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.
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,curl F=∣∣i∂x∂x2j∂y∂xyk∂z∂0∣∣=yk,
curlF⃗⋅n⃗=y,\text{curl}\vec F\cdot \vec n=y,curlF⋅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.S∬curlF⋅ndS=∫01∫01ydydx+∫01∫−10ydydx=∫012y2∣01dx+∫012y2∣−10dx=∫0121dx+∫01(−21)dx=0.
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment