UPSC Maths 2025 Paper 1 Q8b — Solution

15 marks · Section B

Question

Verify Gauss’s divergence theorem for $\vec{F} = [(x^2 - yz)\hat{i} + (y^2 - zx)\hat{j} + (z^2 - xy)\hat{k}]$ , taken over the rectangular parallelopiped $0 \leq x \leq a$ , $0 \leq y \leq b$ , $0 \leq z \leq c$ .

Technique

Apply Gauss’s divergence theorem $\displaystyle\iiint_V (\nabla\cdot\vec F)\,dV = \oiint_S \vec F\cdot\hat n\,dS$ : evaluate the volume integral of the divergence and the total flux through the six faces, and confirm they agree.

Solution

Volume integral of the divergence

$\nabla\cdot\vec F = \frac{\partial}{\partial x}(x^2-yz) + \frac{\partial}{\partial y}(y^2-zx) + \frac{\partial}{\partial z}(z^2-xy) = 2x + 2y + 2z.$ $\iiint_V (2x+2y+2z)\,dV = \int_0^c\!\!\int_0^b\!\!\int_0^a 2(x+y+z)\,dx\,dy\,dz.$ By symmetry, $\displaystyle\iiint_V 2x\,dV = 2\cdot\frac{a^2}{2}\cdot b\cdot c = a^2 bc$ and similarly for $y,z$ , so $\iiint_V (\nabla\cdot\vec F)\,dV = a^2 bc + ab^2 c + abc^2 = abc\,(a + b + c).$

Surface flux through the six faces

Outward normals and surface integrals (only the relevant component of $\vec F$ survives on each face):

Face $x=a$ ( $\hat n=+\hat i$ ), $F_x = a^2 - yz$ : $\int_0^c\!\!\int_0^b (a^2 - yz)\,dy\,dz = a^2 bc - \frac{b^2}{2}\frac{c^2}{2} = a^2 bc - \frac{b^2 c^2}{4}.$
Face $x=0$ ( $\hat n=-\hat i$ ), $F_x = -yz$ , flux $= -\!\int(-yz) = \dfrac{b^2c^2}{4}$ .

Sum of the $x$ -pair: $a^2 bc - \dfrac{b^2c^2}{4} + \dfrac{b^2c^2}{4} = a^2 bc.$
Faces $y=b,\,y=0$ ( $F_y = y^2 - zx$ ): by the same cancellation, sum $= b^2 ac.$
Faces $z=c,\,z=0$ ( $F_z = z^2 - xy$ ): sum $= c^2 ab.$

Total flux: $\oiint_S \vec F\cdot\hat n\,dS = a^2 bc + ab^2 c + abc^2 = abc\,(a+b+c).$

Conclusion

$\iiint_V (\nabla\cdot\vec F)\,dV = abc\,(a+b+c) = \oiint_S \vec F\cdot\hat n\,dS.$ Gauss’s divergence theorem is verified.

Answer

$\boxed{\;\text{Both the volume integral and the surface flux equal } abc\,(a + b + c).\;}$