**Theorem**: (Green's Theorem) Let $ R $ be a simply connected plane region whose boundary is a simple, closed, piecewise smooth curve $ C $ oriented counterclockwise. If $ f(x, y) $ and $ g(x, y) $ are continuous and have continuous first partial derivatives on some open set containing $ R $, then

$ \displaystyle \oint_C f(x, y) dx + g(x, y) dy = \int_R \int \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right) dA $.

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**Problem**: Evaluate $ \displaystyle \oint_C x^2y dx + (y+xy^2) dy $, where $ C $ is the boundary of the region enclosed by $ y = x^2 $ and $ x = y^2 $.

**Solution**: First, verify that this region satisfies all of the requirements for Green's Theorem - indeed, it does. So we may apply the theorem with $ f(x, y) = x^2y $ and $ g(x, y) = y+xy^2 $. From these, we have $ \frac{\partial g}{\partial x} = y^2 $ and $ \frac{\partial f}{\partial y} = x^2 $. Then we obtain

$ \displaystyle \oint_C x^2y dx + (y+xy^2) dy = \int_R \int (y^2-x^2) dA $.

But clearly this integral over the region $ R $ can be represented as $ \displaystyle \int_0^1 \int_{x^2}^{\sqrt{x}} (y^2-x^2) dy dx $, so it remains a matter of calculation to get the answer. First, we evaluate the inner integral to get

$ \displaystyle \int_0^1 \int_{x^2}^{\sqrt{x}} (y^2-x^2) dy dx = \int_0^1 \left[\frac{y^3}{3}-x^2y\right]_{x^2}^{\sqrt{x}} dx = \int_0^1 \left(\frac{x^{3/2}}{3}-x^{5/2}-\frac{x^6}{3}+x^4\right)dx $.

Then finally we have

$ \displaystyle \int_0^1 \left(\frac{x^{3/2}}{3}-x^{5/2}-\frac{x^6}{3}+x^4\right)dx = \left[\frac{2x^{5/2}}{15}-\frac{2x^{7/2}}{7}-\frac{x^7}{21}+\frac{x^5}{5}\right]_0^1 = \frac{2}{15}-\frac{2}{7}-\frac{1}{21}+\frac{1}{5} = 0 $.

QED.

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Comment: To me, Green's Theorem is a very interesting result. It's not at all obvious that a line integral along the boundary of a region is equivalent to an integral of some partial derivatives in the region itself. A simplified proof of the result can be obtained by proving that

$ \displaystyle \oint_C f(x, y) dx = -\int_R \int \frac{\partial f}{\partial y} dA $ and $ \displaystyle \oint_C g(x, y) dy = \int_R \int \frac{\partial g}{\partial x} dA $.

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Practice Problem: Let $ R $ be a plane region with area $ A $ whose boundary is a piecewise smooth simple closed curve $ C $. Show that the centroid $ (\overline{x}, \overline{y}) $ of $ R $ is given by

$ \displaystyle \overline{x} = \frac{1}{2A} \oint_C x^2 dy $ and $ \displaystyle \overline{y} = -\frac{1}{2A} \oint_C y^2 dx $.

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