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Chapter 16: Problem 14

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 16/15.

Step by step solution

01

Define the Region and the Path

The region is bound above by the curve \( y = 3 - x^2 \) and below by the curve \( y = x^4 + 1 \). We need to find the intersection points of these two curves to define the limits of integration. Set \( 3 - x^2 = x^4 + 1 \) and solve for \( x \).
02

Solve for Intersection Points

Rearrange the equation \( 3 - x^2 = x^4 + 1 \) to get \( x^4 + x^2 - 2 = 0 \). Substitute \( z = x^2 \), leading to the quadratic equation \( z^2 + z - 2 = 0 \). Solve for \( z \) to find the square roots for \( x \).
03

Solve the Quadratic Equation

Solve \( z^2 + z - 2 = 0 \) using the quadratic formula: \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For \( a = 1, b = 1, c = -2 \), we find \( z = 1 \) or \( z = -2 \). However, \( z = x^2\), so \( z \geq 0 \). Hence, \( x^2 = 1 \) leading to \( x = \pm 1 \).
04

Find the Circulation using Green's Theorem

Green's Theorem relates a line integral around a closed curve to a double integral over the region it encloses. For \( \mathbf{F}(x, y) = (y + e^x \ln y) \mathbf{i} + \left(\frac{e^x}{y} \right) \mathbf{j} \), the circulation \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} \) equals the double integral over the region \( R \) of \( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \).
05

Calculate the Required Partial Derivatives

Calculate \( \frac{\partial F_2}{\partial x} = \frac{e^x}{y} \) and \( \frac{\partial F_1}{\partial y} = 1 + \frac{e^x}{y} \). The expression under the double integral becomes \( \frac{e^x}{y} - (1 + \frac{e^x}{y}) = -1 \).
06

Set Up and Evaluate the Double Integral

We multiply \(-1\) by the area formula: \[ \iint_R -1 \ dA = - \text{Area of R} \]The limits of integration are from \( x = -1 \) to \( x = 1 \), and \( y \) varies from \( x^4 + 1 \) to \( 3-x^2 \). Evaluate the integral \[ \int_{-1}^{1} \int_{x^4+1}^{3-x^2} -1 \ dy \ dx \].
07

Evaluate the Integral and Find Circulation

First evaluate the inner integral with respect to \( y \), giving \( -[(3-x^2) - (x^4+1)] \). Simplify it to \( x^4 + x^2 - 2 \), leading to \[ \int_{-1}^{1} (-(x^4 + x^2 - 2)) \ dx \]. Compute this integral to find the circulation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Line Integrals with Green's Theorem
A line integral is a powerful concept used to find the cumulative effect of a vector field along a curve. Imagine if you wanted to know how much a force affects an object traveling along a path - the line integral does precisely that.

When applying Green's Theorem, a line integral around a closed curve can be transformed into a double integral over the region enclosed by the curve. This is particularly helpful as it changes a potentially complex problem into a more manageable computation involving partial derivatives.
  • In the provided exercise, the line integral of the vector field \( \mathbf{F} = (y + e^{x} \ln y) \mathbf{i} + \left(\frac{e^{x}}{y}\right) \mathbf{j} \) around the boundary of a region is evaluated using Green's Theorem.
  • This involves finding the closed boundary created by two curves: \( y = 3 - x^2 \) (upper curve) and \( y = x^4 + 1 \) (lower curve).
By using Green's Theorem, the problem which initially appears as a line integral becomes a double integral over the defined region on the \(xy\)-plane. This simplification is crucial, especially in computational mathematics, as it allows you to harness the power of integration over areas rather than just along lines.
Solving Double Integrals
A double integral extends the idea of a single-variable integral to functions of two variables. It calculates the volume under a surface in a specified region in the \(xy\)-plane.

Evaluating a double integral generally involves integrating with respect to one variable and then the other. This process is repeated across the entire region it's covering.
  • In the exercise, the region of interest is defined by finding the intersection points of the curves \( y = 3 - x^2 \) and \( y = x^4 + 1 \).
  • Once these limits are established, the double integral is set up to cover from \( x = -1 \) to \( x = 1 \), with \( y \) varying from \( x^4 + 1 \) to \( 3-x^2 \).
This concept is applied to find the area bounded between the curves, and with the expression for the derivative differences being \(-1\), the computation becomes simpler. Thus, the double integral represents the area of this region multiplied by \(-1\).
Role of Partial Derivatives in Green's Theorem
Partial derivatives play a critical role when applying Green's Theorem. They are used to transform a line integral into a double integral by examining the rate of change of a vector field's components.

For the given vector field \( \mathbf{F} = (y + e^x \ln y) \mathbf{i} + \left(\frac{e^x}{y}\right) \mathbf{j} \), we calculate:
  • \( \frac{\partial F_2}{\partial x} = \frac{e^x}{y} \), which represents the rate of change of the function \( e^x \div y \) with respect to \( x \).
  • \( \frac{\partial F_1}{\partial y} = 1 + \frac{e^x}{y} \), the change of \( y + e^x \ln y \) with respect to \( y \).
The difference \( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \) results in the constant value \(-1\).

By understanding this derivative difference, we set up a double integral with this as the integrand. The simplicity of the result \(-1\) allows easily calculating the circulation as simply \(-1\) times the area of the region. Thus, partial derivatives serve as a bridge in simplifying complex integrals using Green's Theorem.

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Most popular questions from this chapter

If \(S\) is the surface defined by a function \(z=f(x, y)\) that has continuous first partial derivatives throughout a region \(R_{x y}\) in the \(x y\) -plane (Figure \(16.49 ),\) then \(S\) is also the level surface \(F(x, y, z)=0\) of the function \(F(x, y, z)=f(x, y)-z\) . Taking the unit normal to \(R_{x y}\) to be \(\mathbf{p}=\mathbf{k}\) then gives $$ \begin{aligned}|\nabla F|=\left|f_{x} \mathbf{i}+f_{y} \mathbf{j}-\mathbf{k}\right| &=\sqrt{f_{x}^{2}+f_{y}^{2}+1} \\\|\nabla F \cdot \mathbf{p}| &=\left|\left(f_{x} \mathbf{i}+f_{y} \mathbf{j}-\mathbf{k}\right) \cdot \mathbf{k}\right|=|-1|=1 \end{aligned} $$ and $$ \iint_{R_{\mathrm{xy}}} \frac{|\nabla F|}{|\nabla F \cdot \mathbf{p}|} d A=\iint_{R_{\mathrm{xy}}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y $$ Similarly, the area of a smooth surface \(x=f(y, z)\) over a region \(R_{y z}\) in the \(y z\) -plane is $$ A=\iint_{R_{y x}} \sqrt{f_{y}^{2}+f_{z}^{2}+1} d y d z $$ and the area of a smooth \(y=f(x, z)\) over a region \(R_{x z}\) in the \(x z\) -plane is $$ A=\iint_{R_{\mathrm{xz}}} \sqrt{f_{x}^{2}+f_{z}^{2}+1} d x d z $$ Use Equations \((11)-(13)\) to find the area of the surfaces in Exercises \(39-44 .\) The surface cut from the bottom of the paraboloid \(z=x^{2}+y^{2}\) by the plane \(z=3\)

Find the outward flux of the field \(\mathbf{F}=2 x y \mathbf{i}+2 y z \mathbf{j}+2 x z \mathbf{k}\) across the surface of the cube cut from the first octant by the planes \(x=a, y=a, z=a\)

Flow integrals Find the flow of the velocity field \(\mathbf{F}=\) \((x+y) \mathbf{i}-\left(x^{2}+y^{2}\right) \mathbf{j}\) along each of the following paths from \((1,0)\) to \((-1,0)\) in the \(x y\) -plane. a. The upper half of the circle \(x^{2}+y^{2}=1\) b. The line segment from \((1,0)\) to \((-1,0)\) c. The line segment from \((1,0)\) to \((0,-1)\) followed by the line segment from \((0,-1)\) to \((-1,0)\)

Show that if \(\mathbf{F}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k},\) then \(\nabla \times \mathbf{F}=\mathbf{0}\)

Circulation of conservative fields Does Green's Theorem give any information about the circulation of a conservative field? Does this agree with anything else you know? Give reasons for your answer.
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