Chapter 15: Problem 18
Find the counterclockwise circulation of \(\mathbf{F}=\left(y+e^{x} \ln y\right) \mathbf{i}+\) \(\left(e^{x} / y\right) \mathbf{j}\) around the boundary of the region that is bounded above by the curve \(y=3-x^{2}\) and below by the curve \(y=x^{4}+1\)
Short Answer
Expert verified
The counterclockwise circulation is \( \frac{32}{15} \).
Step by step solution
01
Understand the problem
The problem requires finding the circulation of a vector field \( \mathbf{F} \) around a closed curve. The upper boundary of the region is defined by the curve \( y = 3 - x^2 \) and the lower boundary is \( y = x^4 + 1 \).
02
Set up the integral
To find the circulation, we use the line integral \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \). However, it is more efficient to use Green's Theorem, which relates the circulation around a simple closed curve to a double integral over the region \( R \) it encloses.
03
Apply Green's Theorem
According to Green's Theorem, \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dx \, dy \), where \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \). Here, \( M = y + e^x \ln y \) and \( N = e^x / y \).
04
Differentiate \(M\) and \(N\)
Calculate \( \frac{\partial N}{\partial x} = \frac{d}{dx} \left( \frac{e^x}{y} \right) = \frac{e^x}{y} \) since \( y \) is treated as a constant. Calculate \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} \left(y + e^x \ln y \right) = 1 + \frac{e^x}{y} \).
05
Set up the double integral
Subtract the two derivatives: \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = \frac{e^x}{y} - \left(1 + \frac{e^x}{y}\right) = -1 \). Thus, the double integral becomes: \( \iint_{R} -1 \, dx \, dy \).
06
Determine the limits of integration
For \( x \), the region is bounded where the curves intersect: \( 3 - x^2 = x^4 + 1 \). Solve this equation to find the bounds. By solving, the intersection points are \( x = -1 \) and \( x = 1 \).
07
Calculate the area of the region
Integrate \(-1\) over the region. Set the integration limits: \( \int_{-1}^{1} \int_{x^4+1}^{3-x^2} -1 \; dy \; dx \). Integrating with respect to \( y \), we get: \( -[3-x^2 - (x^4 + 1)] \). Simplify and evaluate: \( \int_{-1}^{1} (x^4 - x^2 + 2) \, dx \).
08
Evaluate the integral
Perform the definite integral: \[ \int_{-1}^{1} (x^4 - x^2 + 2) \, dx = \left[ \frac{x^5}{5} - \frac{x^3}{3} + 2x \right]_{-1}^{1} \]. Calculating each term and substituting leads to the final value.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Green's Theorem
Green's Theorem is a powerful tool in vector calculus that connects the circulation of a vector field around a closed curve with a double integral over the region bounded by the curve. This theorem simplifies the computation of certain line integrals by converting them into double integrals. The theorem is applicable to a two-dimensional vector field \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \), where \( C \) is a positively oriented, simple, and closed curve bounding a region \( R \). The relationship given by Green's Theorem is:
- \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dx \, dy \)
Vector Field
A vector field is a mathematical construct where a vector is assigned to every point in a region of space. In the context of our exercise, the vector field \( \mathbf{F} = (y + e^x \ln y) \mathbf{i} + (e^x / y) \mathbf{j} \) assigns a vector to each point \( (x, y) \) in the plane.
- \( M = y + e^x \ln y \) is the component of the vector field in the \( x \)-direction.
- \( N = e^x / y \) is the component in the \( y \)-direction.
Line Integral
A line integral is a type of integral that involves integrating a function over a curve or path. In the context of vector fields, a line integral computes the work done by the vector field as one moves along a curve. For a vector field \( \mathbf{F} = M\mathbf{i} + N\mathbf{j} \) and a curve \( C \), the line integral is given by:
- \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \oint_{C} (M \, dx + N \, dy) \)
Double Integral
A double integral extends the concept of a single integral to functions of two variables. It is used to compute the volume under a surface in three-dimensional space or the area over a region in a two-dimensional space. In the exercise, we use a double integral to calculate the circulation of a vector field around a closed curve by integrating over the region \( R \) bounded by the curves.
- The double integral in our problem is \( \iint_{R} -1 \, dx \, dy \).