# 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?

##### 2 Answers

14

#### Explanation:

The differential

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 = 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

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 pathif 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

# 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

And so , as these evaluate to the same function

# 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

# 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#