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Verify Green's theorem for $\oint(3x^2 - 8y^2)dx + (4y - 6xy)dy$ where c is boundary of the region defined by x = 0, y = 0 & x+ y = 1
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By Green's theorem

$\int\limits_{c} Pdx + Qdy = \int\int\limits_{R}\bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\bigg)dx dy......(A)$

Here $p = 3x^2 - 8y^2 \\ Q = 4y - 6xy$

$\frac{\partial p}{\partial y} = -1by \\ \frac{\partial Q}{\partial x} = -6y$

Consider L.H.S. of Green's theorem

(a) Along C1: y = 0 dy = 0 x varies to 0 to 1

$\int\limits_{c_{1}}Pdx + Qdy = \int\limits_{x = 0}^{1}3x^2 dx = 3\bigg[\frac{x^3}{3}\bigg]_{0}^{1} = 1$

(b) Along C2: y = 1-x, dy = -dx x varies 1 to 0

$\int\limits_{c_{2}}Pdx + Qdy = \int\limits_{x = 1}^{0}3x^2 - 8(1 - x)^2dx - 4[(1 - x) - 6x(1 - x)]dx \\ \therefore \int\limits_{c_{2}}Pdx + Qdy = \bigg[3\frac{x^3}{3} + 8\frac{(1- x)^3}{3} - 4\frac{(1- x)^2}{2}(-1) + 6 \bigg(\frac{x^2}{2} - \frac{x^3}{3}\bigg)\bigg]_{1}^{0} \\ \therefore \int\limits_{c_{2}}Pdx + Qdy = \bigg[0 + \frac{8}{3} + 2 - 1 - \bigg(6*\frac{1}{6}\bigg)\bigg] \\ \therefore \int\limits_{c_{2}}Pdx + Qdy = \bigg[\frac{8}{3} + 2 - 2\bigg] = \frac{8}{3}$

(c) Along C3: x = 0, dx = 0, y varies 1 to 0

$\int\limits_{C_{3}}Pdx + Qdy = \int\limits_{y = 1}^{0} 0 + 4ydy \\ \therefore \int\limits_{C_{3}}Pdx + Qdy = 2[y^2]_{1}^{0} = -2 * 1 \\ \therefore \int\limits_{C_{3}}Pdx + Qdy = -2$

$\therefore \int\limits_{C}Pdx + Qdy = \int\limits_{C_{1}}Pdx + Qdy + \int\limits_{C_{2}}Pdx + Qdy + \int\limits_{C_{3}}Pdx + Qdy = 1 + 8/3 - 2 \\ \int\limits_{C}Pdx + Qdy = 5/3 ..................................................(1)$

Now R.H.S. of equation (A): y: 0 to 1 - x & x: 0 to 1

$\int\int\limits_{R}\bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\bigg)dxdy = \int\limits_{x = 0}^{1}\int\limits_{y = 0}^{1 - x}(-6y)y + 16dxdy \\ \int\int\limits_{R}\bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\bigg)dxdy = \int\limits_{x = 0}^{1}\int\limits_{y = 0}^{1 - x} 10ydxdy \\ \int\int\limits_{R}\bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\bigg)dxdy = \int\limits_{x = 0}^{1}\int\limits_{y = 0}^{1 - x} 10\bigg(\frac{y^2}{2}\bigg)_{0}^{1 - x}dx \\ \int\int\limits_{R}\bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\bigg)dxdy = \int\limits_{x = 0}^{1} \frac{10}{2} (1 - x^2)dx \\ \int\int\limits_{R}\bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\bigg)dxdy = \int\limits_{x = 0}^{1}5\bigg[\frac{(1 - x)^3}{3} (-1)\bigg] \\ \int\int\limits_{R}\bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\bigg)dxdy = -5/3.............................................(2)$

$\therefore$ From (1) and (2) Green's theorem is verified i.e.

$\int\limits_{c} Pdx + Qdy = \int\int\limits_{R}\bigg(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\bigg)dxdy$