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UPSC 2026 Maths Optional Paper 1 Q5e — Step-by-Step Solution 10 marks · Section B
Surface integrals; flux · Vector Analysis · Read the full method →
Question
Given F ⃗ = ( 2 x + 3 y ) i ^ − 4 z j ^ − 5 x k ^ \vec{F} = (2x + 3y)\hat{i} - 4z\hat{j} - 5x\hat{k} F = ( 2 x + 3 y ) i ^ − 4 z j ^ − 5 x k ^ S S S 2 x + 4 y + 4 z = 5 2x + 4y + 4z = 5 2 x + 4 y + 4 z = 5 x = 0 x = 0 x = 0 x = 2 x = 2 x = 2 y = 0 y = 0 y = 0 y = 3 y = 3 y = 3 ∬ S ( ∇ × F ⃗ ) ⋅ n ^ d S \displaystyle\iint_S (\nabla \times \vec{F}) \cdot \hat{n}\, dS ∬ S ( ∇ × F ) ⋅ n ^ d S
Technique
∇ × F ⃗ \nabla\times\vec F ∇ × F constant vector and n ^ \hat n n ^ constant unit normal (the surface is a plane), so the integrand is constant and the whole integral is (constant) × \times × x y xy x y d S = d x d y ∣ n ^ ⋅ k ^ ∣ dS = \dfrac{dx\,dy}{|\hat n\cdot\hat k|} d S = ∣ n ^ ⋅ k ^ ∣ d x d y Stokes’ theorem on the boundary — the agreement is the check.
Solution
Step 1 — Fix the surface and the orientation.
S S S
2 x + 4 y + 4 z = 5 , i.e. z = 5 − 2 x − 4 y 4 , 2x+4y+4z=5,\qquad\text{i.e.}\qquad z = \frac{5-2x-4y}{4}, 2 x + 4 y + 4 z = 5 , i.e. z = 4 5 − 2 x − 4 y , lying over the rectangle
R : 0 ≤ x ≤ 2 , 0 ≤ y ≤ 3 R:\quad 0\le x\le 2,\qquad 0\le y\le 3 R : 0 ≤ x ≤ 2 , 0 ≤ y ≤ 3 in the x y xy x y x = 0 , x = 2 , y = 0 , y = 3 x=0,\ x=2,\ y=0,\ y=3 x = 0 , x = 2 , y = 0 , y = 3 z z z S S S one-to-one onto R R R
The paper does not prescribe an orientation, so we state ours: take n ^ \hat n n ^ , n ^ ⋅ k ^ > 0 \hat n\cdot\hat k>0 n ^ ⋅ k ^ > 0
Step 2 — Compute ∇ × F ⃗ \nabla\times\vec F ∇ × F
With F 1 = 2 x + 3 y F_1=2x+3y F 1 = 2 x + 3 y F 2 = − 4 z F_2=-4z F 2 = − 4 z F 3 = − 5 x F_3=-5x F 3 = − 5 x
∇ × F ⃗ = ∣ i ^ j ^ k ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z 2 x + 3 y − 4 z − 5 x ∣ = i ^ ( ∂ ( − 5 x ) ∂ y − ∂ ( − 4 z ) ∂ z ) − j ^ ( ∂ ( − 5 x ) ∂ x − ∂ ( 2 x + 3 y ) ∂ z ) + k ^ ( ∂ ( − 4 z ) ∂ x − ∂ ( 2 x + 3 y ) ∂ y ) . \nabla\times\vec F=
\begin{vmatrix}
\hat i & \hat j & \hat k\\[2pt]
\dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\[6pt]
2x+3y & -4z & -5x
\end{vmatrix}
= \hat i\!\left(\frac{\partial(-5x)}{\partial y}-\frac{\partial(-4z)}{\partial z}\right)
- \hat j\!\left(\frac{\partial(-5x)}{\partial x}-\frac{\partial(2x+3y)}{\partial z}\right)
+ \hat k\!\left(\frac{\partial(-4z)}{\partial x}-\frac{\partial(2x+3y)}{\partial y}\right). ∇ × F = i ^ ∂ x ∂ 2 x + 3 y j ^ ∂ y ∂ − 4 z k ^ ∂ z ∂ − 5 x = i ^ ( ∂ y ∂ ( − 5 x ) − ∂ z ∂ ( − 4 z ) ) − j ^ ( ∂ x ∂ ( − 5 x ) − ∂ z ∂ ( 2 x + 3 y ) ) + k ^ ( ∂ x ∂ ( − 4 z ) − ∂ y ∂ ( 2 x + 3 y ) ) . Term by term:
i ^ : 0 − ( − 4 ) = 4 , − j ^ : − ( − 5 − 0 ) = + 5 , k ^ : 0 − 3 = − 3. \hat i:\ 0-(-4)=4,\qquad -\hat j:\ -\bigl(-5-0\bigr)=+5,\qquad \hat k:\ 0-3=-3 . i ^ : 0 − ( − 4 ) = 4 , − j ^ : − ( − 5 − 0 ) = + 5 , k ^ : 0 − 3 = − 3. ∇ × F ⃗ = 4 i ^ + 5 j ^ − 3 k ^ (a constant vector). \boxed{\ \nabla\times\vec F = 4\hat i + 5\hat j - 3\hat k\ }\quad\text{(a constant vector).} ∇ × F = 4 i ^ + 5 j ^ − 3 k ^ (a constant vector). Step 3 — The unit normal.
For the plane ϕ = 2 x + 4 y + 4 z = 5 \phi = 2x+4y+4z=5 ϕ = 2 x + 4 y + 4 z = 5 ∇ ϕ = 2 i ^ + 4 j ^ + 4 k ^ \nabla\phi = 2\hat i+4\hat j+4\hat k ∇ ϕ = 2 i ^ + 4 j ^ + 4 k ^ ∣ ∇ ϕ ∣ = 4 + 16 + 16 = 36 = 6 |\nabla\phi| = \sqrt{4+16+16}=\sqrt{36}=6 ∣∇ ϕ ∣ = 4 + 16 + 16 = 36 = 6
n ^ = 2 i ^ + 4 j ^ + 4 k ^ 6 = 1 3 i ^ + 2 3 j ^ + 2 3 k ^ , n ^ ⋅ k ^ = 2 3 > 0. \hat n = \frac{2\hat i + 4\hat j + 4\hat k}{6} = \frac{1}{3}\hat i + \frac{2}{3}\hat j + \frac{2}{3}\hat k,\qquad \hat n\cdot\hat k = \frac23>0 . n ^ = 6 2 i ^ + 4 j ^ + 4 k ^ = 3 1 i ^ + 3 2 j ^ + 3 2 k ^ , n ^ ⋅ k ^ = 3 2 > 0. Step 4 — The (constant) integrand.
( ∇ × F ⃗ ) ⋅ n ^ = ( 4 ) ( 2 ) + ( 5 ) ( 4 ) + ( − 3 ) ( 4 ) 6 = 8 + 20 − 12 6 = 16 6 = 8 3 . (\nabla\times\vec F)\cdot\hat n = \frac{(4)(2) + (5)(4) + (-3)(4)}{6} = \frac{8+20-12}{6} = \frac{16}{6} = \frac{8}{3}. ( ∇ × F ) ⋅ n ^ = 6 ( 4 ) ( 2 ) + ( 5 ) ( 4 ) + ( − 3 ) ( 4 ) = 6 8 + 20 − 12 = 6 16 = 3 8 . Step 5 — Project onto the x y xy x y
d S = d x d y ∣ n ^ ⋅ k ^ ∣ = d x d y 2 / 3 = 3 2 d x d y . dS = \frac{dx\,dy}{|\hat n\cdot\hat k|} = \frac{dx\,dy}{2/3} = \frac32\,dx\,dy . d S = ∣ n ^ ⋅ k ^ ∣ d x d y = 2/3 d x d y = 2 3 d x d y . Hence
∬ S ( ∇ × F ⃗ ) ⋅ n ^ d S = ∬ R 8 3 ⋅ 3 2 d x d y = 4 ∬ R d x d y = 4 ∫ y = 0 3 ∫ x = 0 2 d x d y = 4 ( 2 ) ( 3 ) = 24. \iint_S(\nabla\times\vec F)\cdot\hat n\,dS = \iint_R \frac{8}{3}\cdot\frac32\,dx\,dy = 4\iint_R dx\,dy = 4\int_{y=0}^{3}\!\int_{x=0}^{2} dx\,dy = 4\,(2)(3) = 24 . ∬ S ( ∇ × F ) ⋅ n ^ d S = ∬ R 3 8 ⋅ 2 3 d x d y = 4 ∬ R d x d y = 4 ∫ y = 0 3 ∫ x = 0 2 d x d y = 4 ( 2 ) ( 3 ) = 24. Step 6 — Independent check by Stokes’ theorem.
By Stokes’ theorem the same integral equals ∮ C F ⃗ ⋅ d r ⃗ \oint_C\vec F\cdot d\vec r ∮ C F ⋅ d r C C C S S S anticlockwise as seen from above (this is the sense compatible with the upward n ^ \hat n n ^ C C C R R R
( 0 , 0 ) → ( 2 , 0 ) → ( 2 , 3 ) → ( 0 , 3 ) → ( 0 , 0 ) , (0,0)\to(2,0)\to(2,3)\to(0,3)\to(0,0), ( 0 , 0 ) → ( 2 , 0 ) → ( 2 , 3 ) → ( 0 , 3 ) → ( 0 , 0 ) , with z = 5 − 2 x − 4 y 4 z=\dfrac{5-2x-4y}{4} z = 4 5 − 2 x − 4 y C C C d z = − 1 2 d x − d y dz = -\tfrac12\,dx - dy d z = − 2 1 d x − d y
F ⃗ ⋅ d r ⃗ = ( 2 x + 3 y ) d x − 4 z d y − 5 x d z = ( 2 x + 3 y ) d x − 4 z d y − 5 x ( − 1 2 d x − d y ) , \vec F\cdot d\vec r = (2x+3y)\,dx - 4z\,dy - 5x\,dz
= (2x+3y)\,dx - 4z\,dy - 5x\Bigl(-\tfrac12dx - dy\Bigr), F ⋅ d r = ( 2 x + 3 y ) d x − 4 z d y − 5 x d z = ( 2 x + 3 y ) d x − 4 z d y − 5 x ( − 2 1 d x − d y ) , and using − 4 z = − ( 5 − 2 x − 4 y ) = 2 x + 4 y − 5 -4z = -(5-2x-4y) = 2x+4y-5 − 4 z = − ( 5 − 2 x − 4 y ) = 2 x + 4 y − 5
F ⃗ ⋅ d r ⃗ = ( 9 x 2 + 3 y ) d x + ( 7 x + 4 y − 5 ) d y . \vec F\cdot d\vec r = \left(\frac{9x}{2}+3y\right)dx + \bigl(7x+4y-5\bigr)dy . F ⋅ d r = ( 2 9 x + 3 y ) d x + ( 7 x + 4 y − 5 ) d y . Now the four edges:
edge parametrisation contribution ( 0 , 0 ) → ( 2 , 0 ) (0,0)\to(2,0) ( 0 , 0 ) → ( 2 , 0 ) y = 0 y=0 y = 0 d y = 0 dy=0 d y = 0 x : 0 → 2 x:0\to2 x : 0 → 2 ∫ 0 2 9 x 2 d x = 9 4 [ x 2 ] 0 2 = 9 \displaystyle\int_0^2 \frac{9x}{2}dx = \frac94\bigl[x^2\bigr]_0^2 = 9 ∫ 0 2 2 9 x d x = 4 9 [ x 2 ] 0 2 = 9 ( 2 , 0 ) → ( 2 , 3 ) (2,0)\to(2,3) ( 2 , 0 ) → ( 2 , 3 ) x = 2 x=2 x = 2 d x = 0 dx=0 d x = 0 y : 0 → 3 y:0\to3 y : 0 → 3 ∫ 0 3 ( 9 + 4 y ) d y = 27 + 18 = 45 \displaystyle\int_0^3 (9+4y)\,dy = 27+18 = 45 ∫ 0 3 ( 9 + 4 y ) d y = 27 + 18 = 45 ( 2 , 3 ) → ( 0 , 3 ) (2,3)\to(0,3) ( 2 , 3 ) → ( 0 , 3 ) y = 3 y=3 y = 3 d y = 0 dy=0 d y = 0 x : 2 → 0 x:2\to0 x : 2 → 0 − ∫ 0 2 ( 9 x 2 + 9 ) d x = − ( 9 + 18 ) = − 27 \displaystyle-\int_0^2\left(\frac{9x}{2}+9\right)dx = -(9+18) = -27 − ∫ 0 2 ( 2 9 x + 9 ) d x = − ( 9 + 18 ) = − 27 ( 0 , 3 ) → ( 0 , 0 ) (0,3)\to(0,0) ( 0 , 3 ) → ( 0 , 0 ) x = 0 x=0 x = 0 d x = 0 dx=0 d x = 0 y : 3 → 0 y:3\to0 y : 3 → 0 − ∫ 0 3 ( 4 y − 5 ) d y = − ( 18 − 15 ) = − 3 \displaystyle-\int_0^3 (4y-5)\,dy = -(18-15) = -3 − ∫ 0 3 ( 4 y − 5 ) d y = − ( 18 − 15 ) = − 3
∮ C F ⃗ ⋅ d r ⃗ = 9 + 45 − 27 − 3 = 24. \oint_C\vec F\cdot d\vec r = 9 + 45 - 27 - 3 = 24 . ∮ C F ⋅ d r = 9 + 45 − 27 − 3 = 24. This agrees exactly with Step 5, which confirms both the value and the orientation.
Answer
∇ × F ⃗ = 4 i ^ + 5 j ^ − 3 k ^ , n ^ = 1 3 ( i ^ + 2 j ^ + 2 k ^ ) , ∬ S ( ∇ × F ⃗ ) ⋅ n ^ d S = 24 \boxed{\;\nabla\times\vec F = 4\hat i+5\hat j-3\hat k,\quad \hat n=\tfrac13\bigl(\hat i+2\hat j+2\hat k\bigr),\qquad \iint_S(\nabla\times\vec F)\cdot\hat n\,dS = 24\;} ∇ × F = 4 i ^ + 5 j ^ − 3 k ^ , n ^ = 3 1 ( i ^ + 2 j ^ + 2 k ^ ) , ∬ S ( ∇ × F ) ⋅ n ^ d S = 24 (for the upward normal; the value is − 24 -24 − 24