Question #238812


1. Verify Stock’s theorem for F~ = x^2i+xyj, where s is the area bonded by

a triangle whose vertices are at (0, 0), (1, −1) and (1, 1), respectively.


Expert's answer

curl F⃗=∣i⃗j⃗k⃗∂∂x∂∂y∂∂zx2xy0∣=yk⃗,\text{curl }\vec F=\begin{vmatrix} \vec i& \vec j & \vec k\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ x^2 & xy& 0 \end{vmatrix}=y\vec k,

curlF⃗⋅n⃗=y,\text{curl}\vec F\cdot \vec n=y,

∬ScurlF⃗⋅n⃗dS=∫01∫01ydydx+∫01∫−10ydydx=∫01y22∣01dx+∫01y22∣−10dx=∫0112dx+∫01(−12)dx=0.\underset{S}{\iint }\text{curl}\vec F\cdot \vec ndS =\int_0^1\int_0^1ydydx+\int_0^1\int_{-1}^0ydydx=\int_0^1\frac{y^2}2|_0^1dx+\int_0^1\frac{y^2}2|_{-1}^0dx=\int_0^1\frac 12dx+\int_0^1(-\frac 12)dx=0.


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