Question
The work done by the force ${\vec{F}}=(x^{2}-y^{2})\hat{i}+(x+y)\hat{j}$ in moving a particle along the closed path $C$ containing the curves $x+y=0,x^{2}+y^{2}=16$ and $y = x$ in the first and fourth quadrant is?
The path given is 
I tried to compute it using Green's Theorem
Here $$ M= x²-y²$$ and $$ N= x+y$$ We have , $$ \oint_C{F}.ds =\int \int ( \frac{ \partial{N}}{\partial{x}} - \frac{\partial{M}}{\partial{y}})dA $$ So, $$\oint_C{\vec{F}}.ds = \int \int (1+2y) dA $$ Now converting the problem into polar coordinates by substituting $x= rcos\theta$ and $y= rsin \theta $ And substituting the limitts of $r$ from $0$ to $4$ and $\theta$ from $-\pi/4$ to $\pi/4$ So $$ \oint_C{\vec{F}}.ds = \int_{\theta = -π/4}^{π/4} \int_{r=0}^4(1+2r sin \theta ) r drd\theta$$
$$ = \int_{\theta = -π/4}^{π/4} (\frac{r^2}{2}+\frac{2r³sin \theta }{3})|_0^4 d \theta $$
$$= \int_{\theta = -π/4}^{π/4} 8 + \frac{128}{3} sin \theta d\theta $$
$$ = (8 \theta - \frac{128 cos \theta}{3} )|_{-π/4}^{π/4} $$
Which on solving gave me $4π$ which is not matching any of the options given.There is another approach too, i.e to calculate the line integral of the three curves one by one which is too long process and even, this question has to be solved in less than a minute in the exam.
Why I'm getting it wrong and is this the right approach for this question?
Kindly help. Thank You.
You are doing it correctly. If you want to save time, interpret the double integral using symmetry. $\iint_R 1\,dA$ gives the area of the region $R$, which is $\frac14(\pi\cdot 4^2) = 4\pi$, and $\iint_R y\,dA$ gives $0$ by symmetry (since the region is symmetric about $y=0$). Your answer is correct.
The problem is poorly phrased, as we are not explicitly told the orientation of the curve. But it sounds like you are meant to do traverse those curves in the order given, so you inferred the correct orientation.