I want to calculate $\int_{C}^{} \overrightarrow{F}\overrightarrow{dS}$ for $\overrightarrow{F}(x,y)=(x^2y,x^3y^2)$ and $C$ is the closed curve formed by $y=4$ and $y=x^2$
How can I calculate it? What does $\int_{C}^{} \overrightarrow{F}\overrightarrow{dS}$ mean? I just started to study vector analysis and I have to solve other problems similar to this one. I would really appreciate if someone could solve this one so I can use it as an example to solve the others.
You can calculate the contour integral by parametrization of the curve with a parameter. You can write the path in vectorized form in $\vec{r}(t)=\vec{r}_1(t)+\vec{r}_2(t)$, since $C_1: r_1(t)=\begin{pmatrix}t \\ t^2\end{pmatrix}$ and $C_2: r_2(t)=\begin{pmatrix} t \\ 4\end{pmatrix}$. The parameters value is $t\in[-2,2]$ because of the parabola bordered with $y=4$.
Since the contour integral is additive, you can calculate $\oint \vec{F}\cdot\mathrm{d}\vec{s}=\int_{C_1} \vec{F}\cdot\mathrm{d}\vec{s}+ \int_{C_2} \vec{F}\cdot\mathrm{d}\vec{s}$.
For every Integral you have to do a parametrization of $\vec{F}$. So you replace every $x$ with the $x$-component of the path $\vec{r}_i$ you are integrating along. In the first case, you get $\vec{F}(\vec{r}_1(t))=\begin{pmatrix}t^2 \cdot t\\ t^3\cdot t^4\end{pmatrix}$. $\mathrm{d}\vec{s}$ is the differential path element. You get it by $\mathrm{d}\vec{s}=\frac{\partial \vec{r}}{\partial t}\mathrm{d}t$.
Afterwards you calculate the scalar product $\vec{F}(\vec{r}(t))\cdot\mathrm{d}\vec{s}$ and you get one integral in one line. You can calculate the integral within the limits of $t$. You have to do these steps for $\vec{r}_1(t)$ and $\vec{r}_2(t)$ to get the closed path.