I'd like to verify stokes theorem for the Manifold given by
$$M = \{(x,y)\in\mathbb{R^2}\vert \: x^2+y^2<25, x\geq0\} \cup\{(x,y)\in\mathbb{R^2}\vert \: x = 0, -5<y<5\}$$
in a vector field: $$F = \left(\begin{array}{c}y\,x^2 \\-x\end{array}\right)$$
Therefor I need to show $$Z_1 = \int_{\partial M}\langle F,T\rangle \mathrm{ds} = Z_2 = \int_M\partial_xF_2-\partial_yF_1\,\mathrm{dx\,dy}$$
For the second one I used $$Z_2 = \int_{\frac{-\pi}{2}}^{\frac{-\pi}{2}} \int_0^5(-2\,r\,\cos(\varphi)-r^2\,\cos^2(\varphi))\,r\,\mathrm{dr\,d\varphi} = -\dfrac{500}{3}-\dfrac{625\,\pi}{8}$$
For the first one as approval I used the following parametrization: $$\begin{align}&\gamma_1 = \left(\begin{array}{c}0\\t\end{array}\right)\quad t\in[-5,5],\quad \gamma_2=\left(\begin{array}{c}5\,\cos(t) \\ 5\,\sin(t)\end{array}\right)\quad t\in[-\frac{\pi}{2}, \frac{\pi}{2}]\\\\
&\text{hoping that:} \quad Z_1 = \int \langle F(\gamma_1(t)),{\gamma_1}'(t)\rangle \,\mathrm{dt}+\int \langle F(\gamma_2(t)),{\gamma_2}'(t)\rangle \,\mathrm{dt} = Z_2 \quad \text{but:}\\\\
&F(\gamma_1(t)) = \left(\begin{array}{c}0\\0\end{array}\right) \Rightarrow \int_{\partial_M}\langle \left(\begin{array}{c}0\\0\end{array}\right), \left(\begin{array}{c}0\\1\end{array}\right)\rangle \,\mathrm{dt} = 0 \\\\
&F(\gamma_2(t)) = \left(\begin{array}{c}5^3\,\sin(t)\,\cos^2(t)\\-5\,\cos(t)\end{array}\right) \Rightarrow \int_{\partial M}\langle \left(\begin{array}{c}5^3\,\sin(t)\,\cos^2(t)\\-5\,\cos(t)\end{array}\right),\left(\begin{array}{c}-5\,\sin(t)\\5\,\cos(t)\end{array}\right)\rangle\,\mathrm{dt}\\\\
&\text{and after evaluating:} \quad Z_1 = -\frac{725\,\pi}{8} \neq Z_2
\end{align}$$
Was I parametrizing the border in a wrong order?
Best Answer
You just have a small mistake in calculation of $Z_2$ otherwise your working is all correct.
$ \displaystyle Z_2 = \int_M (\partial_xF_2-\partial_yF_1)\,dx\,dy$
$\vec F = (yx^2, -x)$
So, $\partial_xF_2-\partial_yF_1 = -1 - x^2$
And, $ \displaystyle Z_2 = \int_{-\pi/2}^{\pi/2} \int_0^5 (-1 - r^2 \cos^2\varphi) \ r \ dr \ d\varphi = - \frac{725 \pi}{8}$