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UPSC 2019 Maths Optional Paper 1 Q6b — Step-by-Step Solution 15 marks · Section B
Line integrals · Vector Analysis · asked 8× in 13 yrs · Read the full method →
Question
Find the circulation of F ⃗ \vec F F C C C F ⃗ = ( 2 x + y 2 ) i ^ + ( 3 y − 4 x ) j ^ \vec F=(2x+y^2)\hat i+(3y-4x)\hat j F = ( 2 x + y 2 ) i ^ + ( 3 y − 4 x ) j ^ C C C y = x 2 y=x^2 y = x 2 ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 , 1 ) (1,1) ( 1 , 1 ) y 2 = x y^2=x y 2 = x ( 1 , 1 ) (1,1) ( 1 , 1 ) ( 0 , 0 ) (0,0) ( 0 , 0 )
Technique
Direct parametric line integral on each arc; cross-checked by Green’s theorem on the enclosed region.
Solution
The circulation is ∮ C F ⃗ ⋅ d r ⃗ = ∮ C P d x + Q d y \displaystyle\oint_C\vec F\cdot d\vec r=\oint_C P\,dx+Q\,dy ∮ C F ⋅ d r = ∮ C P d x + Q d y P = 2 x + y 2 , Q = 3 y − 4 x P=2x+y^2,\ Q=3y-4x P = 2 x + y 2 , Q = 3 y − 4 x
C 1 : y = x 2 C_1:\ y=x^2 C 1 : y = x 2 ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 , 1 ) (1,1) ( 1 , 1 ) C 2 : y 2 = x C_2:\ y^2=x C 2 : y 2 = x x = y 2 x=y^2 x = y 2 ( 1 , 1 ) (1,1) ( 1 , 1 ) ( 0 , 0 ) (0,0) ( 0 , 0 )
Step 1 — Integral along C 1 C_1 C 1 y = x 2 y=x^2 y = x 2 x : 0 → 1 x:0\to1 x : 0 → 1
Here d y = 2 x d x dy=2x\,dx d y = 2 x d x
∫ C 1 = ∫ 0 1 [ ( 2 x + x 4 ) + ( 3 x 2 − 4 x ) ( 2 x ) ] d x = ∫ 0 1 ( 2 x + x 4 + 6 x 3 − 8 x 2 ) d x . \int_{C_1}=\int_0^1\Big[(2x+x^4)+(3x^2-4x)(2x)\Big]dx=\int_0^1\big(2x+x^4+6x^3-8x^2\big)dx. ∫ C 1 = ∫ 0 1 [ ( 2 x + x 4 ) + ( 3 x 2 − 4 x ) ( 2 x ) ] d x = ∫ 0 1 ( 2 x + x 4 + 6 x 3 − 8 x 2 ) d x . = [ x 2 + x 5 5 + 6 x 4 4 − 8 x 3 3 ] 0 1 = 1 + 1 5 + 3 2 − 8 3 = 30 + 6 + 45 − 80 30 = 1 30 . =\Big[x^2+\tfrac{x^5}{5}+\tfrac{6x^4}{4}-\tfrac{8x^3}{3}\Big]_0^1=1+\tfrac15+\tfrac32-\tfrac83=\frac{30+6+45-80}{30}=\frac{1}{30}. = [ x 2 + 5 x 5 + 4 6 x 4 − 3 8 x 3 ] 0 1 = 1 + 5 1 + 2 3 − 3 8 = 30 30 + 6 + 45 − 80 = 30 1 . Step 2 — Integral along C 2 C_2 C 2 x = y 2 x=y^2 x = y 2 y : 1 → 0 y:1\to0 y : 1 → 0
Here d x = 2 y d y dx=2y\,dy d x = 2 y d y
∫ C 2 = ∫ 1 0 [ ( 2 y 2 + y 2 ) ( 2 y ) + ( 3 y − 4 y 2 ) ] d y = ∫ 1 0 ( 6 y 3 + 3 y − 4 y 2 ) d y . \int_{C_2}=\int_1^0\Big[(2y^2+y^2)(2y)+(3y-4y^2)\Big]dy=\int_1^0\big(6y^3+3y-4y^2\big)dy. ∫ C 2 = ∫ 1 0 [ ( 2 y 2 + y 2 ) ( 2 y ) + ( 3 y − 4 y 2 ) ] d y = ∫ 1 0 ( 6 y 3 + 3 y − 4 y 2 ) d y . = [ 6 y 4 4 + 3 y 2 2 − 4 y 3 3 ] 1 0 = − ( 3 2 + 3 2 − 4 3 ) = − ( 3 − 4 3 ) = − 5 3 . =\Big[\tfrac{6y^4}{4}+\tfrac{3y^2}{2}-\tfrac{4y^3}{3}\Big]_1^0=-\Big(\tfrac32+\tfrac32-\tfrac43\Big)=-\Big(3-\tfrac43\Big)=-\frac53. = [ 4 6 y 4 + 2 3 y 2 − 3 4 y 3 ] 1 0 = − ( 2 3 + 2 3 − 3 4 ) = − ( 3 − 3 4 ) = − 3 5 . Step 3 — Total circulation
∮ C = 1 30 − 5 3 = 1 − 50 30 = − 49 30 . \oint_C=\frac{1}{30}-\frac53=\frac{1-50}{30}=-\frac{49}{30}. ∮ C = 30 1 − 3 5 = 30 1 − 50 = − 30 49 . Answer
∮ C F ⃗ ⋅ d r ⃗ = − 49 30 . \boxed{\;\oint_C\vec F\cdot d\vec r=-\frac{49}{30}.\;} ∮ C F ⋅ d r = − 30 49 .