Evaluating the Line Integral using Green's Theorem

We need to evaluate the line integral $\int_{C} \left[ (xy+y^{2})dx + x^{2}dy \right]$, where the curve C is bounded by $y=x$ and $y=x^{2}$. This type of line integral around a closed curve can be efficiently evaluated using Green's Theorem.

Green's Theorem states that for a positively oriented, piecewise smooth, simple closed curve C in the plane, and D the connected region bounded by C, the following equality holds:
$ \oint_{C} (L \, dx + M \, dy) = \iint_{D} \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) dA $

Applying Green's Theorem

In this problem, we identify $L = xy + y^2$ and $M = x^2$. The region D is enclosed by the curves $y=x$ and $y=x^2$. These curves intersect when $x = x^2$, which yields $x=0$ and $x=1$. Thus, the region D is defined for $0 \le x \le 1$ and $x^2 \le y \le x$.

First, we calculate the partial derivatives:

• $ \frac{\partial M}{\partial x} = \frac{\partial}{\partial x}(x^2) = 2x $
• $ \frac{\partial L}{\partial y} = \frac{\partial}{\partial y}(xy + y^2) = x + 2y $

Next, we find the difference required for Green's Theorem:

$ \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} = 2x - (x + 2y) = x - 2y $

Calculating the Double Integral

Now, we set up the double integral over the region D:

$ \iint_{D} (x - 2y) \, dA = \int_{0}^{1} \int_{x^2}^{x} (x - 2y) \, dy \, dx $

We evaluate the inner integral with respect to $y$:

$ \int_{x^2}^{x} (x - 2y) \, dy = \left[ xy - y^2 \right]_{y=x^2}^{y=x} $

$ = (x(x) - x^2) - (x(x^2) - (x^2)^2) $

$ = (x^2 - x^2) - (x^3 - x^4) $

$ = 0 - x^3 + x^4 = x^4 - x^3 $

Finally, we evaluate the outer integral with respect to $x$:

$ \int_{0}^{1} (x^4 - x^3) \, dx = \left[ \frac{x^5}{5} - \frac{x^4}{4} \right]_{0}^{1} $

$ = \left( \frac{1^5}{5} - \frac{1^4}{4} \right) - \left( \frac{0^5}{5} - \frac{0^4}{4} \right) $

$ = \frac{1}{5} - \frac{1}{4} - 0 $

$ = \frac{4 - 5}{20} = -\frac{1}{20} $

The value of the line integral is $-\frac{1}{20}$.