Maybe nice

Nice problem! Please do post more problems on vector calculus.

$\int_{\phi}\omega=0$ by Stokes' Theorem since $d\omega=0$ for the given 1-form $\omega$ . Note that this form is defined throughout the disc $M$ enclosed by the circle $\partial M$ . (Recall that Stokes' Theorem states that $\int_{\partial{M}}\omega=\int_{M}d\omega$ )

Using the given parameterization, we see that $\int_{\gamma}\omega=\int_{0}^{2\pi}(\sin^2t+\cos^2t)dt=2\pi$ . Here we cannot use Stokes' Theorem directly since the form is undefined at the origin.

Thus the stated equation is $\boxed{False}$ .

I recognized the integrand as the imaginary part of $\frac{1}{x+iy}$ . Since this function is analytic inside and on the closed loop $\phi$ , the value of the integral is zero (by the Cauchy Integral Theorem). It is not hard to show that the other integral is equal to $2\pi$ (whether you simplify the integrand to 1 or just apply the Cauchy Integral Formula). In other words, the integrals are not equal.