UPSC 2015 Maths Optional Paper 1 Q8c — Step-by-Step Solution

12 marks · Section B

Line integrals · Vector Analysis · asked 8× in 13 yrs · Read the full method →

Question

Evaluate $\displaystyle\int_C e^{-x}(\sin y\,dx+\cos y\,dy)$ , where $C$ is the rectangle with vertices $(0,0),\,(\pi,0),\,(\pi,\pi/2),\,(0,\pi/2)$ .

Technique

Green’s theorem reduces the line integral to a double integral over the rectangle; integrand separates as a product, evaluate each factor.

Solution

Strategy. The integrand is of the form $P\,dx+Q\,dy$ with $P=e^{-x}\sin y$ , $Q=e^{-x}\cos y$ . The contour is a closed curve — use Green’s theorem.

Assume $C$ is traversed counterclockwise (standard orientation).

Step 1 — Green’s theorem

\oint_C P\,dx+Q\,dy=\iint_R\!\!\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)\,dA,

where $R$ is the interior of the rectangle.

Step 2 — Compute partials

$\dfrac{\partial Q}{\partial x}=\dfrac{\partial}{\partial x}(e^{-x}\cos y)=-e^{-x}\cos y$ .

$\dfrac{\partial P}{\partial y}=\dfrac{\partial}{\partial y}(e^{-x}\sin y)=e^{-x}\cos y$ .

\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}=-e^{-x}\cos y-e^{-x}\cos y=-2e^{-x}\cos y.

Step 3 — Double integral

I=\iint_R -2e^{-x}\cos y\,dA=-2\int_0^\pi\!\!\int_0^{\pi/2} e^{-x}\cos y\,dy\,dx.

Separable:

I=-2\left(\int_0^\pi e^{-x}\,dx\right)\left(\int_0^{\pi/2}\cos y\,dy\right).

\int_0^\pi e^{-x}\,dx=[-e^{-x}]_0^\pi=-e^{-\pi}+1=1-e^{-\pi}.

\int_0^{\pi/2}\cos y\,dy=[\sin y]_0^{\pi/2}=1.

I=-2(1-e^{-\pi})\cdot 1=-2(1-e^{-\pi})=2(e^{-\pi}-1).

Answer

\boxed{\;I=2(e^{-\pi}-1).\;}