UPSC 2026 Maths Optional Paper 1 Q7b — Step-by-Step Solution

15 marks · Section B

Gauss divergence theorem · Vector Analysis · asked 10× in 14 yrs · Read the full method →

Question

If $\vec{F} = 4xz\hat{i} - y^2\hat{j} + yz\hat{k}$ , then evaluate $\displaystyle\iint_S \vec{F}\cdot\hat{n}\,dS$ , where $S$ is the surface of a unit cube with two opposite corners at $(0, 0, 0)$ and $(1, 1, 1)$ respectively. Hence verify the divergence theorem.

Technique

Evaluate the flux face by face over the six faces of the cube (on each face the outward unit normal is a constant $\pm\hat i,\pm\hat j,\pm\hat k$ , so $\vec F\cdot\hat n$ is a single component evaluated at $0$ or $1$ ), then compute $\iiint_V \nabla\cdot\vec F\,dV$ independently and compare.

Solution

The cube is $V = \{0\le x\le 1,\ 0\le y\le 1,\ 0\le z\le 1\}$ with boundary $S$ made of six unit squares. On each face the outward normal is constant, so

\iint_S \vec F\cdot\hat n\,dS = \sum_{\text{6 faces}} \iint (\vec F\cdot\hat n)\,dA .

Surface integral — the six faces

Face	Outward $\hat n$	$\vec F\cdot\hat n$ on the face	Contribution
$x=1$	$+\hat i$	$4xz\big	_{x=1} = 4z$
$x=0$	$-\hat i$	$-4xz\big	_{x=0} = 0$
$y=1$	$+\hat j$	$-y^2\big	_{y=1} = -1$
$y=0$	$-\hat j$	$+y^2\big	_{y=0} = 0$
$z=1$	$+\hat k$	$yz\big	_{z=1} = y$
$z=0$	$-\hat k$	$-yz\big	_{z=0} = 0$

Working the three non-zero faces explicitly:

Face $x=1$ ( $\hat n = \hat i$ , $dS = dy\,dz$ ):

\int_0^1\!\!\int_0^1 4z\,dy\,dz = \int_0^1 4z\,dz = \Bigl[2z^2\Bigr]_0^1 = 2 .

Face $y=1$ ( $\hat n = \hat j$ , $dS = dz\,dx$ ):

\int_0^1\!\!\int_0^1 (-1)\,dz\,dx = -1 .

Face $z=1$ ( $\hat n = \hat k$ , $dS = dx\,dy$ ):

\int_0^1\!\!\int_0^1 y\,dx\,dy = \int_0^1 y\,dy = \frac12 .

The three faces through the origin contribute nothing, because on $x=0$ the component $4xz$ vanishes, on $y=0$ the component $-y^2$ vanishes, and on $z=0$ the component $yz$ vanishes.

\therefore\quad \iint_S \vec F\cdot\hat n\,dS = 2 + 0 - 1 + 0 + \frac12 + 0 = \frac{3}{2}.

Volume integral — divergence theorem check

\nabla\cdot\vec F = \frac{\partial}{\partial x}(4xz) + \frac{\partial}{\partial y}(-y^2) + \frac{\partial}{\partial z}(yz) = 4z - 2y + y = 4z - y .

\iiint_V (4z - y)\,dV = \int_0^1\!\!\int_0^1\!\!\int_0^1 (4z-y)\,dx\,dy\,dz = \int_0^1\!\!\int_0^1 (4z-y)\,dy\,dz

= \int_0^1\left[4zy - \frac{y^2}{2}\right]_{y=0}^{y=1} dz = \int_0^1\left(4z - \frac12\right)dz = \Bigl[2z^2 - \tfrac{z}{2}\Bigr]_0^1 = 2 - \frac12 = \frac{3}{2}.

Conclusion

\iint_S \vec F\cdot\hat n\,dS = \frac{3}{2} = \iiint_V \nabla\cdot\vec F\,dV ,

so the divergence theorem is verified for this $\vec F$ and this cube. ( $\vec F$ is a polynomial, hence $C^1$ on the closed cube, and $S$ is a closed piecewise-smooth surface with outward normal — the hypotheses of the theorem hold.)

Answer

\boxed{\;\iint_S \vec F\cdot\hat n\,dS = \frac{3}{2} = \iiint_V \nabla\cdot\vec F\,dV\ \Rightarrow\ \text{divergence theorem verified.}\;}