Use Greens theorem to calculate the line integral H C x 2ydx + (y − 3)dy,, where C is a rectangle with vertices (1, 1), (4, 1), (4, 5) and (1, 5), oriented counterclockwise.
Given,∮C[{M(x)+2y}dx−x2dy]Let m=M(x)+2y and N=−x2∴∂∂y(m)=2 and ∂∂x(N)=−2xGiven, \oint_C[\{M(x)+2y\}dx- x^2dy ]\\ Let \space m= M(x) +2y \space and \space N=-x^2\\ \therefore \frac{\partial }{\partial y}(m)=2 \space and \space \frac{\partial }{\partial x}(N)=-2xGiven,∮C[{M(x)+2y}dx−x2dy]Let m=M(x)+2y and N=−x2∴∂y∂(m)=2 and ∂x∂(N)=−2x
Using the Green theorem
∴∮C[{M(x)+2y}dx−x2dy]=∬D[−2x−2]dA=∫15∫14(−2x−2)dxdy=∫15∫14(−2x−2)dxdy=∫15(−21)dy=−84\therefore \oint_C[\{M(x)+2y\}dx- x^2dy ]=\iint_D[-2x-2]dA\\ =\intop_1^5\int_1^4(-2x-2)dxdy\\ =\int _1^5\int _1^4\left(-2x-2\right)dxdy\\ =\int _1^5\left(-21\right)dy\\ =-84∴∮C[{M(x)+2y}dx−x2dy]=∬D[−2x−2]dA=∫15∫14(−2x−2)dxdy=∫15∫14(−2x−2)dxdy=∫15(−21)dy=−84
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