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UPSC 2017 Maths Optional Paper 1 Q8c-ii — Step-by-Step Solution 8 marks · Section B
Line integrals · Vector Analysis · asked 8× in 13 yrs · Read the full method →
Question
Using Green’s theorem, evaluate the ∫ C F ⃗ ( r ⃗ ) ⋅ d r ⃗ \displaystyle\int_C \vec F(\vec r)\cdot d\vec r ∫ C F ( r ) ⋅ d r counterclockwise where F ⃗ ( r ⃗ ) = ( x 2 + y 2 ) i ^ + ( x 2 − y 2 ) j ^ \vec F(\vec r)=(x^2+y^2)\hat i+(x^2-y^2)\hat j F ( r ) = ( x 2 + y 2 ) i ^ + ( x 2 − y 2 ) j ^ and d r ⃗ = d x i ^ + d y j ^ d\vec r=dx\,\hat i+dy\,\hat j d r = d x i ^ + d y j ^ and the curve C C C is the boundary of the region R = { ( x , y ) ∣ 1 ≤ y ≤ 2 − x 2 } R=\{(x,y)\mid 1\le y\le 2-x^2\} R = {( x , y ) ∣ 1 ≤ y ≤ 2 − x 2 } .
Technique
Green’s theorem ∮ P d x + Q d y = ∬ ( Q x − P y ) d A \oint P\,dx+Q\,dy=\iint(Q_x-P_y)\,dA ∮ P d x + Q d y = ∬ ( Q x − P y ) d A with Q x − P y = 2 x − 2 y Q_x-P_y=2x-2y Q x − P y = 2 x − 2 y over the parabola–line region.
Solution
Step 1 — Set up Green’s theorem
With P = x 2 + y 2 P=x^2+y^2 P = x 2 + y 2 and Q = x 2 − y 2 Q=x^2-y^2 Q = x 2 − y 2 , Green’s theorem gives
∮ C P d x + Q d y = ∬ R ( ∂ Q ∂ x − ∂ P ∂ y ) d A . \oint_C P\,dx+Q\,dy=\iint_R\Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big)\,dA. ∮ C P d x + Q d y = ∬ R ( ∂ x ∂ Q − ∂ y ∂ P ) d A .
∂ Q ∂ x = 2 x , ∂ P ∂ y = 2 y ⟹ ∂ Q ∂ x − ∂ P ∂ y = 2 x − 2 y . \frac{\partial Q}{\partial x}=2x,\qquad \frac{\partial P}{\partial y}=2y\quad\Longrightarrow\quad \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=2x-2y. ∂ x ∂ Q = 2 x , ∂ y ∂ P = 2 y ⟹ ∂ x ∂ Q − ∂ y ∂ P = 2 x − 2 y .
Step 2 — Describe the region
R = { 1 ≤ y ≤ 2 − x 2 } R=\{1\le y\le 2-x^2\} R = { 1 ≤ y ≤ 2 − x 2 } : bounded below by the horizontal line y = 1 y=1 y = 1 and above by the parabola y = 2 − x 2 y=2-x^2 y = 2 − x 2 . They meet where 2 − x 2 = 1 ⇒ x 2 = 1 ⇒ x = ± 1 2-x^2=1\Rightarrow x^2=1\Rightarrow x=\pm1 2 − x 2 = 1 ⇒ x 2 = 1 ⇒ x = ± 1 . So x ∈ [ − 1 , 1 ] x\in[-1,1] x ∈ [ − 1 , 1 ] and, for each x x x , y y y runs from 1 1 1 to 2 − x 2 2-x^2 2 − x 2 .
Step 3 — Evaluate the double integral
∬ R ( 2 x − 2 y ) d A = ∫ − 1 1 ∫ 1 2 − x 2 ( 2 x − 2 y ) d y d x . \iint_R(2x-2y)\,dA=\int_{-1}^{1}\int_{1}^{2-x^2}(2x-2y)\,dy\,dx. ∬ R ( 2 x − 2 y ) d A = ∫ − 1 1 ∫ 1 2 − x 2 ( 2 x − 2 y ) d y d x .
Inner integral:
∫ 1 2 − x 2 ( 2 x − 2 y ) d y = [ 2 x y − y 2 ] 1 2 − x 2 = 2 x ( 2 − x 2 ) − ( 2 − x 2 ) 2 − ( 2 x − 1 ) . \int_{1}^{2-x^2}(2x-2y)\,dy=\Big[2xy-y^2\Big]_{1}^{2-x^2}=2x(2-x^2)-(2-x^2)^2-\big(2x-1\big). ∫ 1 2 − x 2 ( 2 x − 2 y ) d y = [ 2 x y − y 2 ] 1 2 − x 2 = 2 x ( 2 − x 2 ) − ( 2 − x 2 ) 2 − ( 2 x − 1 ) .
The 2 x ( 2 − x 2 ) − 2 x 2x(2-x^2)-2x 2 x ( 2 − x 2 ) − 2 x part is odd in x x x and integrates to 0 0 0 over [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] . The remaining even part is
− ( 2 − x 2 ) 2 + 1 = − ( 4 − 4 x 2 + x 4 ) + 1 = − 3 + 4 x 2 − x 4 . -(2-x^2)^2+1=-(4-4x^2+x^4)+1=-3+4x^2-x^4. − ( 2 − x 2 ) 2 + 1 = − ( 4 − 4 x 2 + x 4 ) + 1 = − 3 + 4 x 2 − x 4 .
∫ − 1 1 ( − 3 + 4 x 2 − x 4 ) d x = 2 ∫ 0 1 ( − 3 + 4 x 2 − x 4 ) d x = 2 [ − 3 x + 4 x 3 3 − x 5 5 ] 0 1 = 2 ( − 3 + 4 3 − 1 5 ) . \int_{-1}^{1}(-3+4x^2-x^4)\,dx=2\int_{0}^{1}(-3+4x^2-x^4)\,dx=2\Big[-3x+\frac{4x^3}{3}-\frac{x^5}{5}\Big]_0^1=2\Big(-3+\frac43-\frac15\Big). ∫ − 1 1 ( − 3 + 4 x 2 − x 4 ) d x = 2 ∫ 0 1 ( − 3 + 4 x 2 − x 4 ) d x = 2 [ − 3 x + 3 4 x 3 − 5 x 5 ] 0 1 = 2 ( − 3 + 3 4 − 5 1 ) .
= 2 ⋅ − 45 + 20 − 3 15 = 2 ⋅ − 28 15 = − 56 15 . =2\cdot\frac{-45+20-3}{15}=2\cdot\frac{-28}{15}=-\frac{56}{15}. = 2 ⋅ 15 − 45 + 20 − 3 = 2 ⋅ 15 − 28 = − 15 56 .
Answer
∮ C F ⃗ ⋅ d r ⃗ = − 56 15 . \boxed{\;\oint_C\vec F\cdot d\vec r=-\frac{56}{15}.\;} ∮ C F ⋅ d r = − 15 56 .