## Wednesday, June 6, 2007

### Colorful! Topic: Calculus.

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