Problem 29 - Thomas Calculus

Question

Let \(C\) be the boundary of a region on which Green's Theorem holds. Use Green's Theorem to calculate $$\begin{array}{l}{\text { a. } \oint_{C} f(x) d x+g(y) d y} \\ {\text { b. } \oint \text { ky } d x+h x d y \quad(k \text { and } h \text { constants) }}\end{array}$$

Step-by-Step Solution

Answer

a. 0; b. \\(h-k\\) times the area of \\(R\\).

Step 1: Understanding Green's Theorem

Green's Theorem relates a line integral around a simple closed curve \(C\) to a double integral over the plane region \(R\) bounded by \(C\). It is stated as \( \oint_{C} (P \, dx + Q \, dy) = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \), where \(P\) and \(Q\) are functions of \(x\) and \(y\). We'll apply this theorem to both parts of the exercise.

Step 2: Solve Part (a) Using Green's Theorem

For part (a), the line integral is \( \oint_{C} f(x) \, dx + g(y) \, dy \). Here, \(P(x, y) = f(x)\) and \(Q(x, y) = g(y)\). To use Green’s Theorem, compute the partial derivatives: \( \frac{\partial Q}{\partial x} = 0 \) and \( \frac{\partial P}{\partial y} = 0 \). Thus, the double integral becomes \( \iint_{R} \left( 0 - 0 \right) \, dA = 0 \). Therefore, \( \oint_{C} f(x) \, dx + g(y) \, dy = 0 \).

Step 3: Solve Part (b) Using Green's Theorem

For part (b), the line integral is \( \oint_{C} ky \, dx + h x \, dy \) with constants \(k\) and \(h\). Here, \(P(x, y) = ky\) and \(Q(x, y) = hx\). Compute the partial derivatives: \( \frac{\partial Q}{\partial x} = h \) and \( \frac{\partial P}{\partial y} = k \). Therefore, \( \iint_{R} \left( h - k \right) \, dA \) becomes \((h-k) \iint_{R} \, dA\), which equals \((h-k) \times \text{{Area}}(R)\). Thus, \( \oint_{C} ky \, dx + h x \, dy = (h-k) \times \text{{Area}}(R) \).

Key Concepts

Line Integrals

Line integrals are a way of integrating a function along a curve. Imagine you are walking along a path and at every step, you measure something like temperature. To find out the total experience of temperature along the path, you would use a line integral. In mathematical terms, a line integral \( \oint_{C} F \, ds \) sums up the values of a field \( F \) along a curve \( C \).

For a given function consisting of two components, like \( F(x, y) = P(x, y) \, dx + Q(x, y) \, dy \), the line integral calculates the accumulation of the field over path \( C \). It's especially useful in physics and engineering where one needs to understand the circulation and flux of forces along paths.

Double Integrals

Double integrals extend the concept of a single integral to two dimensions. Instead of finding the area under a curve, double integrals calculate the volume under a surface over a given region. Imagine a blanket laid down on a hilly terrain. By using a double integral \(( \iint_{R} f(x, y) \, dA )\) over a region \( R \), you can calculate the total mass of the blanket if the function \( f(x, y) \) represents density.

With Green’s Theorem, double integrals help translate line integrals into a more manageable area calculation when dealing with closed curves. You effectively switch from measuring around the boundary to measuring over the area enclosed by the boundary, which can simplify complex integral calculations significantly.

Partial Derivatives

Partial derivatives represent how a function changes as one of several variables changes while keeping the others constant. Think of a surface described by a function of two variables, \( f(x, y) \).

By taking a partial derivative like \( \frac{\partial f}{\partial x} \), you measure how the surface changes as \( x \) increases while \( y \) stays the same. This is like moving along slices of the surface. In the context of Green's Theorem, partial derivatives are crucial. They form the difference \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \) that transforms line integrals into double integrals, connecting the boundary with the region it encloses.

• \( \frac{\partial Q}{\partial x} \) focuses on how the flow changes parallel to the \( x \)-axis.
• \( \frac{\partial P}{\partial y} \) focuses on changes parallel to the \( y \)-axis.

Knowing these values is key to applying the theorem correctly.