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UPSC 2020 Maths Optional Paper 1 Q7a — Step-by-Step Solution 20 marks · Section B
Stokes' theorem · Vector Analysis · asked 10× in 14 yrs · Read the full method →
Question
Verify the Stokes’ theorem for the vector field F ⃗ = x y i ^ + y z j ^ + x z k ^ \vec F=xy\hat i+yz\hat j+xz\hat k F = x y i ^ + y z j ^ + x z k ^ S S S z = 1 − x 2 z=1-x^2 z = 1 − x 2 0 ≤ x ≤ 1 0\le x\le 1 0 ≤ x ≤ 1 − 2 ≤ y ≤ 2 -2\le y\le 2 − 2 ≤ y ≤ 2 S S S
Technique
Curl computation; surface integral over a graph z = g ( x , y ) z=g(x,y) z = g ( x , y ) n ^ d S = ( − g x , − g y , 1 ) d x d y \hat n\,dS=(-g_x,-g_y,1)\,dx\,dy n ^ d S = ( − g x , − g y , 1 ) d x d y
Solution
Stokes’ theorem: ∬ S ( ∇ × F ⃗ ) ⋅ n ^ d S = ∮ ∂ S F ⃗ ⋅ d r ⃗ \displaystyle\iint_S(\nabla\times\vec F)\cdot\hat n\,dS=\oint_{\partial S}\vec F\cdot d\vec r ∬ S ( ∇ × F ) ⋅ n ^ d S = ∮ ∂ S F ⋅ d r
Step 1 — Curl
∇ × F ⃗ = ∣ i ^ j ^ k ^ ∂ x ∂ y ∂ z x y y z x z ∣ = ( ∂ y ( x z ) − ∂ z ( y z ) ) i ^ + ( ∂ z ( x y ) − ∂ x ( x z ) ) j ^ + ( ∂ x ( y z ) − ∂ y ( x y ) ) k ^ \nabla\times\vec F=\begin{vmatrix}\hat i&\hat j&\hat k\\[2pt]\partial_x&\partial_y&\partial_z\\[2pt] xy&yz&xz\end{vmatrix}
=\big(\partial_y(xz)-\partial_z(yz)\big)\hat i+\big(\partial_z(xy)-\partial_x(xz)\big)\hat j+\big(\partial_x(yz)-\partial_y(xy)\big)\hat k ∇ × F = i ^ ∂ x x y j ^ ∂ y y z k ^ ∂ z x z = ( ∂ y ( x z ) − ∂ z ( y z ) ) i ^ + ( ∂ z ( x y ) − ∂ x ( x z ) ) j ^ + ( ∂ x ( y z ) − ∂ y ( x y ) ) k ^ = ( 0 − y ) i ^ + ( 0 − z ) j ^ + ( 0 − x ) k ^ = − y i ^ − z j ^ − x k ^ . =(0-y)\hat i+(0-z)\hat j+(0-x)\hat k=-y\,\hat i-z\,\hat j-x\,\hat k . = ( 0 − y ) i ^ + ( 0 − z ) j ^ + ( 0 − x ) k ^ = − y i ^ − z j ^ − x k ^ . Step 2 — Surface integral
The surface is z = g ( x , y ) = 1 − x 2 z=g(x,y)=1-x^2 z = g ( x , y ) = 1 − x 2 R = { 0 ≤ x ≤ 1 , − 2 ≤ y ≤ 2 } R=\{0\le x\le1,\ -2\le y\le2\} R = { 0 ≤ x ≤ 1 , − 2 ≤ y ≤ 2 } upward orientation,
n ^ d S = ( − g x , − g y , 1 ) d x d y = ( 2 x , 0 , 1 ) d x d y . \hat n\,dS=(-g_x,\,-g_y,\,1)\,dx\,dy=(2x,\,0,\,1)\,dx\,dy . n ^ d S = ( − g x , − g y , 1 ) d x d y = ( 2 x , 0 , 1 ) d x d y . On S S S z = 1 − x 2 z=1-x^2 z = 1 − x 2 ∇ × F ⃗ = ( − y , − ( 1 − x 2 ) , − x ) \nabla\times\vec F=(-y,\,-(1-x^2),\,-x) ∇ × F = ( − y , − ( 1 − x 2 ) , − x )
( ∇ × F ⃗ ) ⋅ n ^ d S = [ ( − y ) ( 2 x ) + ( − ( 1 − x 2 ) ) ( 0 ) + ( − x ) ( 1 ) ] d x d y = ( − 2 x y − x ) d x d y . (\nabla\times\vec F)\cdot\hat n\,dS=\big[(-y)(2x)+(-(1-x^2))(0)+(-x)(1)\big]dx\,dy=(-2xy-x)\,dx\,dy . ( ∇ × F ) ⋅ n ^ d S = [ ( − y ) ( 2 x ) + ( − ( 1 − x 2 )) ( 0 ) + ( − x ) ( 1 ) ] d x d y = ( − 2 x y − x ) d x d y . Integrate:
∬ S ( ∇ × F ⃗ ) ⋅ n ^ d S = ∫ − 2 2 ∫ 0 1 ( − 2 x y − x ) d x d y . \iint_S(\nabla\times\vec F)\cdot\hat n\,dS=\int_{-2}^{2}\!\!\int_0^1(-2xy-x)\,dx\,dy . ∬ S ( ∇ × F ) ⋅ n ^ d S = ∫ − 2 2 ∫ 0 1 ( − 2 x y − x ) d x d y . Inner (∫ 0 1 x d x = 1 2 \int_0^1 x\,dx=\tfrac12 ∫ 0 1 x d x = 2 1 ∫ 0 1 ( − 2 x y − x ) d x = − 2 y ⋅ 1 2 − 1 2 = − y − 1 2 \displaystyle\int_0^1(-2xy-x)dx=-2y\cdot\tfrac12-\tfrac12=-y-\tfrac12 ∫ 0 1 ( − 2 x y − x ) d x = − 2 y ⋅ 2 1 − 2 1 = − y − 2 1 ∫ − 2 2 ( − y − 1 2 ) d y = [ − y 2 2 − y 2 ] − 2 2 = ( − 2 − 1 ) − ( − 2 + 1 ) = − 3 − ( − 1 ) = − 2. \displaystyle\int_{-2}^2\Big(-y-\tfrac12\Big)dy=\big[-\tfrac{y^2}{2}-\tfrac{y}{2}\big]_{-2}^2=(-2-1)-(-2+1)=-3-(-1)=-2. ∫ − 2 2 ( − y − 2 1 ) d y = [ − 2 y 2 − 2 y ] − 2 2 = ( − 2 − 1 ) − ( − 2 + 1 ) = − 3 − ( − 1 ) = − 2.
∬ S ( ∇ × F ⃗ ) ⋅ n ^ d S = − 2. \iint_S(\nabla\times\vec F)\cdot\hat n\,dS=-2 . ∬ S ( ∇ × F ) ⋅ n ^ d S = − 2. Step 3 — Boundary line integral
The upward normal makes ∂ S \partial S ∂ S counterclockwise as seen from above : in the x y xy x y
( 0 , − 2 ) → ( 1 , − 2 ) → ( 1 , 2 ) → ( 0 , 2 ) → ( 0 , − 2 ) , (0,-2)\to(1,-2)\to(1,2)\to(0,2)\to(0,-2), ( 0 , − 2 ) → ( 1 , − 2 ) → ( 1 , 2 ) → ( 0 , 2 ) → ( 0 , − 2 ) , with z = 1 − x 2 z=1-x^2 z = 1 − x 2 F ⃗ ⋅ d r ⃗ = x y d x + y z d y + x z d z \vec F\cdot d\vec r=xy\,dx+yz\,dy+xz\,dz F ⋅ d r = x y d x + y z d y + x z d z
Edge 1: x = t , y = − 2 , z = 1 − t 2 , t : 0 → 1 x=t,\,y=-2,\,z=1-t^2,\ t:0\to1 x = t , y = − 2 , z = 1 − t 2 , t : 0 → 1 d y = 0 dy=0 d y = 0 d z = − 2 t d t dz=-2t\,dt d z = − 2 t d t
∫ 0 1 [ ( t ) ( − 2 ) d t + ( t ) ( 1 − t 2 ) ( − 2 t ) d t ] = ∫ 0 1 ( − 2 t − 2 t 2 + 2 t 4 ) d t = − 1 − 2 3 + 2 5 = − 19 15 . \int_0^1\Big[(t)(-2)\,dt+(t)(1-t^2)(-2t)\,dt\Big]=\int_0^1\big(-2t-2t^2+2t^4\big)dt=-1-\tfrac23+\tfrac25=-\tfrac{19}{15}. ∫ 0 1 [ ( t ) ( − 2 ) d t + ( t ) ( 1 − t 2 ) ( − 2 t ) d t ] = ∫ 0 1 ( − 2 t − 2 t 2 + 2 t 4 ) d t = − 1 − 3 2 + 5 2 = − 15 19 . Edge 2: x = 1 , y = t , z = 1 − 1 = 0 , t : − 2 → 2 x=1,\,y=t,\,z=1-1=0,\ t:-2\to2 x = 1 , y = t , z = 1 − 1 = 0 , t : − 2 → 2 d x = 0 , d z = 0 dx=0,\,dz=0 d x = 0 , d z = 0 y z d y = t ⋅ 0 ⋅ d t = 0 yz\,dy=t\cdot0\cdot dt=0 y z d y = t ⋅ 0 ⋅ d t = 0
∫ − 2 2 0 d t = 0. \int_{-2}^2 0\,dt=0 . ∫ − 2 2 0 d t = 0. Edge 3: x = t , y = 2 , z = 1 − t 2 , t : 1 → 0 x=t,\,y=2,\,z=1-t^2,\ t:1\to0 x = t , y = 2 , z = 1 − t 2 , t : 1 → 0 d y = 0 dy=0 d y = 0 d z = − 2 t d t dz=-2t\,dt d z = − 2 t d t
∫ 1 0 [ ( t ) ( 2 ) d t + ( t ) ( 1 − t 2 ) ( − 2 t ) d t ] = ∫ 1 0 ( 2 t − 2 t 2 + 2 t 4 ) d t = − [ t 2 − 2 3 t 3 + 2 5 t 5 ] 0 1 = − ( 1 − 2 3 + 2 5 ) = − 11 15 . \int_1^0\Big[(t)(2)\,dt+(t)(1-t^2)(-2t)\,dt\Big]=\int_1^0\big(2t-2t^2+2t^4\big)dt=-\big[t^2-\tfrac23t^3+\tfrac25t^5\big]_0^1=-\Big(1-\tfrac23+\tfrac25\Big)=-\tfrac{11}{15}. ∫ 1 0 [ ( t ) ( 2 ) d t + ( t ) ( 1 − t 2 ) ( − 2 t ) d t ] = ∫ 1 0 ( 2 t − 2 t 2 + 2 t 4 ) d t = − [ t 2 − 3 2 t 3 + 5 2 t 5 ] 0 1 = − ( 1 − 3 2 + 5 2 ) = − 15 11 . Edge 4: x = 0 , y = t , z = 1 − 0 = 1 , t : 2 → − 2 x=0,\,y=t,\,z=1-0=1,\ t:2\to-2 x = 0 , y = t , z = 1 − 0 = 1 , t : 2 → − 2 d x = 0 dx=0 d x = 0 d z = 0 dz=0 d z = 0 y z d y = t ⋅ 1 d t yz\,dy=t\cdot1\,dt y z d y = t ⋅ 1 d t
∫ 2 − 2 t d t = [ t 2 2 ] 2 − 2 = 2 − 2 = 0. \int_2^{-2} t\,dt=\big[\tfrac{t^2}{2}\big]_2^{-2}=2-2=0 . ∫ 2 − 2 t d t = [ 2 t 2 ] 2 − 2 = 2 − 2 = 0. Sum:
∮ ∂ S F ⃗ ⋅ d r ⃗ = − 19 15 + 0 − 11 15 + 0 = − 30 15 = − 2. \oint_{\partial S}\vec F\cdot d\vec r=-\tfrac{19}{15}+0-\tfrac{11}{15}+0=-\tfrac{30}{15}=-2 . ∮ ∂ S F ⋅ d r = − 15 19 + 0 − 15 11 + 0 = − 15 30 = − 2. Step 4 — Conclusion
Answer
∬ S ( ∇ × F ⃗ ) ⋅ n ^ d S = − 2 = ∮ ∂ S F ⃗ ⋅ d r ⃗ \boxed{\ \iint_S(\nabla\times\vec F)\cdot\hat n\,dS=-2=\oint_{\partial S}\vec F\cdot d\vec r\ } ∬ S ( ∇ × F ) ⋅ n ^ d S = − 2 = ∮ ∂ S F ⋅ d r