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Chapter 8: Problem 6

Verify Greens theorem for the indicated region \(D\) and boundary \(a D\), and functions \(P\) and \(Q\). $$\begin{array}{l} D=\left[0, \frac{\pi}{2}\right] \times\left[0, \frac{\pi}{2}\right], \quad P(x, y)=\sin x \\ Q(x, y)=\cos y \end{array}$$

Short Answer

Expert verified
Applying Green's Theorem, both the line and double integrals equal zero, confirming the theorem.

Step by step solution

01

Understand 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 states that \( \oint_{C} (P \, dx + Q \, dy) = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dx \, dy \). Here, \(P(x, y) = \sin x\) and \(Q(x, y) = \cos y\).
02

Calculate Partial Derivatives

Calculate \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \). For \(Q(x, y) = \cos y\), \( \frac{\partial Q}{\partial x} = 0\) because it depends solely on \(y\). For \(P(x, y) = \sin x\), \( \frac{\partial P}{\partial y} = 0\) because it depends solely on \(x\).
03

Evaluate Double Integral

Substitute into the formula: \( \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dx \, dy = \iint_{D} (0 - 0) \, dx \, dy = \iint_{D} 0 \, dx \, dy = 0\). As the integrand is zero, the double integral evaluates to zero.
04

Evaluate Line Integral

Parameterize the boundary of the region \(D\) which is a square defined by \((0, 0), (\frac{\pi}{2}, 0), (\frac{\pi}{2}, \frac{\pi}{2}), (0, \frac{\pi}{2})\). Calculate the line integral around each boundary segment and sum the results.
05

Segment Integrations - Horizontal Bottom Edge

For the bottom edge from \((0, 0)\) to \( (\frac{\pi}{2}, 0) \), parameterize by \(x = t, y = 0\) with \(0 \leq t \leq \frac{\pi}{2}\). The line integral is \( \int_{0}^{\frac{\pi}{2}} \sin t \, dt = [-\cos t]_{0}^{\frac{\pi}{2}} = (0 + 1) = 1\).
06

Segment Integrations - Vertical Right Edge

For the right edge from \((\frac{\pi}{2}, 0)\) to \( (\frac{\pi}{2}, \frac{\pi}{2}) \), parameterize by \(x = \frac{\pi}{2}, y = t\) with \(0 \leq t \leq \frac{\pi}{2}\). The line integral is \( \int_{0}^{\frac{\pi}{2}} \cos t \, dt = [\sin t]_{0}^{\frac{\pi}{2}} = (1 - 0) = 1\).
07

Segment Integrations - Horizontal Top Edge

For the top edge from \((\frac{\pi}{2}, \frac{\pi}{2})\) to \( (0, \frac{\pi}{2}) \), parameterize by \(x = t, y = \frac{\pi}{2}\) with \(\frac{\pi}{2} \geq t \geq 0\). The line integral is \( \int_{\frac{\pi}{2}}^{0} \sin t \, dt = 1\), but since direction is reversed, the result is \(-1\).
08

Segment Integrations - Vertical Left Edge

For left edge from \((0, \frac{\pi}{2})\) to \( (0, 0) \), parameterize by \(x = 0, y = t\) with \(\frac{\pi}{2} \geq t \geq 0\). The line integral is \( \int_{\frac{\pi}{2}}^{0} \cos t \, dt = 1\), but since direction is reversed, the result is \(-1\).
09

Sum Line Integrals

Sum the line integrals for each segment: \(1 + 1 - 1 - 1 = 0\).
10

Conclusion

The line integral around the square path is zero. Comparing this with the double integral which was also zero, Green's Theorem confirms the calculations as both the line integral and the double integral match.

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental concept in calculus where we focus on a function of several variables. Each variable in such a function has its own "partial derivative"
Line Integral
A line integral is a type of integral where we are essentially integrating along a curve or path within a field.
Double Integral
Double integrals are used to integrate over a two-dimensional area or region.
Parameterization
Parameterization is the process of defining a curve, surface, or geometric construct in terms of one or more variables.

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Most popular questions from this chapter

Show that any two potential functions for a vector field on \(\mathrm{R}^{3}\) differ at most by a constant.

If \(C\) is a closed curve that is the boundary of a surface \(S\) and \(f\) and \(g\) are \(C^{2}\) functions, show that (a) \(\int_{C} f \nabla g \cdot d \mathbf{s}=\iint_{S}(\nabla f \times \nabla g) \cdot d \mathbf{S}\) (b) \(\int_{C}(f \nabla g+g \nabla f) \cdot d \mathbf{s}=0\)

Prove the following generalization of Green's theorem: Let \(D\) be a region in the \(x y\) plane with boundary a finite number of oriented simple closed curves. Suppose that by means of a finite number of line segments parallel to the coordinate axes, \(D\) can be decomposed into a finite number of simple regions \(D_{i}\) with the boundary of each \(D_{i}\) oriented counterclockwise (see Figure 8.1 .5 ). Then if \(P\) and \(Q\) are of class \(C^{1}\) on \(D,\) $$ \iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) d x d y=\int_{\partial D} P d x+Q d y $$ where \(\partial D\) is the oriented boundary of \(D .\) (HINT: Apply Green's theorem to each \(D_{i} .\) )

If \(\mathbf{V}=A(x, y, z) \mathbf{i}+B(x, y, z) \mathbf{j}+C(x, y, z) \mathbf{k}\) is a vector field on \(K \subset \mathbb{R}^{3}\), define the operation Form \(_{2}:\) Vector Fields \(\rightarrow\) 2-forms by Form \(_{2}(\mathbf{V})=A d y d z+B d z d x+C d x d y\) (a) Show that Form \(_{2}\left(\alpha \mathbf{V}_{1}+\mathbf{V}_{2}\right)=\alpha\) Form \(_{2}\left(\mathbf{V}_{1}\right)+\) Form \(_{2}\left(\mathbf{V}_{2}\right)\) where \(\alpha\) is a real number. (b) Show that Form \(_{2}(\text { curl } \mathbf{V})=d \omega,\) where \(\omega=A d x+B d y+C d z\)

Let \(D\) be the region in the \(x y\) plane lying between the curves \(y=x^{2}+4\) and \(y=2 x^{2} .\) Describe the boundary \(\partial D\) as a piece wise smooth curve, oriented counterclockwise.
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