Chapter 16: Problem 4
In Exercises \(1-4,\) verify the conclusion of Green's Theorem by evaluating both sides of Equations \((3)\) and \((4)\) for the field \(\mathbf{F}=M \mathbf{i}+N \mathbf{j}\) . Take the domains of integration in each case to be the disk \(R : x^{2}+y^{2} \leq a^{2}\) and its bounding circle \(C : \mathbf{r}=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq 2 \pi\) $$\mathbf{F}=-x^{2} y \mathbf{i}+x y^{2} \mathbf{j}$$
Short Answer
Step by step solution
Understand Green's Theorem
Identify M and N
Compute \( \frac{\partial N}{\partial x} \) and \( \frac{\partial M}{\partial y} \)
Express the Double Integral
Evaluate the Double Integral
Evaluate the Line Integral
Confirm Equality of Integrals
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Line Integrals and Green's Theorem
In Green's Theorem, the line integral is considered around a closed curve on a plane. This theorem provides a fascinating link between the line integral around a curve and the double integral over the region bounded by the curve. The equation for Green's Theorem can be written as: \[ \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \]
Here:
- \( C \) is the closed curve.
- \( \mathbf{F} \) is a vector field, which can be written as \( M \mathbf{i} + N \mathbf{j} \).
- \( d\mathbf{r} \) represents a small segment of the curve.
Understanding Double Integrals
In the context of Green's Theorem, we consider a double integral over a disk-shaped region \( R \), specifically evaluating the expression \( (x^2 + y^2) \) when it’s transformed into polar coordinates. The switch to polar coordinates is done because they are simpler to work with for circular domains:\[ \iint_{R} (x^2 + y^2) \, dA = \iint_{R} r^2 \, dr \, d\theta \]
- \( r \) is the radius from the center of the circle.
- \( \theta \) goes from 0 to \( 2\pi \), covering the whole circle.
Exploring Vector Fields
For the vector field \( \mathbf{F} = -x^2 y \mathbf{i} + x y^2 \mathbf{j} \), we break it into its respective \( x \) and \( y \) components:
- \( M = -x^2 y \)
- \( N = x y^2 \)
Vector fields are not just imaginary concepts; they underpin many phenomena in real life, from electrical circuits to fluid dynamics, offering insights into interactions and behaviors at multiple levels.