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UPSC 2022 Maths Optional Paper 1 Q6c — Step-by-Step Solution 15 marks · Section B
Line integrals · Vector Analysis · asked 8× in 13 yrs · Read the full method →
Question
Verify Green’s theorem in the plane for ∮ C ( 3 x 2 − 8 y 2 ) d x + ( 4 y − 6 x y ) d y \displaystyle\oint_C(3x^2-8y^2)\,dx+(4y-6xy)\,dy ∮ C ( 3 x 2 − 8 y 2 ) d x + ( 4 y − 6 x y ) d y C C C x = 0 , y = 0 , x + y = 1 x=0,\,y=0,\,x+y=1 x = 0 , y = 0 , x + y = 1
Technique
Direct computation of both sides of Green’s theorem: double integral over triangle (using vertical strips); line integral parametrised over three sides (with CCW orientation).
Solution
Region: Triangle with vertices ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 , 0 ) (1,0) ( 1 , 0 ) ( 0 , 1 ) (0,1) ( 0 , 1 )
Green’s theorem: ∮ C P d x + Q d y = ∬ R ( ∂ Q ∂ x − ∂ P ∂ y ) d A \oint_C P\,dx+Q\,dy=\iint_R\!\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)dA ∮ C P d x + Q d y = ∬ R ( ∂ x ∂ Q − ∂ y ∂ P ) d A
Here P = 3 x 2 − 8 y 2 P=3x^2-8y^2 P = 3 x 2 − 8 y 2 Q = 4 y − 6 x y Q=4y-6xy Q = 4 y − 6 x y
Step 1 — Compute the double integral
∂ Q ∂ x = − 6 y \dfrac{\partial Q}{\partial x}=-6y ∂ x ∂ Q = − 6 y ∂ P ∂ y = − 16 y \dfrac{\partial P}{\partial y}=-16y ∂ y ∂ P = − 16 y
∂ Q ∂ x − ∂ P ∂ y = − 6 y − ( − 16 y ) = 10 y \dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}=-6y-(-16y)=10y ∂ x ∂ Q − ∂ y ∂ P = − 6 y − ( − 16 y ) = 10 y
Double integral over triangle:
∬ R 10 y d A = 10 ∫ 0 1 ∫ 0 1 − x y d y d x = 10 ∫ 0 1 ( 1 − x ) 2 2 d x = 5 ∫ 0 1 ( 1 − x ) 2 d x . \iint_R 10y\,dA=10\int_0^1\!\!\int_0^{1-x}y\,dy\,dx=10\int_0^1\dfrac{(1-x)^2}{2}\,dx=5\int_0^1(1-x)^2\,dx. ∬ R 10 y d A = 10 ∫ 0 1 ∫ 0 1 − x y d y d x = 10 ∫ 0 1 2 ( 1 − x ) 2 d x = 5 ∫ 0 1 ( 1 − x ) 2 d x . Substitute u = 1 − x u=1-x u = 1 − x 5 ∫ 1 0 u 2 ( − d u ) = 5 ∫ 0 1 u 2 d u = 5 / 3 5\int_1^0 u^2(-du)=5\int_0^1 u^2\,du=5/3 5 ∫ 1 0 u 2 ( − d u ) = 5 ∫ 0 1 u 2 d u = 5/3
So double integral = 5 / 3 =5/3 = 5/3
Step 2 — Compute the line integral
Traverse the triangle counterclockwise: ( 0 , 0 ) → ( 1 , 0 ) → ( 0 , 1 ) → ( 0 , 0 ) (0,0)\to(1,0)\to(0,1)\to(0,0) ( 0 , 0 ) → ( 1 , 0 ) → ( 0 , 1 ) → ( 0 , 0 )
Side 1: ( 0 , 0 ) → ( 1 , 0 ) (0,0)\to(1,0) ( 0 , 0 ) → ( 1 , 0 ) y = 0 , d y = 0 y=0,\;dy=0 y = 0 , d y = 0 x : 0 → 1 x:0\to 1 x : 0 → 1
∮ S 1 = ∫ 0 1 [ 3 x 2 − 0 ] d x + ∫ 0 = ∫ 0 1 3 x 2 d x = x 3 ∣ 0 1 = 1 \oint_{S_1}=\int_0^1[3x^2-0]\,dx+\int 0=\int_0^1 3x^2\,dx=x^3\big|_0^1=1 ∮ S 1 = ∫ 0 1 [ 3 x 2 − 0 ] d x + ∫ 0 = ∫ 0 1 3 x 2 d x = x 3 0 1 = 1
Side 2: ( 1 , 0 ) → ( 0 , 1 ) (1,0)\to(0,1) ( 1 , 0 ) → ( 0 , 1 ) Parametrise: x = 1 − t , y = t x=1-t,\;y=t x = 1 − t , y = t t : 0 → 1 t:0\to 1 t : 0 → 1 d x = − d t , d y = d t dx=-dt,\;dy=dt d x = − d t , d y = d t
P = 3 ( 1 − t ) 2 − 8 t 2 P=3(1-t)^2-8t^2 P = 3 ( 1 − t ) 2 − 8 t 2 Q = 4 t − 6 ( 1 − t ) t = 4 t − 6 t + 6 t 2 = − 2 t + 6 t 2 Q=4t-6(1-t)t=4t-6t+6t^2=-2t+6t^2 Q = 4 t − 6 ( 1 − t ) t = 4 t − 6 t + 6 t 2 = − 2 t + 6 t 2
∮ S 2 = ∫ 0 1 [ P ( − d t ) + Q d t ] = ∫ 0 1 ( − P + Q ) d t \oint_{S_2}=\int_0^1[P(-dt)+Q\,dt]=\int_0^1(-P+Q)\,dt ∮ S 2 = ∫ 0 1 [ P ( − d t ) + Q d t ] = ∫ 0 1 ( − P + Q ) d t
− P + Q = − 3 ( 1 − t ) 2 + 8 t 2 − 2 t + 6 t 2 = − 3 ( 1 − t ) 2 + 14 t 2 − 2 t -P+Q=-3(1-t)^2+8t^2-2t+6t^2=-3(1-t)^2+14t^2-2t − P + Q = − 3 ( 1 − t ) 2 + 8 t 2 − 2 t + 6 t 2 = − 3 ( 1 − t ) 2 + 14 t 2 − 2 t
Expand ( 1 − t ) 2 = 1 − 2 t + t 2 (1-t)^2=1-2t+t^2 ( 1 − t ) 2 = 1 − 2 t + t 2 − 3 + 6 t − 3 t 2 -3+6t-3t^2 − 3 + 6 t − 3 t 2
So − P + Q = − 3 + 6 t − 3 t 2 + 14 t 2 − 2 t = − 3 + 4 t + 11 t 2 -P+Q=-3+6t-3t^2+14t^2-2t=-3+4t+11t^2 − P + Q = − 3 + 6 t − 3 t 2 + 14 t 2 − 2 t = − 3 + 4 t + 11 t 2
∫ 0 1 ( − 3 + 4 t + 11 t 2 ) d t = − 3 + 2 + 11 / 3 = − 1 + 11 / 3 = 8 / 3 \int_0^1(-3+4t+11t^2)\,dt=-3+2+11/3=-1+11/3=8/3 ∫ 0 1 ( − 3 + 4 t + 11 t 2 ) d t = − 3 + 2 + 11/3 = − 1 + 11/3 = 8/3
Side 3: ( 0 , 1 ) → ( 0 , 0 ) (0,1)\to(0,0) ( 0 , 1 ) → ( 0 , 0 ) x = 0 , d x = 0 x=0,\;dx=0 x = 0 , d x = 0 y : 1 → 0 y:1\to 0 y : 1 → 0
Q = 4 y − 0 = 4 y Q=4y-0=4y Q = 4 y − 0 = 4 y
∮ S 3 = ∫ 0 d x + ∫ 1 0 4 y d y = 4 ⋅ y 2 / 2 ∣ 1 0 = 2 ( 0 − 1 ) = − 2 \oint_{S_3}=\int 0\,dx+\int_1^0 4y\,dy=4\cdot y^2/2\big|_1^0=2(0-1)=-2 ∮ S 3 = ∫ 0 d x + ∫ 1 0 4 y d y = 4 ⋅ y 2 /2 1 0 = 2 ( 0 − 1 ) = − 2
Total line integral:
∮ C = 1 + 8 / 3 + ( − 2 ) = − 1 + 8 / 3 = 5 / 3. \oint_C=1+8/3+(-2)=-1+8/3=5/3. ∮ C = 1 + 8/3 + ( − 2 ) = − 1 + 8/3 = 5/3. Step 3 — Compare
Both line integral and double integral equal 5 / 3 5/3 5/3 Green’s theorem verified.
Answer
∮ C P d x + Q d y = ∬ R ( ∂ Q ∂ x − ∂ P ∂ y ) d A = 5 3 . \boxed{\;\oint_C P\,dx+Q\,dy=\iint_R\!\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)dA=\dfrac{5}{3}.\;} ∮ C P d x + Q d y = ∬ R ( ∂ x ∂ Q − ∂ y ∂ P ) d A = 3 5 .