← 2017 Paper 1

UPSC 2017 Maths Optional Paper 1 Q8c-ii — Step-by-Step Solution

8 marks · Section B

Line integrals · Vector Analysis · asked 8× in 14 yrs · Read the full method →

Question

Using Green’s theorem, evaluate the ∫CF⃗(r⃗)⋅dr⃗\displaystyle\int_C \vec F(\vec r)\cdot d\vec r counterclockwise where F⃗(r⃗)=(x2+y2)i^+(x2−y2)j^\vec F(\vec r)=(x^2+y^2)\hat i+(x^2-y^2)\hat j and dr⃗=dx i^+dy j^d\vec r=dx\,\hat i+dy\,\hat j and the curve CC is the boundary of the region R={(x,y)∣1≤y≤2−x2}R=\{(x,y)\mid 1\le y\le 2-x^2\}.

Technique

Green’s theorem ∮P dx+Q dy=∬(Qx−Py) dA\oint P\,dx+Q\,dy=\iint(Q_x-P_y)\,dA with Qx−Py=2x−2yQ_x-P_y=2x-2y over the parabola–line region.

Solution

Step 1 — Set up Green’s theorem

With P=x2+y2P=x^2+y^2 and Q=x2−y2Q=x^2-y^2, Green’s theorem gives

∮CP dx+Q dy=∬R(∂Q∂x−∂P∂y) dA.\oint_C P\,dx+Q\,dy=\iint_R\Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big)\,dA. ∂Q∂x=2x,∂P∂y=2y⟹∂Q∂x−∂P∂y=2x−2y.\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.

Step 2 — Describe the region

R={1≤y≤2−x2}R=\{1\le y\le 2-x^2\}: bounded below by the horizontal line y=1y=1 and above by the parabola y=2−x2y=2-x^2. They meet where 2−x2=1⇒x2=1⇒x=±12-x^2=1\Rightarrow x^2=1\Rightarrow x=\pm1. So x∈[−1,1]x\in[-1,1] and, for each xx, yy runs from 11 to 2−x22-x^2.

Step 3 — Evaluate the double integral

∬R(2x−2y) dA=∫−11∫12−x2(2x−2y) dy dx.\iint_R(2x-2y)\,dA=\int_{-1}^{1}\int_{1}^{2-x^2}(2x-2y)\,dy\,dx.

Inner integral:

∫12−x2(2x−2y) dy=[2xy−y2]12−x2=2x(2−x2)−(2−x2)2−(2x−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).

The 2x(2−x2)−2x2x(2-x^2)-2x part is odd in xx and integrates to 00 over [−1,1][-1,1]. The remaining even part is

−(2−x2)2+1=−(4−4x2+x4)+1=−3+4x2−x4.-(2-x^2)^2+1=-(4-4x^2+x^4)+1=-3+4x^2-x^4. ∫−11(−3+4x2−x4) dx=2∫01(−3+4x2−x4) dx=2[−3x+4x33−x55]01=2(−3+43−15).\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). =2⋅−45+20−315=2⋅−2815=−5615.=2\cdot\frac{-45+20-3}{15}=2\cdot\frac{-28}{15}=-\frac{56}{15}.

Answer

  ∮CF⃗⋅dr⃗=−5615.  \boxed{\;\oint_C\vec F\cdot d\vec r=-\frac{56}{15}.\;}
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