Answer
$-\dfrac{\pi}{4}$
Work Step by Step
Green's Theorem states that: $\int_C fdx+g dy=\iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA$
Here, we have $f=x^2y ; g =0$
Now, $\dfrac{\partial g}{\partial x} =0$ and $\dfrac{\partial f}{\partial y}=x^2$
Therefore,
$ \iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA=\iint_D -x^2 \ dx \ dy$
Now, in polar co-ordinates, we have
$\int_C x^2y \ dx=\int_0^{2 \pi}\int_0^{1} -r^2(r) \ d\theta \ dr \\=\int_0^{2 \pi}\int_0^{1} -r^3 \cos^2 \theta \ d\theta \ dr \\=[\int_0^{2 \pi} \cos^2 \theta \ d\theta] \times [-\int_0^{1} r^3 \ dr] \\=[\dfrac{\theta}{2}+\dfrac{\sin 2 \theta}{4}]_0^{2 \pi} \times [\dfrac{-r^4}{4}]_0^1$
Thus, $\int_C x^2y \ dx =-\dfrac{\pi}{4}$
