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Chapter 6: Problem 4

Use Green's theorem [formula }(9.7)] to evaluate the given integrals. \(\int_{C} e^{x} \cos y d x-e^{x} \sin y d y,\) where \(C\) is the broken line from \(\bar{A}=(\ln 2,0)\) to \(D=(0,1)\) and then from \(D\) to \(B=\) \((-\ln 2,0) .\) Hint: Apply Green's theorem to the integral around the closed curve \(A D B A\)

Short Answer

Expert verified
The integral of \(\int_{C} e^{x} \cos y dx - e^{x} \sin y dy\) over the closed curve CDA is zero.

Step by step solution

01

Understand Green's Theorem

Green's Theorem relates a line integral around a simple closed curve to a double integral over the plane region bounded by the curve. The theorem states that for a positively oriented, piecewise smooth, simple closed curve C enclosing a region D, \[ \oint_{C} P dx + Q dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \]Here, identify P and Q from the given integral.
02

Identify P and Q

In the given integral \(\int_{C} \left(e^x \cos y \right) dx - \left(e^x \sin y \right) dy \), we can identify \[P = e^x \cos y \]\[Q = -e^x \sin y\]
03

Compute partial derivatives

Calculating the partial derivatives required for Green's theorem:\[ \frac{\partial Q}{\partial x} = \frac{\partial (-e^x \sin y)}{\partial x} = -e^x \sin y \] \[ \frac{\partial P}{\partial y} = \frac{\partial (e^x \cos y)}{\partial y} = -e^x \sin y\]
04

Apply Green’s theorem

Apply the values in Green’s theorem: \[ \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA = \iint_{D} \left( -e^x \sin y + e^x \sin y \right) dA = \iint_{D} 0 \ dA \]Since the integrand is zero, the double integral is also zero.
05

Conclusion

Therefore, the integral over the closed curve CDA is zero. Since the integrand is zero:\[ \oint_{C} e^x \cos y dx - e^x \sin y dy = 0 \]

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Key Concepts

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

Line Integrals
Line integrals are a generalization of integrals to functions over a curve. Instead of integrating over a simple interval, you evaluate the integral along a path or curve. In the case of Green's Theorem, the line integral involves the components of a vector field along a closed curve.
  • Example: Consider the curve C: From \( (\text{ln} 2, 0) \) to \( (0, 1) \) and then from \( (0, 1) \) to \((- \text{ln} 2, 0) \).
  • The given integrated function is \(\begin{vmatrix} e^x \cos y \ dx - e^x \sin y \ dy \end{vmatrix}\).
Here, P = \( e^x \cos y \) and Q = \(- e^x \sin y \). These functions describe the vector field over the curve C. By transforming this line integral into a double integral over the region enclosed by C, Green's theorem simplifies the computing process significantly.
Double Integrals
Double integrals extend the concept of a single integral to functions of two variables over a region in the plane. Instead of summing slices of the area under a curve, double integrals accumulate the value of a function over a two-dimensional region.
  • For Green's Theorem, double integrals convert line integrals into more straightforward area-based calculations.
  • The general form of a double integral is \(\begin{vmatrix} \int_{a}^{b} \int_{c}^{d} f(x,y)\ dx \ dy \end{vmatrix} \), where \(D\) is the region in the plane.
Applying this to the problem:
Identify \( D \) as the region enclosed by the curve ADBA. According to Green's theorem, the given line integral is transformed to \(\begin{vmatrix} \iint_{D} \left( - e^x \sin y + e^x \sin y \right) \ dA \end{vmatrix}\).Since the integrand becomes zero over the region D, the resulting double integral evaluates to zero.
Partial Derivatives
Partial derivatives measure the rate of change of a multivariable function with respect to one variable while keeping the other variables constant. They are fundamental to concepts such as gradient, divergence, and curl in vector fields. For Green's Theorem, partial derivatives are used to transform the line integral into a double integral.
  • Example: Given P = \(\begin{vmatrix} e^x \cos y \end{vmatrix}\) and Q = \(\begin{vmatrix} - e^x \sin y \end{vmatrix}\), we calculate \(\frac{\partial Q}{\partial x} \) and \(\frac{\partial P}{\partial y} \).

Step Calculation:
  1. \(\frac{\partial Q}{\partial x} = \frac{\partial (-e^x \sin y)}{\partial x} = - e^x \sin y \)
  2. \(\frac{\partial P}{\partial y} = \frac{\partial (e^x \cos y)}{\partial y} = -e^x \sin y \)

Both partial derivatives yield the same negative sine term. Subtracting these gives zero, making the double integral vanish when integrated over region D.

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Most popular questions from this chapter

Use either Stokes' theorem or the divergence theorem to evaluate each of the following integrals in the easiest possible way. \(\iint_{\text {surface } \sigma} \operatorname{curl}\left(x^{2} \mathbf{i}+z^{2} \mathbf{j}-y^{2} \mathbf{k}\right) \cdot \mathbf{n} d \sigma,\) where \(\sigma\) is the part of the surface \(z=4-x^{2}-y^{2}\) above the \((x, y)\) plane.

Verify that each of the following force fields is conservative. Then find, for each, a scalar potential \(\phi\) such that \(\mathbf{F}=-\nabla \phi\). $$\mathbf{F}=-k \mathbf{r}, \mathbf{r}=\mathbf{i} x+\mathbf{j} y+\mathbf{k} z, \quad k=\mathrm{const.}$$

Consider a uniform distribution of total mass \(m^{\prime}\) over a spherical shell of radius The potential energy \(\phi\) of a mass \(m\) in the gravitational field of the spherical shell is. $$\phi=\left\\{\begin{array}{ll}\text { const. } & \text { if } m \text { is inside the spherical shell, } \\ -\frac{C m^{\prime}}{r} & \text { if } m \text { is outside the spherical shell, where } r \text { is the distance } \\\ & \text { from the center of the sphere to } m, \text { and } C \text { is a constant. }\end{array}\right.$$ Assuming that the earth is a spherical ball of radius \(R\) and constant density, find the potential and the force on a mass \(m\) outside and inside the earth. Evaluate the constants in terms of the acceleration of gravity \(g,\) to get \(\mathbf{F}=-\frac{m g R^{2}}{r^{2}} \mathbf{e}_{r}, \quad\) and \(\quad \phi=-\frac{m g R^{2}}{r}\) \(m\) outside the earth; \(\mathbf{F}=-\frac{m g r}{R} \mathbf{e}_{r}, \quad\) and \(\quad \phi=\frac{m g}{2 R}\left(r^{2}-3 R^{2}\right)\) \(m\) inside the earth.

Starting from the point \((1,1),\) in what direction does the function \(\phi=x^{2}-y^{2}+2 x y\) decrease most rapidly?

Evaluate each integral in the simplest way possible. \(\oint \mathbf{V} \cdot d \mathbf{r}\) around the boundary of the square with vertices (1,0),(0,1),(-1,0),(0,-1) if \(\mathbf{V}=x^{2} \mathbf{i}+5 x \mathbf{j}\).
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