Chapter 15: Problem 10
Use Green's Theorem to evaluate the integral. In each exercise, assume that the curve \(C\) is oriented counterclockwise.\(\oint_{C} x^{2} y d x-y^{2} x d y\), where \(C\) is the boundary of the region in the first quadrant, enclosed between the coordinate axes and the circle \(x^{2}+y^{2}=16\)
Short Answer
Expert verified
The integral evaluates to \(-32\pi\).
Step by step solution
01
Recall Green's Theorem
Green's Theorem relates a line integral around a simple closed curve \( C \) to a double integral over the plane region \( D \) bounded by \( C \). It is stated as follows: \( \oint_{C} (L \, dx + M \, dy) = \iint_{D} \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) \, dA \).
02
Identify L and M
Given the line integral \( \oint_{C} x^{2}y \, dx - y^{2}x \, dy \), identify the functions: \( L = x^{2}y \) and \( M = -y^{2}x \).
03
Compute Partial Derivatives
Calculate the partial derivatives: \( \frac{\partial M}{\partial x} = \frac{\partial}{\partial x}(-y^{2}x) = -y^{2} \) and \( \frac{\partial L}{\partial y} = \frac{\partial}{\partial y}(x^{2}y) = x^{2} \).
04
Set Up the Double Integral
Substitute into Green's Theorem: \( \iint_{D} \left( -y^{2} - x^{2} \right) \, dA \). The region \( D \) is in the first quadrant and is enclosed by the coordinate axes and the circle \( x^{2} + y^{2} = 16 \).
05
Convert to Polar Coordinates
Since the region \( D \) is circular, use polar coordinates: \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \) with \( r: 0 \rightarrow 4 \) and \( \, \theta: 0 \rightarrow \frac{\pi}{2} \). In polar coordinates, the area element \( dA = r \, dr \, d\theta \).
06
Evaluate the Double Integral
Substitute the expressions in polar coordinates and set up the double integral: \( -\iint_{D} (r^2) \, r \, dr \, d\theta = -\int_{0}^{\pi/2} \int_{0}^{4} r^3 \, dr \, d\theta \). First, integrate with respect to \( r \): \( \int_{0}^{4} r^3 \, dr = \left[ \frac{r^4}{4} \right]_{0}^{4} = \left[ \frac{256}{4} \right] = 64 \).
07
Integrate with respect to \( \theta \)
Now, integrate with respect to \( \theta \): \( -\int_{0}^{\pi/2} 64 \, d\theta = -\left[ 64\theta \right]_{0}^{\pi/2} = -\left[ 64 \frac{\pi}{2} \right] = -32\pi \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Line Integrals
Line integrals are a fascinating type of integral in calculus that generalize the notion of integrating functions along a line or curve. Unlike standard integrals that summarize an area's function along a straight line, line integrals can account for the influence of a function along a path that bends and twists in any dimension. In the context of Green's Theorem, line integrals have essential applications, particularly in vector fields. They can be written as:
- Line integral of a scalar field: Considered over a scalar function along a curve, often denoted as \( \int_C f(x,y) \, ds \).
- Line integral of a vector field: This involves a vector field often written as \( \oint_C \mathbf{F} \cdot d\mathbf{r} \), where \(d\mathbf{r}\) is a differential vector.
Exploring Polar Coordinates
Polar coordinates are an alternative method of representing points in a plane, especially useful for circular or rotational regions. Unlike Cartesian coordinates, which use \( (x, y) \), polar coordinates represent points using a radius \( r \) and an angle \( \theta \):
- \( x = r \cos(\theta) \)
- \( y = r \sin(\theta) \)
Mastering Double Integrals
Double integrals extend the concept of an integral to two dimensions, allowing for the computation of areas and volumes under surfaces. They are represented as \( \iint_{D} f(x, y) \, dA \), where \(D\) is a region in the \(xy\)-plane:
- They can be used to calculate the mass of a region if the density is non-uniform.
- They help in finding the total rainfall over a particular area if the rainfall distribution varies.