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UPSC 2025 Maths Optional Paper 1 Q8b — Step-by-Step Solution 15 marks · Section B
Gauss divergence theorem · Vector Analysis · asked 9× in 13 yrs · Read the full method →
Question
Verify Gauss’s divergence theorem for F ⃗ = [ ( x 2 − y z ) i ^ + ( y 2 − z x ) j ^ + ( z 2 − x y ) k ^ ] \vec{F} = [(x^2 - yz)\hat{i} + (y^2 - zx)\hat{j} + (z^2 - xy)\hat{k}] F = [( x 2 − y z ) i ^ + ( y 2 − z x ) j ^ + ( z 2 − x y ) k ^ ] 0 ≤ x ≤ a 0 \leq x \leq a 0 ≤ x ≤ a 0 ≤ y ≤ b 0 \leq y \leq b 0 ≤ y ≤ b 0 ≤ z ≤ c 0 \leq z \leq c 0 ≤ z ≤ c
Technique
Apply Gauss’s divergence theorem ∭ V ( ∇ ⋅ F ⃗ ) d V = ∯ S F ⃗ ⋅ n ^ d S \displaystyle\iiint_V (\nabla\cdot\vec F)\,dV = \oiint_S \vec F\cdot\hat n\,dS ∭ V ( ∇ ⋅ F ) d V = ∬ S F ⋅ n ^ d S
Solution
Volume integral of the divergence
∇ ⋅ F ⃗ = ∂ ∂ x ( x 2 − y z ) + ∂ ∂ y ( y 2 − z x ) + ∂ ∂ z ( z 2 − x y ) = 2 x + 2 y + 2 z . \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. ∇ ⋅ F = ∂ x ∂ ( x 2 − y z ) + ∂ y ∂ ( y 2 − z x ) + ∂ z ∂ ( z 2 − x y ) = 2 x + 2 y + 2 z . ∭ V ( 2 x + 2 y + 2 z ) d V = ∫ 0 c ∫ 0 b ∫ 0 a 2 ( x + y + z ) d x d y d z . \iiint_V (2x+2y+2z)\,dV = \int_0^c\!\!\int_0^b\!\!\int_0^a 2(x+y+z)\,dx\,dy\,dz. ∭ V ( 2 x + 2 y + 2 z ) d V = ∫ 0 c ∫ 0 b ∫ 0 a 2 ( x + y + z ) d x d y d z . By symmetry, ∭ V 2 x d V = 2 ⋅ a 2 2 ⋅ b ⋅ c = a 2 b c \displaystyle\iiint_V 2x\,dV = 2\cdot\frac{a^2}{2}\cdot b\cdot c = a^2 bc ∭ V 2 x d V = 2 ⋅ 2 a 2 ⋅ b ⋅ c = a 2 b c y , z y,z y , z
∭ V ( ∇ ⋅ F ⃗ ) d V = a 2 b c + a b 2 c + a b c 2 = a b c ( a + b + c ) . \iiint_V (\nabla\cdot\vec F)\,dV = a^2 bc + ab^2 c + abc^2 = abc\,(a + b + c). ∭ V ( ∇ ⋅ F ) d V = a 2 b c + a b 2 c + ab c 2 = ab c ( a + b + c ) . Surface flux through the six faces
Outward normals and surface integrals (only the relevant component of F ⃗ \vec F F
Face x = a x=a x = a (n ^ = + i ^ \hat n=+\hat i n ^ = + i ^ F x = a 2 − y z F_x = a^2 - yz F x = a 2 − y z
∫ 0 c ∫ 0 b ( a 2 − y z ) d y d z = a 2 b c − b 2 2 c 2 2 = a 2 b c − b 2 c 2 4 . \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}. ∫ 0 c ∫ 0 b ( a 2 − y z ) d y d z = a 2 b c − 2 b 2 2 c 2 = a 2 b c − 4 b 2 c 2 .
Face x = 0 x=0 x = 0 (n ^ = − i ^ \hat n=-\hat i n ^ = − i ^ F x = − y z F_x = -yz F x = − y z = − ∫ ( − y z ) = b 2 c 2 4 = -\!\int(-yz) = \dfrac{b^2c^2}{4} = − ∫ ( − y z ) = 4 b 2 c 2
Sum of the x x x a 2 b c − b 2 c 2 4 + b 2 c 2 4 = a 2 b c . a^2 bc - \dfrac{b^2c^2}{4} + \dfrac{b^2c^2}{4} = a^2 bc. a 2 b c − 4 b 2 c 2 + 4 b 2 c 2 = a 2 b c .
Faces y = b , y = 0 y=b,\,y=0 y = b , y = 0 (F y = y 2 − z x F_y = y^2 - zx F y = y 2 − z x = b 2 a c . = b^2 ac. = b 2 a c .
Faces z = c , z = 0 z=c,\,z=0 z = c , z = 0 (F z = z 2 − x y F_z = z^2 - xy F z = z 2 − x y = c 2 a b . = c^2 ab. = c 2 ab .
Total flux:
∯ S F ⃗ ⋅ n ^ d S = a 2 b c + a b 2 c + a b c 2 = a b c ( a + b + c ) . \oiint_S \vec F\cdot\hat n\,dS = a^2 bc + ab^2 c + abc^2 = abc\,(a+b+c). ∬ S F ⋅ n ^ d S = a 2 b c + a b 2 c + ab c 2 = ab c ( a + b + c ) . Conclusion
∭ V ( ∇ ⋅ F ⃗ ) d V = a b c ( a + b + c ) = ∯ S F ⃗ ⋅ n ^ d S . \iiint_V (\nabla\cdot\vec F)\,dV = abc\,(a+b+c) = \oiint_S \vec F\cdot\hat n\,dS. ∭ V ( ∇ ⋅ F ) d V = ab c ( a + b + c ) = ∬ S F ⋅ n ^ d S . Gauss’s divergence theorem is verified.
Answer
Both the volume integral and the surface flux equal a b c ( a + b + c ) . \boxed{\;\text{Both the volume integral and the surface flux equal } abc\,(a + b + c).\;} Both the volume integral and the surface flux equal ab c ( a + b + c ) .