Question Number 189057 by neinhaltsieger369 last updated on 11/Mar/23
$$\: \\ $$$$\:\mathrm{Help}! \\ $$$$\: \\ $$$$\:\mathrm{Evaluate}\:\:\mathrm{the}\:\:\mathrm{following}\:\:\mathrm{integral}\:\:\mathrm{usings}\:\:\mathrm{Green}\:\mathrm{theorem}: \\ $$$$\: \\ $$$$\:\oint\mathrm{4xy}{d}\mathrm{x}\:\:+\:\:\mathrm{x}^{\mathrm{2}} {d}\mathrm{y} \\ $$$$\: \\ $$$$\:\mathrm{Where}\:\:{C}\:\:\mathrm{is}\:\:\mathrm{the}\:\:\mathrm{square}\:\:\mathrm{of}\:\:\mathrm{vertices}\:\:\left(\mathrm{0},\mathrm{0}\right),\:\left(\mathrm{0},\mathrm{2}\right),\:\left(\mathrm{2},\mathrm{0}\right)\:\:\mathrm{and}\:\:\left(\mathrm{2},\mathrm{2}\right). \\ $$$$\: \\ $$
Commented by mr W last updated on 11/Mar/23
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Commented by neinhaltsieger369 last updated on 11/Mar/23
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Commented by mr W last updated on 11/Mar/23
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Commented by Ar Brandon last updated on 11/Mar/23
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Commented by Ar Brandon last updated on 11/Mar/23
To evaluate the given integral using Green's Theorem, we need to find the curl of the given vector field F = (4xy, x^2).
So, ∂F₂/∂x = 2x and ∂F₁/∂y = 4x. Thus, curl(F) = ∂F₂/∂x - ∂F₁/∂y = -2x.
Now, we can apply Green's Theorem which states that for a vector field F = (P, Q) and a simple closed curve C enclosing a region R, the line integral of F along C is equal to the double integral of curl(F) over the region R, i.e.,
∮C (P dx + Q dy) = ∬R ( ∂Q/∂x - ∂P/∂y ) dA
In our case, P = 4xy and Q = x^2, so we have:
∮C (4xy dx + x^2 dy) = ∬R (-2x) dA
The region R enclosed by the curve C is a square with vertices (0,0), (0,2), (2,0) and (2,2).
We can express the double integral over R as an iterated integral as follows:
∬R (-2x) dA = ∫[0,2] ∫[0,2] (-2x) dy dx
Integrating with respect to y first, we get:
∫[0,2] (-2x) dy = -2xy ∣[0,2] = -4x
Then, we integrate with respect to x:
∫[0,2] -4x dx = -2x^2 ∣[0,2] = -8
Therefore, the line integral of F along C is:
∮C (4xy dx + x^2 dy) = ∬R (-2x) dA = -8
Hence, the value of the given line integral using Green's Theorem is -8.
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Commented by neinhaltsieger369 last updated on 12/Mar/23
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