# Show that the line integral int_{-1,2}^{3,1}(y^(2)+2xy)dx+(x^(2)+2xy)dy is independent of path and evaluate it?

Apr 8, 2018

14

#### Explanation:

The differential

$M \mathrm{dx} + N \mathrm{dy} = \left({y}^{2} + 2 x y\right) \mathrm{dx} + \left({x}^{2} + 2 x y\right) \mathrm{dy}$
$q \quad = {y}^{2} \mathrm{dx} + 2 x y \mathrm{dy} + {x}^{2} \mathrm{dy} + 2 x y \mathrm{dx}$
$q \quad = {y}^{2} \mathrm{dx} + x d \left({y}^{2}\right) + {x}^{2} \mathrm{dy} + y d \left({x}^{2}\right)$
$q \quad = d \left(x {y}^{2} + {x}^{2} y\right)$

Since this is an exact differential, the integral depends only on the endpoints and not on the path.

Thus

${\int}_{\left(- 1 , 2\right)}^{\left(3 , 1\right)} \left[{y}^{2} + 2 x y\right] \mathrm{dx} + \left[{x}^{2} + 2 x y\right] \mathrm{dy}$
$q \quad = {\int}_{\left(- 1 , 2\right)}^{\left(3 , 1\right)} d \left(x {y}^{2} + {x}^{2} y\right) = {\left(x {y}^{2} + {x}^{2} y\right)}_{\left(- 1 , 2\right)}^{\left(3 , 1\right)}$
$q \quad = \left(3 \times {1}^{2} + {3}^{2} \times 1\right) - \left({\left(- 1\right)}^{2} \times 2 + \left(- 1\right) \times {2}^{2}\right)$
$q \quad = 12 - \left(- 2\right) = 14$

Apr 8, 2018

${\int}_{\left(- 1 , 2\right)}^{\left(3 , 1\right)} \setminus \left({y}^{2} + 2 x y\right) \setminus \mathrm{dx} + \left({x}^{2} + 2 x y\right) \setminus \mathrm{dy} = 14$

#### Explanation:

We wish to show that the line integral:

$S = {\int}_{\left(- 1 , 2\right)}^{\left(3 , 1\right)} \setminus \left({y}^{2} + 2 x y\right) \setminus \mathrm{dx} + \left({x}^{2} + 2 x y\right) \setminus \mathrm{dy}$

is independent of path and evaluate it.

We can utilize a fundamental theorem of multivariable calculus:

If $M \left(x y , y\right)$ and $N \left(x , y\right)$ have continuous first partial derivatives on a simply connected region $D$ then line integral over a path $C$

${\int}_{C} \setminus M \left(x , y\right) \setminus \mathrm{dx} + N \left(x , y\right) \setminus \mathrm{dy}$

is independent of the path of $C$ in $D$ if and only if

$\frac{\partial M}{\partial y} = \left(\partial N\right) \left(/ \partial x\right)$

For the given line integral, we can define

 {: (M(x,y) = y^2 + 2xy, => (partial M)/(partial y) = 2y+2x), (N(x,y) = x^2+2xy, => (partial B)/(partial x) = 2x+2y) :}

Then observing that $M$ and $N$ satisfy the above conditions we conclude that the line integral is independent of path. QED.

In order to evaluate the line integral we utilize two further theorem:

If $\boldsymbol{\underline{F}} \left(x , y\right) = M \left(x , y\right) \boldsymbol{\underline{\hat{i}}} + N \left(x y . y\right) \boldsymbol{\underline{\hat{j}}}$ is continuous on an open connected region then the line integral ${\int}_{C} \boldsymbol{\underline{F}} \cdot d \boldsymbol{\underline{r}}$ is independent of path if any only if $\boldsymbol{\underline{F}} \left(x , y\right) = \boldsymbol{\nabla} f \left(x , y\right)$ for some function $f$

and:

If $\boldsymbol{\underline{F}} \left(x , y\right) = M \left(x , y\right) \boldsymbol{\underline{\hat{i}}} + N \left(x , y\right) \boldsymbol{\underline{\hat{j}}}$ is continuous on an open connected region $D$ and $C$ is a piecewise smooth curve in $D$ with endpoints $A \left({x}_{1} , {y}_{1}\right)$ and $B \left({x}_{2} , {y}_{2}\right)$ then:

${\int}_{C} \setminus M \left(x , y\right) \setminus \mathrm{dx} + N \left(x , y\right) \setminus \mathrm{dy} = {\int}_{\left({x}_{1} , {y}_{1}\right)}^{\left({x}_{2} , {y}_{2}\right)} \boldsymbol{\underline{F}} \cdot d \boldsymbol{\underline{r}}$
$\text{ } = {\left[f \left(x , y\right)\right]}_{\left({x}_{1} , {y}_{1}\right)}^{\left({x}_{2} , {y}_{2}\right)}$
$\text{ } = f \left({x}_{2} , {y}_{2}\right) - f \left({x}_{1} , {y}_{1}\right)$

Thus noting that we have established that the line integral is independent of path we apply the first theorem and seek a function $f$ such that

$\boldsymbol{\nabla} f \left(x , y\right) = M \left(x , y\right) \boldsymbol{\underline{\hat{i}}} + N \left(x y . y\right) \boldsymbol{\underline{\hat{j}}}$
$\text{ } = \left({y}^{2} + 2 x y\right) \boldsymbol{\underline{\hat{i}}} + N \left({x}^{2} + 2 x y\right) \boldsymbol{\underline{\hat{j}}}$

Thus we require (by integrating) that

$\frac{\partial f}{\partial x} = {y}^{2} + 2 x y \implies f = x {y}^{2} + {x}^{2} y + u \left(y\right)$
$\frac{\partial f}{\partial y} = {x}^{2} + 2 x y \implies f = {x}^{2} y + x {y}^{2} + v \left(x\right)$

Where $u \left(y\right)$ is an arbitrary function of $y$ alone and $v \left(x\right)$ is an arbitrary function of $x$ alone (the equivalent of the constants of integration for single variable calculus).

And so , as these evaluate to the same function $f$, we require that

$x {y}^{2} + {x}^{2} y + u \left(y\right) = {x}^{2} y + x {y}^{2} + v \left(x\right)$

$\therefore \textcolor{b l u e}{\cancel{x {y}^{2}}} + \textcolor{g r e e n}{\cancel{{x}^{2} y}} + u \left(y\right) = \textcolor{g r e e n}{\cancel{{x}^{2} y}} + \textcolor{b l u e}{\cancel{x {y}^{2}}} + v \left(x\right)$

$\therefore u \left(y\right) = v \left(x\right)$

And so we can arbitrarily choose $u \left(y\right) = v \left(x\right) = 0$, giving us the desired function, $f$:

$f \left(x , y\right) = x {y}^{2} + {x}^{2} y$

We can now readily evaluate the line integral using the last theorem:

$S = {\int}_{\left(- 1 , 2\right)}^{\left(3 , 1\right)} \setminus \left({y}^{2} + 2 x y\right) \setminus \mathrm{dx} + \left({x}^{2} + 2 x y\right) \setminus \mathrm{dy}$

$\setminus \setminus = {\left[x {y}^{2} + {x}^{2} y\right]}_{\left(- 1 , 2\right)}^{\left(3 , 1\right)}$

$\setminus \setminus = \left\{\left(3\right) {\left(1\right)}^{2} + {\left(3\right)}^{2} \left(1\right)\right\} - \left\{\left(- 1\right) {\left(2\right)}^{2} + {\left(- 1\right)}^{2} \left(2\right)\right\}$

$\setminus \setminus = \left(3 + 9\right) - \left(- 4 + 2\right)$
$\setminus \setminus = 12 - \left(- 2\right)$
$\setminus \setminus = 14$