← 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 13 yrs · Read the full method →

Question

Using Green’s theorem, evaluate the CF(r)dr\displaystyle\int_C \vec F(\vec r)\cdot d\vec r counterclockwise where F(r)=(x2+y2)i^+(x2y2)j^\vec F(\vec r)=(x^2+y^2)\hat i+(x^2-y^2)\hat j and dr=dxi^+dyj^d\vec r=dx\,\hat i+dy\,\hat j and the curve CC is the boundary of the region R={(x,y)1y2x2}R=\{(x,y)\mid 1\le y\le 2-x^2\}.

Technique

Green’s theorem Pdx+Qdy=(QxPy)dA\oint P\,dx+Q\,dy=\iint(Q_x-P_y)\,dA with QxPy=2x2yQ_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=x2y2Q=x^2-y^2, Green’s theorem gives

CPdx+Qdy=R(QxPy)dA.\oint_C P\,dx+Q\,dy=\iint_R\Big(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\Big)\,dA. Qx=2x,Py=2yQxPy=2x2y.\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={1y2x2}R=\{1\le y\le 2-x^2\}: bounded below by the horizontal line y=1y=1 and above by the parabola y=2x2y=2-x^2. They meet where 2x2=1x2=1x=±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 2x22-x^2.

Step 3 — Evaluate the double integral

R(2x2y)dA=1112x2(2x2y)dydx.\iint_R(2x-2y)\,dA=\int_{-1}^{1}\int_{1}^{2-x^2}(2x-2y)\,dy\,dx.

Inner integral:

12x2(2x2y)dy=[2xyy2]12x2=2x(2x2)(2x2)2(2x1).\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(2x2)2x2x(2-x^2)-2x part is odd in xx and integrates to 00 over [1,1][-1,1]. The remaining even part is

(2x2)2+1=(44x2+x4)+1=3+4x2x4.-(2-x^2)^2+1=-(4-4x^2+x^4)+1=-3+4x^2-x^4. 11(3+4x2x4)dx=201(3+4x2x4)dx=2[3x+4x33x55]01=2(3+4315).\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). =245+20315=22815=5615.=2\cdot\frac{-45+20-3}{15}=2\cdot\frac{-28}{15}=-\frac{56}{15}.

Answer

  CFdr=5615.  \boxed{\;\oint_C\vec F\cdot d\vec r=-\frac{56}{15}.\;}
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