Use Green's Theorem to compute the value of the line integral $\int_{\gamma}ydx + x^2dy$

Question

Use Green's Theorem to compute the value of the line integral $\int_{\gamma}ydx + x^2dy$, where $\gamma$ is the circle given by $g(t) = (\cos t, \sin t), 0 \leq t \leq 2\pi$

I have

$$\begin{align} \int_{\gamma} Fdx + Gdy = \int_{0}^{2\pi}(\sin t + \cos^2 t)dt = \bigg[\cos t - \sin^2t \bigg]_{0}^{2\pi} = (1-0) - (1-0) = 0 \end{align}$$

I get my answer as 0, but my textbook says the answer is $-\pi$.

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I don't think I used the fact that $\gamma$ is a circle correctly, but not really sure what went wrong. Any help would be appreciated, thanks

Answer 1

$d(ydx+x^2dy)=dy\wedge dx+2xdx\wedge dy=dx\wedge dy$. Green theorem implies that $\int_\gamma ydx+x^2dy=\int_{B(0,1)}dx\wedge dy=\pi$ the later is the area of the ball of radius $1$ in the plane.

Answer 2

By Green's theorem, we have

$$ \iint_{\text{disk}} 2x-1 dA$$

The integral of the first term is $0$ because it is odd, and the region has symmetry across the line $x=0$. So we are left with an integral that equals negative the area of the circle, $-\pi$

Answer 3

Made a change to polar coordinates, $x= r \cos \theta; y=r \sin \theta $ with $r∈[0,1]$ and $\theta ∈ [0,2 \pi]$ and as $dA=dxdy$,the Jacobian of the transformation $x=rcosθ$; $y=rsinθ$ is $r$ so $dxdy=rdrdθ$

\begin{align} \int_0^{2\pi} \int_0^1 (2r \cos \theta-1) \, r dr \, d\theta = & \int_0^{2\pi} \Big[\frac{2}{3} r^3 \cos \theta-\frac{r^2}{2}\Big]_0^1 \, d\theta\\ = & \int_0^{2\pi} \Big[\frac{2}{3} \cos \theta-\frac{1}{2}\Big]_0^1 \, d\theta\\ = & \Big[\frac{2}{3} \sin \theta- \frac{1}{2} \theta\ \Big]_0^1=- \pi\\ \end{align}