Answer
\[
\boxed{\oint_C (x^2 - y^2)\,dx + x\,dy = 9\pi.}
\]
Work Step by Step
\[
\oint_C (x^2 - y^2)\,dx + x\,dy,
\]
where \(C\) is the circle \(x^2 + y^2 = 9\) oriented counterclockwise.
\textbf{Solution:}
By Green’s Theorem,
\[
\oint_C P\,dx + Q\,dy = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA
\]
where \(P = x^2 - y^2\) and \(Q = x\).
Then
\[
\frac{\partial Q}{\partial x} = 1, \quad \frac{\partial P}{\partial y} = -2y,
\]
so
\[
\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 + 2y.
\]
Hence
\[
\oint_C (x^2 - y^2)\,dx + x\,dy = \iint_R (1 + 2y)\, dA.
\]
Convert to polar coordinates:
\[
x = r\cos\theta, \quad y = r\sin\theta, \quad dA = r\,dr\,d\theta,
\]
and since \(C\) is \(x^2 + y^2 = 9\), we have \(0 \le r \le 3\).
\[
\iint_R (1 + 2y)\,dA = \int_0^{2\pi} \int_0^3 (1 + 2r\sin\theta)\, r\, dr\, d\theta.
\]
Compute the inner integral:
\[
\int_0^3 (1 + 2r\sin\theta)\, r\, dr = \int_0^3 (r + 2r^2\sin\theta)\, dr
= \frac{9}{2} + 18\sin\theta.
\]
\[
\int_0^3 (1 + 2r\sin\theta)\, r\, dr = \int_0^3 (r + 2r^2\sin\theta)\, dr=\int_0^3 rdr+2\sin\theta\ \int_0^3 r^2dr=\vert \frac{r^2}{2} \vert_{0}^{3}+2\sin{\theta} \vert \frac{r^3}{3}\vert_{0}^{3}=\frac{9}{2}-0+2\sin{\theta}(\frac{3^3}{3}-0)=\frac{9}{2}+2\sin{\theta}(\frac{27}{3})=\frac{9}{2}+2(\sin{\theta})(9)
= \frac{9}{2} + 18\sin\theta.
\]
Now integrate with respect to \(\theta\):
\[
\int_0^{2\pi} \left(\frac{9}{2} + 18\sin\theta\right) d\theta=\frac{9}{2}\int_0^{2\pi} d\theta+18\int_0^{2\pi}\sin{\theta} d\theta=\frac{9}{2}\vert \theta \vert_{0}^{2\pi}-18\vert\cos{\theta}\vert_{0}^{2\pi}=\frac{9}{2}(2\pi)-18(\cos{0}-\cos{2\pi})
=9\pi -18(1-1)=9\pi-18(0)=9\pi-0= 9\pi.
\]
\[
\boxed{\oint_C (x^2 - y^2)\,dx + x\,dy = 9\pi.}
\]
