# Using Green's Theorem to evaluate the area enclosed by a line and curve

Question Use Green's Theorem to evaluate the line integral $$\oint_{C} (x^2 + xy) dx + x y^2 dy$$ where C is the boundary region trapped by the line $y = 2x$ and the curve $y = -2x^2$.

Solution $$\oint_{C} (P\, dx+Q\, dy) = \iint_D dx\,dy \: \left ( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)$$

$$\iint (y^2 - x) dx dy$$ I know I need to get boundaries for the double integral but I don't know how to go about getting them.

• have you tried x from -1 to 0 and y from 2x to -2x^2? – Olba12 Apr 3 '16 at 13:54
• Have you tried plotting it out? – Kenny Lau Apr 3 '16 at 14:28
• Plotted it out and and used your boundaries Olba12. I got an answer of 19/42. Thanks – Ncartoon Apr 3 '16 at 16:07

That is a picture of a graph of those two functions. I imagine you could draw that.

$$\int\int_D y^2 - x \space dydx$$

Now $$D= \{ (x,y) \in \mathbb{R}^2 | \space 2x \le y \le -2x^2 \space , -1 \le x \le 0 \}$$

Why $0, -1$ because $2x=-2x^2 \space => \space 2x^2+2x=0$

Solutions for that are $-1,0$.

Ok we now have the borders so we can caluclate.

$$\int_{-1}^{0}\int_{2x}^{-2x^2} y^2-x \space dydx =\int_{-1}^{0} (\frac{y^3}{3} -xy)|_{2x}^{-2x^2}dx=\int_{-1}^{0} -\frac{8x^6}{3}-\frac{8x^3}{3}-x(-2x^2-2x) dx$$

$$=\int_{-1}^{0} -\frac{8}{3}(x^6+x^3)+2(x^3+x^2)dx=-\frac{8}{3}(\frac{x^7}{7}+\frac{x^4}{4})|_{-1}^{0} + 2(\frac{x^4}{4}+\frac{x^3}{3})|_{-1}^{0}$$

$$=-\frac{8}{3}(\frac{1}{7}-\frac{1}{4})+2(-\frac{1}{4}+\frac{1}{3})=\frac{2}{7} +\frac{1}{6}=\frac{19}{42}$$