Problem 18 - Calculus: Early Transcendental Functions

Question

Use Green's Theorem to evaluate the indicated line integral. \(\oint_{C} 4 y d x+y^{3} d y+z^{4} d z,\) where \(C\) is \(x^{2}+y^{2}=4\) in the plane \(z=0\)

Step-by-Step Solution

Answer

\(-16\pi\)

Step 1: Identify the form of line integral

We can see that the line integral appears in the format \(\oint_C P dx + Q dy + R dz\), where \(P = 4y\), \(Q = y^3\) and \(R = z^4\). But since C is in the plane \(z = 0\), R does not contribute to the value of the line integral and can be left out. Our integral thus reduces to \(\oint_{C} 4 y d x+y^{3} d y\).

Step 2: Apply Green's Theorem

Green's theorem states that a line integral over a simple closed curve C is equal to the double integral over the region D enclosed by C. Therefore, we can write our line integral as a double integral over the circle of radius 2 using Green's theorem: \(\oint_{C} 4 y d x + y^3 dy = \int\int_D (dQ/dx - dP/dy) dA\), where \(dQ/dx\) is the derivative of \(Q = y^3\) with respect to x and \(dP/dy\) is the derivative of \(P = 4y\) with respect to y.

Step 3: Evaluate dQ/dx and dP/dy

Calculate the partial derivatives to find \(dQ/dx = 0\) and \(dP/dy = 4\). Then substitute these into the double integral expression: \(\int\int_D (0 - 4) dA\).

Step 4: Evaluate the double integral

The double integral over the region D becomes just integral of negative 4 over the region D. Because D is a circle of radius 2, its area is \(\pi r^2 = 4\pi\). So multiply negative 4 by this area to get \(-16\pi\).

Key Concepts

Line Integral

A line integral extends the concept of definite integrals to functions along a curve or path in a vector field. It is particularly useful in physics and engineering to calculate quantities like work done by a force field along a path.

For the given exercise, we have the line integral:

• \( \oint_{C} 4 y \, dx + y^3 \, dy + z^4 \, dz \)

This integral is examined over the path \( C \), specifically over a circle where \( x^2 + y^2 = 4 \) in the plane \( z = 0 \). For such a path, the component \( z^4 \, dz \) diminishes because \( z = 0 \) simplifies it to:

• \( \oint_{C} 4 y \, dx + y^3 \, dy \)

This reduction clarifies our focus to the interaction between the \( P \) and \( Q \) components of the vector field within the plane.

In summary, a line integral like this measures the accumulated effect (or circulation) of the vector field components \( 4y \) and \( y^3 \) along the path \( C \).

Double Integral

Green's Theorem facilitates the transformation of line integrals into double integrals over a region \( D \) enclosed by the path \( C \). This conversion is vital in simplifying computations.

According to Green's Theorem:

• \( \oint_{C} P \, dx + Q \, dy = \int\int_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \)

In this context, \( D \) is the interior of the circle \( x^2 + y^2 = 4 \). The transformation delivers a double integral over the circular region.

Computing the double integral involves determining the area of the circle, which simplifies the complexity of our calculation. The given formula captures the integral over the entire area adding structural depth to the line integral's inherent path-based approach.
Embracing double integrals through Green's Theorem grants an efficient method to transitioning from path-oriented issues to ones concerning area, rendering complexity into simplicity. In this scenario, with the area of the circle being \( 4\pi \) (as calculated by \( \pi r^2 \)), the double integral resolves to \( -16\pi \).

Partial Derivatives

Partial derivatives capture the change of a function concerning one variable while keeping others constant. They form the core of transforming between differential and integral calculus forms, especially in multi-variable contexts.

Within Green's Theorem application, we require:

• \( \frac{\partial Q}{\partial x} = 0 \)
• \( \frac{\partial P}{\partial y} = 4 \)

Here, applying partial differentiation simplifies the transformation of the line integral into a clean, computable double integral. The derivative of \( Q = y^3 \) with respect to \( x \) equals zero, emphasizing how the vector field's effect on the \( x \)-axis does not change. Conversely, \( \frac{\partial P}{\partial y} \) gives a constant \( 4 \), portraying uniform contribution across the \( y \)-axis.

Partial derivatives function as an essential tool in rendering multivariable calculus manageable, leading to clearer integral forms in problem solving. Thus, these derivatives are indispensable in translating dynamics between curves and areas to analyze fields effectively.