Question

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

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

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.

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 \[P = e^x \cos y \]\[Q = -e^x \sin y\]

Compute partial derivatives

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

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.

Conclusion

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

Key concepts

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.