Problem 190 - Calculus Volume 3

Question

Use Green's theorem to evaluate line integral \(\oint_{C}\left(y-\ln \left(x^{2}+y^{2}\right)\right) d x+\left(2 \arctan \frac{y}{x}\right) d y,\) where \(C\) is the positively oriented circle \((x-2)^{2}+(y-3)^{2}=1\).

Step-by-Step Solution

Answer

The value of the line integral is \(-\pi\).

Step 1: Identify Components

Green's Theorem relates a line integral around a simple curve to a double integral over the region it encloses. It is given by the formula \[ \oint_{C} (P \, dx + Q \, dy) = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \]. In this problem, identify \(P = y - \ln(x^2 + y^2)\) and \(Q = 2 \arctan \left( \frac{y}{x} \right)\).

Step 2: Compute Partial Derivatives

Calculate the partial derivatives needed for Green's Theorem. Calculate \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\). For \(Q = 2 \arctan \left( \frac{y}{x} \right)\), use the chain rule to find \(\frac{\partial Q}{\partial x} = -\frac{2y}{x^2 + y^2}\). For \(P = y - \ln(x^2 + y^2)\), \(\frac{\partial P}{\partial y} = 1 - \frac{2y}{x^2 + y^2}\).

Step 3: Plug into Green's Theorem

Substitute the partial derivatives into Green's Theorem: \[ \iint_{R} \left( \frac{-2y}{x^2 + y^2} - \left(1 - \frac{2y}{x^2 + y^2}\right) \right) \, dA = \iint_{R} \left(-1\right) \, dA \].

Step 4: Calculate the Area of the Region

The region \(R\) is the area enclosed by the circle \((x-2)^2 + (y-3)^2 = 1\). The area of a circle with radius 1 is \(\pi\).

Step 5: Evaluate the Double Integral

Since the integrand is constant \(-1\), the double integral becomes \(\iint_{R} -1 \, dA = -\int_{R} 1 \, dA = -\pi\).

Step 6: Conclude the Calculation

Substitute the result of the area calculation into the formula: the line integral \( \oint_{C}\) is \(-\pi\).

Key Concepts

Line Integral

Line integrals are an essential component when dealing with vector fields along a curve. They help calculate the total influence of a vector field across a path. Imagine walking along a path with varying wind forces. The line integral computes how much work the wind does on you during the walk.
Here's why they matter:

• Line integrals can help explore how connected points in a field are.
• They allow us to measure the cumulative effect of a field along a curve.
• In applications, they translate real-world problems like calculating work done by a force.

By breaking a path into small intervals, we can sum up tiny effects across each interval and calculate the line integral using calculus. Understanding line integrals is crucial in applying Green's theorem, as shown in the original exercise!

Partial Derivatives

Partial derivatives are about seeing how a function changes as one of several variables changes, keeping others constant. They're foundational in multivariable calculus.
Think of it as:

• Understanding terrain slopes where you only change direction in one axis, leaving others untouched.
• The change in function solely due to changes in one input, while ignoring the rest.

In Green's theorem, partial derivatives help differentiate components of a vector field. For example, from the original exercise, partial derivatives like \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \) quantify how specific vector field components change.

Circle Area Calculation

The area calculation of a circle is a straightforward yet powerful concept. In math and physics, knowing the area helps quantify the space enclosed by circular boundaries.
Here's why it's important:

• Calculates space interactions, such as chemical reaction areas or target distances.
• A fundamental skill for further integrals involving circular paths.

In the original solution, the circle described as \((x-2)^2 + (y-3)^2 = 1\) represents a circle of radius 1 centered at (2, 3). The area is simply \(\pi\) as the formula \( \pi r^2 \) applies, with a radius of 1.

Double Integral

The double integral enables calculating the sum of values across two-dimensional regions. It's a method to integrate over an area, opposed to just along a line.
Think of it as:

• Generalizing single integrals to account for surfaces spanning on a two-dimensional plane.
• Summing values like mass, temperature, or density, distributed across surfaces.

In applying Green's Theorem, the double integral \( \iint_{R} (-1) \, dA \) is evaluated over the cir clear region \(R\) in the original exercise. By understanding double integrals, we can convert complex problems like calculating the line integral into manageable computations.