Find the flux through the boundary of the rectangle 0≤x≤3,0≤y≤4for fluid flowing along the vector field ⟨x5+2,ycos(3x)⟩.The Flux Form of Green's Theorem:Suppose that →F(x,y)=⟨P(x,y),Q(x,y)⟩ is a vector field over a domain in R2. Then the divergence of →F is defined to bediv→F=∂P/∂x+∂Q/∂yIf R is a region in the domain of →F whose boundary is the counterclockwise simple closed curve C, then the flux form of Green's theorem states that the flux integral∮C→F⋅→ndr=∮C(Pdy−Qdx)]can be computed as follows:∮C→F⋅→ndr=∬Rdiv→FdA.



Answer :

The flux through the boundary of the rectangle 0≤x≤3,0≤y≤4for fluid flowing along the vector field ⟨x5+2,ycos(3x)⟩ is 45.

To use the flux form of Green's theorem, we need to compute the divergence of the vector field ⟨x^5+2, ycos(3x)⟩. The divergence is given by the expression

div ⟨[tex]x^5+2[/tex], ycos(3x)⟩ = ∂P/∂x + ∂Q/∂y

where P(x,y) = [tex]x^5+2[/tex] and Q(x,y) = ycos(3x). Substituting these expressions, we have

div ⟨[tex]x^5+2[/tex] ycos(3x)⟩ =[tex](5x^4)[/tex] [tex]+ 0 = 5x^4[/tex]

The region R is the rectangle 0 ≤ x ≤ 3, 0 ≤ y ≤ 4, so the flux integral can be computed as follows:

∮C →F⋅→ndr = ∬R div ⟨[tex]x^5+2[/tex], ycos(3x)⟩ dA

[tex]= \int \int R (5x^4) dA\\= \int_0^3 ∫0^4 (5x^4) dy dx\\= \int_0^3 (5x^4) ∫0^4 dy dx\\= \int_0^3 (5x^4)(4) dx\\= (5/5) \int_0^3 (x^4) dx\\= (5/5) [(x^5/5)]^3_0\\= (5/5)(3^5/5 - 0)\\= 243/5[/tex]

Therefore, the flux through the boundary of the rectangle is 243/5.

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