# Use Stokes' Theorem to compute oint_C 1/2 z^2 dx + (xy)dy+2020dz, where C is the triangle with vertices at(1,0,0),(0,2,0), and (0,0,2) traversed in the order.

Use Stokes' Theorem to compute ${\oint }_{C}\frac{1}{2}{z}^{2}dx+\left(xy\right)dy+2020dz$, where C is the triangle with vertices at(1,0,0),(0,2,0), and (0,0,2) traversed in the order.
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Step 1
The aim is to find ${\oint }_{C}\frac{1}{2}{z}^{2}dx+xydy+2020dz$ where C is the triangle.
(1, 0, 0), (0, 2, 0), (0, 0, 2).
Step 2
Given the stoke’s theorem,

Where S is $\frac{x}{1}+\frac{y}{2}+\frac{z}{2}=1$
Then curl $F=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)i+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)j+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)k$
Here $F=\frac{1}{2}{Z}^{2}i+xyj+2020k$
$curlF=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right)i+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)=zj+yk$
If the surface S is,
$\int {\int }_{S}F.dS=\int {\int }_{D}-P\left(\frac{\partial g}{\partial x}\right)-Q\left(\frac{\partial g}{\partial y}\right)+RdA$
Where $F=Pi+Qj+Rk$.
Where, $z=2-y-2x.$
Then using the stoke's theorem,

Substitute $z=2-y-2x$,
$\int \int CurlF\int {\int }_{D}-2+2x+2y-2x-2ydA$
$=-2\int {\int }_{D}dA$
Hence, the area of the triangle is $\frac{base×height}{2}=\frac{1}{2}$.