Question

Prove Theorem 9.7 .2 by completing the following argument. According to Green's theorem in the plane, if \(C\) is a sufficiently smooth simple closed curve, and if \(F\) and \(G\) are continuous and have continuous first partial derivatives, then $$ \int_{C}[F(x, y) d y-G(x, y) d x]=\iint_{R}\left[F_{x}(x, y)+G_{y}(x, y)\right] d A $$ where \(C\) is traversed counterclockwise and \(R\) is the region enclosed by \(C .\) Assume that \(x=\phi(t), y=\psi(t)\) is a solution of the system ( 15) that is periodic with period \(T\). Let \(C\) be the closed curve given by \(x=\phi(t), y=\psi(t)\) for \(0 \leq t \leq T\). Show that for this curve the line integral is zero. Then show that the conclusion of Theorem 9.7 .2 must follow.

Short answer

#tag_title#Step 2: Show that the line integral is zero#tag_content# To show that the line integral is zero, we need to show that the given parameterized solution satisfies: $$ \int_{0}^{T}\left[F(\phi(t), \psi(t))\frac{d\psi(t)}{dt}-G(\phi(t), \psi(t))\frac{d\phi(t)}{dt}\right] dt = 0 $$ We know that the parameterized solution $(x, y) = (\phi(t), \psi(t))$ is a periodic solution to the system. Thus, we have: $$ \begin{cases} \frac{d\phi}{dt} = F(\phi(t), \psi(t)) \\ \frac{d\psi}{dt} = G(\phi(t), \psi(t)) \end{cases} $$ Substituting this into the integral, we get: $$ \int_{0}^{T}\left[F(\phi(t), \psi(t))G(\phi(t), \psi(t))-G(\phi(t), \psi(t))F(\phi(t), \psi(t))\right] dt $$ Simplifying the integrand gives: $$ \int_{0}^{T}\left[F(\phi(t), \psi(t))G(\phi(t), \psi(t))-F(\phi(t), \psi(t))G(\phi(t), \psi(t))\right] dt = 0 $$ Since the integrand is identically zero, the line integral is indeed zero. #tag_title#Step 3: Show the conclusion of Theorem 9.7.2 follows#tag_content# Since the line integral of our parameterized periodic solution is zero, we can now attempt to show how this demonstrates the conclusion of Theorem 9.7.2. Theorem 9.7.2 states that if a sufficiently smooth simple closed curve C and two continuous functions, F and G, are given with continuous first partial derivatives, then the line integral of F and G along curve C is zero. We have already shown that the line integral of our parameterized periodic solution is zero. By Green's theorem, the double integral over the region R enclosed by curve C is zero as well: $$ \iint_{R}\left[F_{x}(x, y)+G_{y}(x, y)\right] d A = 0 $$ The zero line integral implies that the contribution to the total integral from all points inside the region R balances out. This means that there must be some equal and opposite contributions within the region, proving the conclusion of Theorem 9.7.2.

Step-by-step solution

Compute the line integral using Green's theorem

Using Green's theorem, the integral along the curve C can be written as: $$ \int_{C}[F(x, y) d y-G(x, y) d x]=\iint_{R}\left[F_{x}(x, y)+G_{y}(x, y)\right] d A $$ We are given a parameterized solution for the system \((x, y) = (\phi(t), \psi(t))\). So, let's transform the integral in terms of t: $$ \int_{0}^{T}\left[F(\phi(t), \psi(t))\frac{d\psi(t)}{dt}-G(\phi(t), \psi(t))\frac{d\phi(t)}{dt}\right] dt $$

Key concepts

Line Integral

A line integral is a technique to integrate a function along a curve. Imagine tracing a path through a field where each point has a value. The line integral accumulates the values along this path. It's key in physics for calculating work done by a force along a path.
In the context of Green's Theorem, a line integral relates to the circulation of a vector field around a closed curve. You evaluate it over a curve defined by two functions, say \(F(x, y)\) and \(G(x, y)\). The integral then becomes: \[ \int_{C}[F(x, y) d y-G(x, y) d x] \] Here, "C" is a curve in the plane, and "dy" and "dx" represent small changes along y and x. Green's Theorem connects this line integral with a region's area, simplifying many calculations.

Parameterized Solution

A parameterized solution lets us express a curve using parameters. Typically, these parameters can be a single variable, such as \(t\), to represent multiple variables \( (x, y)\). This is especially useful in calculus and physics when describing complex paths.
For example, a curve might be represented as \(x = \phi(t)\) and \(y = \psi(t)\). Here, \(\phi(t)\) and \(\psi(t)\) describe how "x" and "y" change over time. It turns a two-variable problem into a single-variable one, which simplifies analysis and integrations. \[ \int_{0}^{T}\left[F(\phi(t), \psi(t))\frac{d\psi(t)}{dt}-G(\phi(t), \psi(t))\frac{d\phi(t)}{dt}\right] dt \] This specific transformation allows us to solve the line integral along the parameterized curve.

Partial Derivatives

Partial derivatives measure how a function changes as one variable changes, while keeping others constant. In multivariable calculus, they are foundational for analyzing functions with several variables. They're like a flashlight showing how one dimension varies at a time.
For a function \(F(x, y)\), the partial derivative with respect to "x" would be \(F_x(x, y)\) and with respect to "y" would be \(F_y(x, y)\).
These derivatives form a key part of Green's Theorem. They help bridge changes along the curve to changes over the area: \[ \iint_{R}\left[F_x(x, y) + G_y(x, y)\right] dA \] Understanding partial derivatives enables analysis over curves and surfaces, crucial in physics and engineering applications.

Closed Curve

A closed curve is a loop that starts and ends at the same point without crossing itself. Think of it as a rubber band layed on a plane. Closed curves are fundamental for applying Green's Theorem.
They enclose an area (like a circle or an ellipse) that makes it possible to explore the relationship between a line integral around the curve and a double integral over the region it encloses.
In Green's Theorem, such closed curves allow us to derive the equivalence between the circulation around "C" and the area "R" it outlines: \[ \int_{C}[...] = \iint_{R}[...] \] Traversing the curve counterclockwise is standard, maintaining consistency in calculations, particularly in physics.