Question

For a simple closed curve \(C\) in the plane show by Green's theorem that the area inclosed is $$ A=\frac{1}{2} \oint_{C}(x d y-y d x) $$

Short answer

By Green's theorem and the choice of \(L = -y\) and \(M = x\), the area is \(A = \frac{1}{2} \oint_{C} (x dy - y dx)\).

Step-by-step solution

Understand Green's Theorem

Recall Green's theorem, which states that for a positively oriented, simple closed curve \(C\) in the plane and a region \(D\) enclosed by \(C\), \[\oint_{C} L dx + M dy = \iint_{D} \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) dA \]

Choose Functions L and M

Choose \(L = -y\) and \(M = x\). This choice will help in deriving the formula for the area. Now rewrite Green's theorem using these functions: \[ \oint_{C} -y dx + x dy = \iint_{D} \left( \frac{\partial x}{\partial x} - \frac{\partial (-y)}{\partial y} \right) dA \]

Simplify the Area Integral

Now compute the partial derivatives: \( \frac{\partial x}{\partial x} = 1 \) and \( \frac{\partial (-y)}{\partial y} = -1 \). Substitute these values back into the integral: \[ \oint_{C} -y dx + x dy = \iint_{D} \left( 1 - (-1) \right) dA = \iint_{D} 2 dA \]

Express Area in Terms of the Integral

Observe that the integral of \(2 dA\) over the region \(D\) is simply twice the area of \(D\): \[ \iint_{D} 2 dA = 2 A \]. Divide by 2 to solve for the area, \(A\): \[ \oint_{C} (-y dx + x dy) = 2A \Rightarrow A = \frac{1}{2} \oint_{C} (x dy - y dx) \]

Key concepts

Line Integrals

In calculus, line integrals are an essential concept used to evaluate the integral of a function along a curve. They help calculate quantities such as work done by a force field on a particle moving along a path, or fluid flow across a boundary.
The line integral of a vector field \(\mathbf{F}(x,y) = M(x,y)\hat{i} + L(x,y)\hat{j}\) along a curve \(C\) is given by:
\[ \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (M dx + L dy) \]
Here, \(dx\) and \(dy\) represent infinitesimal changes along the x and y directions, respectively.
Green's Theorem connects line integrals with double integrals, providing a powerful tool for converting boundary integrals into area integrals.
Using Green's theorem makes it easier to compute areas enclosed by simple closed curves.

Area Computation

Green's Theorem is instrumental in calculating the area enclosed by a curve. It allows for a direct computation from the line integral around the boundary.
Recall the area formula derived using Green's theorem:
\[ A = \frac{1}{2} \oint_{C} (x dy - y dx) \]
This formula computes the area enclosed by the curve \(C\) by integrating around its boundary.
The beauty of this approach lies in its simplicity, especially for irregular shapes, as it reduces a complex area integral into a straightforward boundary integral.
By carefully choosing functions \(L = -y\) and \(M = x\), Green's theorem transforms a line integral into an easily solvable double integral.

Simple Closed Curve

A simple closed curve is a curve in the plane that does not intersect itself and forms a closed loop. Examples include circles, ellipses, and the boundaries of complex polygons.
Simple closed curves are essential in the application of Green's Theorem, as the theorem requires the curve to be positively oriented and closed.
For any such curve \(C\) enclosing a region \(D\), Green's theorem states:
\[ \oint_{C} (L dx + M dy) = \iint_{D} \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) dA \]
This foundational theorem converts complex line integrals around the boundary into double integrals over the region, simplifying calculations of various physical properties, such as area.