Inactive Tutor answered • 07/20/24
To compute the flux of the vector field ⟨x^5,−xy^4⟩ out of the given rectangle using Green's Theorem, we can use the following steps:
Green's Theorem relates a line integral around a simple closed curve C to a double integral over the plane region D bounded by C. For a vector field F=⟨P(x,y),Q(x,y)⟩ Green's Theorem states:
∮CF⋅nds=∬D(∂x∂Q−∂y∂P)dA
where n is the outward-pointing unit normal vector to the curve C
Given: F=⟨P,Q⟩=⟨x5,−xy4⟩,we need to compute:
∂Q/∂x=∂(−xy4)∂x=−y4
∂P/∂y=∂y∂(x5)=0.
So the double integral becomes:
∬D(∂Q/∂x−∂P/∂y) dA=∬D(−y4−0) dA=∬D−y4 dA
D is the rectangle with vertices (0,0), (4,0), (4,1), and (0,1).
Setting up the integral:
∬D−y4dA= -∫04 ∫01y4dydx
Compute the integral and you are done!
![{\displaystyle {\begin{aligned}&{\underline {\text{Problem}}}\colon \\&{\text{Using Green's Theorem, compute the flux of the vector field }}\langle P,Q\rangle =\langle x^{5},-xy^{4}\rangle {\text{ out of the rectangle }}R\\&{\text{with vertices }}(0,0),(4,0),(4,1),{\text{ and }}(0,1).\\\\&{\underline {\text{Solution}}}\colon \\&{\text{If }}(x(t),y(t)){\text{ describes the (counterclockwise) boundary }}\partial R,{\text{ the tangent is }}\langle {\dot {x}},{\dot {y}}\rangle ,{\text{ and the outward }}\\&{\text{unit normal vector is }}{\tfrac {1}{\sqrt {{\dot {x}}^{2}+{\dot {y}}^{2}}}}\langle {\dot {y}},-{\dot {x}}\rangle ={\tfrac {dt}{ds}}\langle {\dot {y}},-{\dot {x}}\rangle ={\tfrac {1}{ds}}\langle dy,-dx\rangle .\\&\qquad {\begin{aligned}{\text{Flux}}=\int _{\partial R}\langle P,Q\rangle \cdot {\tfrac {1}{ds}}\langle dy,-dx\rangle \,ds&=\int _{\partial R}P\,dy-Q\,dx\\&=\int _{R}(P_{x}+Q_{y})\,dA&&{\text{[Green's Theorem]}}\\&=\int _{0}^{4}\int _{0}^{1}{\bigl (}5x^{4}-4xy^{3}{\bigr )}\,dy\,dx\\&=\int _{0}^{4}{\bigl (}5x^{4}-x{\bigr )}\,dx=1016\end{aligned}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/290a12606202cc36f3066a2e71e68286f09e2f0b)
