frac{\pi}{768}.$

Disk $D$ ($z=0$, $\hat n=-\hat k$): $\vec F\cdot\hat n=-z^2y^2|_{z=0}=0$.

$\iint_S=\frac{\pi}{768}-0=\boxed{\;\frac{\pi}{768}.\;}$

Divergence theorem verified: both sides $=\pi/768$.

2025 Paper 1, 2025-P1-Q8b (15 marks)

Verify Gauss’s theorem for $\vec F=(x^2-yz,y^2-zx,z^2-xy)$ over the box $0\le x\le a$, $0\le y\le b$, $0\le z\le c$.

Volume. $\nabla\cdot\vec F=2x+2y+2z$: $\iiint_V(2x+2y+2z)\,dV=a^2bc+ab^2c+abc^2=abc(a+b+c).$

Surface (6 faces). The $-yz$, $-zx$, $-xy$ cross-terms cancel in face-pairs (e.g., face $x=a$: $\int(a^2-yz)\,dy\,dz=a^2bc-b^2c^2/4$; face $x=0$: $\int(0-yz)\cdot(-1)\,dy\,dz=+b^2c^2/4$; sum $=a^2bc$). By symmetry across the three pairs: $\text{Total flux}=a^2bc+ab^2c+abc^2=abc(a+b+c).\quad\text{✓}$

$\boxed{\;\text{Both sides}=abc(a+b+c).\;}$

Common Traps

• Always state the theorem when the question says “state and verify” — losing this preamble costs marks.
• For the cylinder (2019), the bottom face $z=0$ contributes zero because $\vec F\cdot(-\hat k)=-z^2|_{z=0}=0$; it must still be accounted for.
• Symmetry kills odd-function integrals: $\int_0^{2\pi}\sin\theta\,d\theta=0$ eliminates any $y$-dependent term in the volume integral; $\int_0^{2\pi}\sin^3\theta\,d\theta=0$ eliminates $\sin^3$ on a lateral surface.
• Open surface (2023 hemisphere): add the disk, apply the theorem, subtract the disk’s contribution. The disk gives zero here because the third component vanishes at $z=0$.
• For the box (2025): the cross-terms $(-yz,-zx,-xy)$ do not contribute to the divergence; they cancel in surface pairs. Don’t forget this cancellation when summing the six face integrals.

Marks-Aware Writing

9-12-mark questions (2017, 2018): Compute divergence (one line), state the theorem, evaluate the volume integral (two lines), box the answer. No surface integral needed for a pure-flux question.

15-mark questions (2013, 2019, 2022, 2024, 2025): For flux: divergence + separable volume integral, clearly written. For verification: state the theorem, labelled volume integral (show symmetry argument for odd terms), all faces listed in a short table (face $|$ $\hat n$ $|$ $\vec F\cdot\hat n$ $|$ integral), state equality.

20-mark questions (2021, 2023): Full verification with 4–5 boundary faces, coordinate transformation written out, every Wallis-type integral cited. For 2023: close-the-surface argument written explicitly, spherical coordinates with all limits stated, disk contribution checked.

Practice Set