Question

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?

Answer 1

14

Explanation:

The differential

`Mdx+Ndy = (y^2+2xy)dx+(x^2+2xy)dy`
`qquad = y^2dx+2xydy+x^2dy+2xydx`
`qquad = y^2dx+x d(y^2)+x^2dy+yd(x^2)`
`qquad =d(xy^2+x^2y)`

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

Thus

`int_{(-1,2)}^{(3,1)}[y^(2)+2xy]dx+[x^(2)+2xy]dy`
`qquad = int_{(-1,2)}^{(3,1)} d(xy^2+x^2y) = (xy^2+x^2y)_{(-1,2)}^{(3,1)}`
`qquad = (3times1^2+3^2times 1)-((-1)^2times 2+(-1)times 2^2)`
`qquad = 12-(-2) = 14`

Answer 2

`int_{(-1,2)}^{(3,1)} \ (y^2 + 2xy) \ dx + (x^2+2xy) \ dy = 14`

Explanation:

We wish to show that the line integral:

`S = int_{(-1,2)}^{(3,1)} \ (y^2 + 2xy) \ dx + (x^2+2xy) \ dy`

is independent of path and evaluate it.

We can utilize a fundamental theorem of multivariable calculus:

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

`int_C \ M(x,y) \ dx + N(x,y) \ dy`

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

`(partial M)/(partial y) = (partial N) (/partial x)`

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 `bb(ulF)(x,y) = M(x,y)bb(ul hat i) + N(xy.y) bb(ul hat j)` is continuous on an open connected region then the line integral `int_C bb(ul F) * d bb(ul r)` is independent of path if any only if `bb ul(F)(x,y) = bb( grad)f(x,y)` for some function `f`

and:

If `bb(ulF)(x,y) = M(x,y)bb(ul hat i) + N(x,y) bb(ul hat j)` is continuous on an open connected region `D` and `C` is a piecewise smooth curve in `D` with endpoints `A(x_1,y_1)` and `B(x_2,y_2)` then:

`int_C \ M(x,y) \ dx + N(x,y) \ dy = int_{(x_1,y_1)}^{(x_2,y_2)} bb(ul F) * d bb(ul r)`
`" " = [ f(x,y) ]_{(x_1,y_1)}^{(x_2,y_2)}`
`" " = f(x_2,y_2) -f(x_1,y_1)`

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

`bb( grad)f(x,y) = M(x,y)bb(ul hat i) + N(xy.y) bb(ul hat j)`
`" " = (y^2 + 2xy)bb(ul hat i) + N(x^2+2xy) bb(ul hat j)`

Thus we require (by integrating) that

`(partial f)/(partial x) = y^2 + 2xy => f =xy^2+x^2y + u(y)`
`(partial f)/(partial y) = x^2 + 2xy => f =x^2y+xy^2 + v(x)`

Where `u(y)` is an arbitrary function of `y` alone and `v(x)` 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

`xy^2+x^2y + u(y) =x^2y+xy^2 + v(x)`

`:. color(blue)cancel(xy^2) + color(green)cancel(x^2y) + u(y) = color(green)cancel(x^2y) + color(blue)cancel(xy^2) + v(x)`

`:. u(y) = v(x)`

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

`f(x,y) = xy^2+x^2y`

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

`S = int_{(-1,2)}^{(3,1)} \ (y^2 + 2xy) \ dx + (x^2+2xy) \ dy`

`\ \ = [xy^2+x^2y]_{(-1,2)}^{(3,1)}`

`\ \ = {(3)(1)^2+(3)^2(1)} - {(-1)(2)^2+(-1)^2(2)}`

`\ \ = (3+9)-(-4+2)`
`\ \ = 12-(-2)`
`\ \ = 14`