91Ó°ÊÓ

Consider the vector field below:

The goal is to calculate the integral${âˆ\neg }_{S}curl\mathbf{F}(x,y,z)·\mathbf{n}dS$, with the following surface$S$definition:

The surface$S$is the region of the unit sphere$x^2+y^2+z^2=1$above the plane$z=\frac{1}{2}$with the normal vector pointing upwards.

A vector field's curl$\mathbf{F}(x,y,z)=F_1(x,y,z)\mathbf{i}+F_2(x,y,z)\mathbf{j}+F_3(x,y,z)\mathbf{k}\mathbf{}$is defined as follows:

After that, the vector field's curl will be,

$$\begin{gathered} curl\mathit{F}(x,y,z)=\left|\begin{array}{ccc}\mathit{i}& \mathit{j}& \mathit{k}\\ \frac{\mathrm{Ï‘}}{\mathrm{Ï‘}x}& \frac{\mathrm{Ï‘}}{\mathrm{Ï‘}y}& \frac{\mathrm{Ï‘}}{\mathrm{Ï‘}z}\\ 5y& 5x& x^2\end{array}\right| \\ =\left[\frac{\mathrm{Ï‘}\left(z^2\right)}{\mathrm{Ï‘}y}-\frac{\mathrm{Ï‘}\left(5x\right)}{\mathrm{Ï‘}z}\right]\mathit{i}-\left[\frac{\mathrm{Ï‘}\left(z^2\right)}{\mathrm{Ï‘}x}-\frac{\mathrm{Ï‘}\left(5y\right)}{\mathrm{Ï‘}z}\right]\mathit{j}+\left[\frac{\mathrm{Ï‘}\left(5x\right)}{\mathrm{Ï‘}x}-\frac{\mathrm{Ï‘}\left(5y\right)}{\mathrm{Ï‘}y}\right]\mathit{k} \\ =\left[0-0\right]\mathit{i}-\left[0-0\right]\mathit{j}+\left[5-5\right]\mathit{k} \\ =0\mathit{i}-0\mathit{j}+0\mathit{k} \\ =\mathbf{0} \end{gathered}$$

The value of $curl\mathbf{F}(x,y,z)·\mathbf{n}\mathbf{}$will then be,

$$\begin{gathered} curl\mathit{F}(x,y,z)·\mathit{n}=\mathbf{0}\mathbf{·}\mathit{n} \\ =0 \end{gathered}$$

The required integral will then be,

$$\begin{gathered} ∫{∫}_{S}curl\mathit{F}(x,y,z)·\mathit{n}dS=∫{∫}_{S}0dS \\ =0 \end{gathered}$$

As a result, the necessary integral is:

${âˆ\neg }_{S}curl\mathbf{F}(x,y,z)·\mathbf{n}dS=0$