91Ó°ÊÓ

The given expression is .${âˆ\circledR }_{c}ydx+zdy+xdz$

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines it magnitude

Given .${âˆ\circledR }_{c}ydx+zdy+xdz$

The surface is given$x+y=2,{\text{x}}^{\text{2}}{\text{+y}}^{\text{2}}{\text{+z}}^{\text{2}}\text{= 2}\left(\text{x + y}\right)$

The equation of the given curve c is found below.

$x^2+y^2+z^2=2(x+y)\text{and}x+y=2.$

It is the curve of intersection of the sphere $x^2+y^2+z^2=2(x+y)$and the Plane

$x+y=2.$

Write in another form.

$x^2+y^2+z^2=4\left(sphere\right)$and$x+y=2.\left(Plane\right)$

Solve further.

$\begin{array}{rcl}{âˆ\circledR }_{c}ydx+zdy+xdz& =& {∫}_c(yi+zj+xk)·d\mathrm{γ}\\ & =& {âˆ\neg }_{S}curl(4i+zj+xk)·\hat{h}ds\end{array}$

Use Stokes’s theorem where S is the surface of the given circle and C is the boundary.

$Cuxl(yi+zj+xk)=\left|\begin{array}{ccc}i& j& k\\ dldx& dldy& ∂l_{∂z}\\ y& z& x\end{array}\right|$

Solve further.

$\begin{array}{rcl}\mathrm{ϕ}& =& x+4-2\\ ∇d& =& i+j\\ |∇\mathrm{ϕ}|& =& \sqrt{1+1}\\ & =& \sqrt{2}\end{array}$

Therefore, it becomes as shown below.

$\begin{array}{rcl}\hat{h}& =& \frac{∇\mathrm{ϕ}}{|∇\mathrm{ϕ}|}\\ & =& \frac{i+j}{{\hat{v}}^2}\end{array}$

$\begin{array}{rcl}curlF·n& =& -(i+j+k)·\frac{(i+j)}{\sqrt{2}}\\ & =& -\sqrt{2}\end{array}$

Find the given line integral.

$\begin{array}{rcl}âˆ\neg -\sqrt{2}dS& =& -\sqrt{2}×\text{(Area of the circle}C\\ & =& -\sqrt{2}\left(\mathrm{Ï€}·{\left(2\right)}^2\right)\\ & =& -4\sqrt{2}\mathrm{Ï€}\end{array}$