Question

In terms of the parameters \(a\) and \(b\), how is the value of \(\oint_{C}\) ay \(d x+b x d y\) related to the area of the region enclosed by \(C,\) assuming counterclockwise orientation of \(C ?\)

Short answer

Answer: The relationship between the line integral and the area enclosed by the curve C is given by the formula \(\oint_{C} (ay \, dx + bx \, dy) = (b - a)A\).

Step-by-step solution

Understand the Green's theorem

Green's theorem states that for a positively oriented, simple smooth curve C with a plane region D enclosed by C: $$\oint_{C} (P \, dx + Q \, dy) = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \, dA$$ where P and Q are functions of x and y that have continuous first partial derivatives on an open region containing D.

Identify P, Q and find their partial derivatives

In our given integral, \(P(x, y) = ay\) and \(Q(x, y) = bx\). Now we'll find their partial derivatives with respect to x and y: \(\frac{\partial P}{\partial y} = \frac{\partial (ay)}{\partial y} = a\) \(\frac{\partial Q}{\partial x} = \frac{\partial (bx)}{\partial x} = b\)

Apply Green's theorem

Now, we'll use the Green's theorem. According to the theorem: $$\oint_{C} (ay \, dx + bx \, dy) = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \, dA$$ Using our calculated values for the partial derivatives, $$\oint_{C} (ay \, dx + bx \, dy) = \iint_{D} (b - a) \, dA$$

Find the relationship between the integral and the area

We now have a double integral to evaluate. Since the integrand \((b - a)\) is a constant, we can take it out of the integral and write: $$\oint_{C} (ay \, dx + bx \, dy) = (b - a) \iint_{D} dA$$ The remaining integral simply represents the area of the region D enclosed by C, which we denote as A: $$A = \iint_{D} dA$$

Write the final relation

Now we can substitute A back into the equation: $$\oint_{C} (ay \, dx + bx \, dy) = (b - a)A$$ So, the value of the integral \(\oint_{C} ay \, dx + bx \, dy\) is related to the area A of the region enclosed by the curve C by the relation: $$\oint_{C} (ay \, dx + bx \, dy) = (b - a)A$$

Key concepts

Line Integrals

Imagine walking along a curve while pulling a cart: the effort you exert is influenced by the path you take and the properties of the field you're walking through. In mathematics, this concept is represented by line integrals. It's a means to integrate a function over a curve, which can be a path in a field, much like how our cart encounters different ground on its journey.

For our exercise, the line integral is expressed as \(\oint_{C} ay \, dx + bx \, dy\), where \( C \) represents the curve we 'walk' along, and \( ay \, dx + bx \, dy \) represents the 'effort' depending on the position (y or x). Essentially, the line integral helps us measure how much 'work' is done along a certain path in a vector field.

Partial Derivatives

When you focus on variable behavior while holding everything else constant, partial derivatives come into play. They are essentials in fields like economics, where understanding the effect of changing one factor—like the price of a good—on demand, while keeping all other factors constant, can be crucial for analysis.

In the exercise, we identify functions \( P(x, y) = ay \) and \( Q(x, y) = bx \) and take their partial derivatives. These partial derivatives \( \frac{\partial P}{\partial y} = a \) and \( \frac{\partial Q}{\partial x} = b \) represent the rate at which the functions \(P\) and \(Q\) change as \(x\) and \(y\) change individually. This concept is pivotal for applying Green's theorem, which relates the line integral around a closed curve to a double integral over the region it encloses.

Double Integrals

Double integrals extend the concept of an integral to two-dimensional spaces, allowing us to calculate volumes under surfaces, or, as in our exercise, the area of a plane region.

Think of a double integral as a way to add up small pieces of area (\(dA\)) in two dimensions to find the total area (\(A\)). In the context of Green's theorem, after substituting our partial derivatives, we integrate over the entire region \(D\) enclosed by the curve \(C\) to find this area. The result is a multiple of the difference \(b - a\), representing how parameters \(a\) and \(b\) scale the area of the region enclosed by \(C\).

Plane Region Area

The area of a plane region can be thought of like tiling a floor: you calculate the size of the 'tiles' and then sum them up to cover the entire 'floor' or region. In calculus, we use integration to do this mathematically.

By applying Green's theorem in our exercise, we reduce the problem of finding the area enclosed by a curve to a simpler problem: calculating a double integral over that same region. In essence, \(A = \iint_{D} dA\) is the way we express the total area of that 'tiled floor', which leads us to understand that the original line integral over curve \(C\) gives us direct information about this area, scaled by the difference of \(b\) and \(a\).