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Apply Green's Theorem to evaluate the integrals. \(\oint_{C}(6 y+x) d x+(y+2 x) d y\) \(C :\) The circle \((x-2)^{2}+(y-3)^{2}=4\)

Short Answer

Expert verified
The integral evaluates to \(-16\pi\).

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. The theorem states: \( \oint_{C} (P\, dx + Q\, dy) = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \). In this problem, we identify \( P(x, y) = 6y + x \) and \( Q(x, y) = y + 2x \).
02

Compute Partial Derivatives

To apply Green's Theorem, we need \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \).Calculate:\[\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(y + 2x) = 2\]\[\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(6y + x) = 6\]
03

Set Up the Double Integral

Substitute the partial derivatives into the double integral formula from Green's Theorem:\[\iint_{D} \left(2 - 6 \right) \, dA = \iint_{D} (-4) \, dA\]
04

Evaluate the Area Integral

The region \( D \) is the disk described by the circle \( (x-2)^2 + (y-3)^2 = 4 \), having a radius of 2. The area of this disk is:\[\text{Area of } D = \pi \times (2)^2 = 4\pi\]Thus, the integral becomes:\[\iint_{D} (-4) \, dA = -4 \times 4\pi = -16\pi\]
05

Conclusion

By applying Green's Theorem and evaluating the resulting integral, the value of the line integral around the closed curve \( C \) is \( -16\pi \).

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Key Concepts

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

Line Integral
A line integral is a way of integrating functions along a curve or path in the plane or space. It is often used to calculate things like work done by a force field on a moving object. In the context of Green's Theorem, a line integral is expressed as:
  • \( \oint_{C}(P \, dx + Q \, dy) \),
where \( C \) is a closed curve, and \( P \) and \( Q \) are functions of \( x \) and \( y \).
The circle in our original problem is such a curve, denoted \( C \). The line integral is calculated around this closed curve, which in many cases represents the boundary of a region.
Green's Theorem helps us transform this line integral into a more manageable form, using a double integral over the region enclosed by \( C \).
Double Integral
Double integrals are used to integrate over two-dimensional regions. In the context of Green's Theorem, they help us find the integral over a region within a plane. Typically, a double integral is represented as:
  • \( \iint_{D} f(x, y) \, dA \),
where \( D \) is the region we are integrating over and \( f(x, y) \) is the function to be integrated.
For the specific application of Green's Theorem, we use the double integral to evaluate the expression \( \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \).
By converting the line integral around the curve into a double integral over the region, the problem becomes finding the area under a surface defined by this difference in derivatives. In the example exercise, the double integral simplifies and evaluates as \(-4 \times \text{Area of } D\), leading to the final result.
Partial Derivatives
Partial derivatives are pivotal in multivariable calculus, allowing us to examine how a function changes as only one of its variables changes. When dealing with functions of two variables like \( P(x, y) \) and \( Q(x, y) \), each has partial derivatives with respect to \( x \) and \( y \):
  • \( \frac{\partial P}{\partial y} \) is the derivative of \( P \) with respect to \( y \)
  • \( \frac{\partial Q}{\partial x} \) is the derivative of \( Q \) with respect to \( x \)
In Green’s Theorem, we need these to calculate the difference \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \).
This expression measures how the vector field described by \( P \) and \( Q \) behaves over the region bounded by the curve.
For our example function, \( \frac{\partial Q}{\partial x} = 2 \) and \( \frac{\partial P}{\partial y} = 6 \), lead us to determine that this difference is \(-4\), simplifying our double integral expression.
Closed Curve
A closed curve in mathematics is a path that starts and ends at the same point without crossing itself. In two-dimensional space, this curve encloses a region. These types of curves are essential when applying Green's Theorem because:
  • The theorem applies specifically to simple closed curves,
  • Allowing the transformation of a line integral to a double integral,
The curve \( C \) in our original problem is given by the equation \((x - 2)^2 + (y - 3)^2 = 4\), indicating a circle centered at \((2, 3)\) with radius 2.
This particular curve helps define the boundary of the region \( D \) for the double integral.
Such regions and curves assist in examining the behavior of vector fields. By evaluating a line integral around a closed curve, Green's Theorem enables us to gain insights into the vector field enclosed by the curve itself.

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Most popular questions from this chapter

Zero circulation Let \(f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}\) . Show that the clockwise circulation of the field \(\mathbf{F}=\nabla f\) around the circle \(x^{2}+y^{2}=a^{2}\) in the \(x y\) -plane is zero \begin{equation} \begin{array}{l}{\text { a. by taking } \mathbf{r}=(a \cos t) \mathbf{i}+(a \sin t) \mathbf{j}, 0 \leq t \leq 2 \pi, \text { and inte- }} \\ {\text { grating } \mathbf{F} \cdot d \mathbf{r} \text { over the circle. }} \\ {\text { b. by applying Stokes' Theorem. }}\end{array} \end{equation}

Let \(R\) be a region in the \(x y\) -plane that is bounded by a piecewise smooth simple closed curve \(C\) and suppose that the moments of inertia of \(R\) about the \(x\) - and \(y\) -axes are known to be \(I_{x}\) and \(I_{y}\) . Evaluate the integral $$\oint_{C} \nabla\left(r^{4}\right) \cdot \mathbf{n} d s$$ where \(r=\sqrt{x^{2}+y^{2}},\) in terms of \(I_{x}\) and \(I_{y}\)

Use a CAS and Green's Theorem to find the counterclockwise circulation of the field \(\mathbf{F}\) around the simple closed curve \(C\). Perform the following CAS steps. a. Plot \(C\) in the \(x y\)-plane. b. Determine the integrand \((\partial N / \partial x)-(\partial M / \partial y)\) for the tangential form of Green's Theorem. c. Determine the (double integral) limits of integration from your plot in part (a) and evaluate the curl integral for the circulation. \(\mathbf{F}=x^{-1} e^{y} \mathbf{i}+\left(e^{y} \ln x+2 x\right) \mathbf{j},\) \(C :\) The boundary of the region defined by \(y=1+x^{4}(\) below \()\) and \(y=2\) (above)

If \(\mathbf{F}=M \mathbf{i}+N \mathbf{j}+P \mathbf{k}\) is a differentiable vector field, we define the notation \(\mathbf{F} \cdot \nabla\) to mean $$M \frac{\partial}{\partial x}+N \frac{\partial}{\partial y}+P \frac{\partial}{\partial z}$$ For differentiable vector fields \(\mathbf{F}_{1}\) and \(\mathbf{F}_{2},\) verify the following identities. $$\begin{array}{l}{\text { a. } \nabla \times\left(\mathbf{F}_{1} \times \mathbf{F}_{2}\right)=\left(\mathbf{F}_{2} \cdot \nabla\right) \mathbf{F}_{1}-\left(\mathbf{F}_{1} \cdot \nabla\right) \mathbf{F}_{2}+\left(\nabla \cdot \mathbf{F}_{2}\right) \mathbf{F}_{1}-} \\\ {\quad\left(\nabla \cdot \mathbf{F}_{1}\right) \mathbf{F}_{2}} \\ {\mathbf{b} . \nabla\left(\mathbf{F}_{1} \cdot \mathbf{F}_{2}\right)=\left(\mathbf{F}_{1} \cdot \nabla\right) \mathbf{F}_{2}+\left(\mathbf{F}_{2} \cdot \nabla\right) \mathbf{F}_{1}+\mathbf{F}_{1} \times\left(\nabla \times \mathbf{F}_{2}\right)+} \\ {\mathbf{F}_{2} \times\left(\nabla \times \mathbf{F}_{1}\right)}\end{array}$$

Green's first formula Suppose that \(f\) and \(g\) are scalar functions with continuous first- and second-order partial derivatives throughout a region \(D\) that is bounded by a closed piecewise smooth surface \(S .\) Show that $$\iint_{S} f \nabla g \cdot \mathbf{n} d \sigma=\iiint_{D}\left(f \nabla^{2} g+\nabla f \cdot \nabla g\right) d V \quad \quad\quad(10)$$ Equation \((10)\) is Green's first formula. (Hint: Apply the Divergence Theorem to the field \(\mathbf{F}=f \nabla g . )\)

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