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Question Number 189057 by neinhaltsieger369 last updated on 11/Mar/23

$$$$$$\mathrm{Help}!$$$$$$$$\mathrm{Evaluate}\mathrm{the}\mathrm{following}\mathrm{integral}\mathrm{usings}\mathrm{Green}\mathrm{theorem}:$$$$$$$$\oint 4\mathrm{xy}d\mathrm{x}+{\mathrm{x}}^{2}d\mathrm{y}$$$$$$$$\mathrm{Where}C\mathrm{is}\mathrm{the}\mathrm{square}\mathrm{of}\mathrm{vertices}(0,0),(0,2),(2,0)\mathrm{and}(2,2).$$$$$$

Commented by mr W last updated on 11/Mar/23

$$Nein!$$$$Halt!$$$$Sieger!$$

Commented by neinhaltsieger369 last updated on 11/Mar/23

$$Help!$$

Commented by mr W last updated on 11/Mar/23

$$Ihaveexpectedthat{you}^{\prime }llcontinue$$$$with$$$$\mathit{H}\mathit{i}\mathit{l}\mathit{f}\mathit{e}!$$

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. <<Chat GPT>>

Commented by neinhaltsieger369 last updated on 12/Mar/23

$$\mathrm{Thank}\mathrm{you}$$

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