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punniya.v

Answered question

2022-01-19

please provide answer for this question

nick1337

Expert2022-02-01Added 777 answers

Green Theorem:

$\oint_{c} M (x, y) d x + N (x, y) d y = \int \int_{R} (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} d x d y)$

$\vec{F} (x, y) = 2 x (x + y) \vec{i} + (x^{2} + x y + y^{2}) \vec{j}$

$\vec{r} = x \vec{i} + y \vec{j}$

$d \vec{r} = d x \vec{i} + d y \vec{j}$

$\vec{F} \cdot d \vec{r} = 2 x (x, y), N (x, y) = x^{2} + x y + y^{2}$

Now

$M = 2 x^{2} + 2 x y$

$\frac{\partial M}{\partial y} = 0 + 2 x = 2 x$

$\frac{\partial N}{\partial x} = 2 x + y + 0 = 2 x + y$

$\int \int_{R} (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} d x d y = \int \int_{R} (2 x + y - 2 x) d x d y$

$= \int \int_{R} (y) d x d y$

Let us construct horizontal strip which moves one end on $x = 0$ and other end on $x = 1$ and this strip moves $y = 0$ to $y = 1$

$\int \int_{R} (y) d x d y = \int_{y = 0}^{1} (\int_{y = 0}^{1} y d x) d y$

$= \int_{y = 0}^{1} (x y)_{0}^{1} d y$

$= \int_{y = 0}^{1} y d y$

$= (\frac{y^{2}}{2})_{0}^{1}$

$\frac{1}{2} (1 - 0)$

$= \frac{1}{2}$

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