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Chapter 18: Problem 13

Use Green's theorem to evaluate the line integral. \(\oint_{c}(x+y) d x+\left(y+x^{2}\right) d y\) \(C\) is the boundary of the region between the circles \(x^{2}+y^{2}=1\) and \(x^{2}+y^{2}=4\)

Short Answer

Expert verified
The value of the line integral is \(-3\pi\).

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: \[ \oint_C (M\,dx + N\,dy) = \iint_D \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right)\,dA \] where \(M = x+y\) and \(N = y+x^2\) in this problem.
02

Compute Partial Derivatives

For applying Green's Theorem, compute \(\frac{\partial N}{\partial x}\) and \(\frac{\partial M}{\partial y}\). We have \(N = y + x^2\), so \(\frac{\partial N}{\partial x} = 2x\). For \(M = x + y\), \(\frac{\partial M}{\partial y} = 1\).
03

Set Up the Double Integral

Insert the partial derivatives into the double integral: \[ \iint_D \left( 2x - 1 \right)\,dA. \] Here, \(D\) is the annular region between the circles \(x^2 + y^2 = 1\) and \(x^2 + y^2 = 4\).
04

Convert to Polar Coordinates

Due to symmetry, convert to polar coordinates \((r, \theta)\) where \(x = r\cos\theta\) and \(y = r\sin\theta\). The Jacobian of this transformation is \(r\).\[ \iint_D (2r\cos\theta - 1)r\,dr\,d\theta. \]
05

Define Limits of Integration

The region \(D\) is between the circles of radius 1 and 2. The limits for \(r\) are from 1 to 2, and for \(\theta\) from 0 to \(2\pi\). Thus, the integral becomes \[ \int_0^{2\pi} \int_1^2 (2r^2 \cos\theta - r) \, dr \, d\theta. \]
06

Evaluate the Inner Integral

Evaluate \(\int_1^2 (2r^2 \cos\theta - r) \, dr\). This gives \[ \left. \, \left(\frac{2r^3}{3}\cos\theta - \frac{r^2}{2}\right) \right|_1^2 = \left(\frac{16}{3}\cos\theta - 2\right) - \left(\frac{2}{3}\cos\theta - \frac{1}{2}\right) = \frac{14}{3}\cos\theta - \frac{3}{2}. \]
07

Evaluate the Outer Integral

Solve \(\int_0^{2\pi} \left(\frac{14}{3}\cos\theta - \frac{3}{2}\right) \, d\theta\). Note that \(\int_0^{2\pi} \cos\theta \, d\theta = 0\), therefore the integral evaluates to \(-3\pi\). We get: \[ \int_0^{2\pi} \left( \frac{14}{3}\cos\theta - \frac{3}{2} \right) \, d\theta = \int_0^{2\pi} \left(-\frac{3}{2} \right) \, d\theta = -3\pi. \]

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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 type of integral that finds the value of a function along a curve or path. In essence, you integrate over a line rather than over an area or a volume. In the context of Green's Theorem, a line integral is calculated along a closed path, known as the boundary of a region.
  • The line integral, \(\oint_{C} (M\,dx + N\,dy)\), computes the circulation or flux around curve \(C\).
  • The components \(M\) and \(N\) belong to the vector field, representing how the field behaves along the path.
In practical terms, the line integral helps calculate the total effect along \(C\), whether it is movement, flow, or another factor. This concept plays a valuable role in linking different areas of calculus, like in Green's Theorem, where it connects to double integrals.
Double Integral
The double integral is an extension of a single integral, applied over a two-dimensional area or region. In Green's theorem, the line integral around a closed curve is equal to the double integral over the region it encloses.
  • The double integral, \(\iint_D \,dA\), sums up values over an area \(D\), which is bounded by curve \(C\).
  • This integration considers changes in both the x and y-directions simultaneously, hence its 'double' nature.
In problems involving Green's Theorem, double integrals transform the problem from a curve-based to an area-based perspective. By switching the focus from curves to areas, double integrals make calculations simpler and more connected to physical properties like area and force.
Polar Coordinates
Polar coordinates provide an alternative to Cartesian coordinates, using distances and angles instead of x and y positions. They are particularly useful for problems with circular or rotational symmetry, simplifying the integration process.
  • In polar coordinates, a point is denoted as \( (r, \theta) \), where \(r\) is the radius and \(\theta\) is the angle.
  • The transformation from Cartesian to polar coordinates uses \(x = r\cos\theta\) and \(y = r\sin\theta\).
Green's Theorem often involves circular regions, making polar coordinates a more natural fit. They simplify the integration bounds, transforming complex regions into simpler radial ranges. When using polar coordinates, remember to include the Jacobian \(r\) in your integrals, accounting for changes in area element sizes.
Calculus
Calculus is the mathematical language used to understand change, motion, and the accumulation of quantities. It includes both derivative and integral calculus, which are applied to a variety of fields, from physics to engineering.
  • Integral calculus focuses on accumulation, helping calculate areas, volumes, and totals over time or space.
  • Derivative calculus focuses on rates of change, helping to understand how quantities change instantaneously.
In higher dimensions, calculus extends these ideas through tools like multiple integrals and partial derivatives, as used in Green's Theorem. Concepts such as line integrals and double integrals stretch calculus beyond simple one-dimensional cases, offering solutions to more complex problems found in real-world applications.

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Most popular questions from this chapter

Use Green's theorem to evaluate the line integral. \(\oint_{c} y^{2} d x+x^{2} d y\) \(C\) is the boundary of the region bounded by the semicircle \(y=\sqrt{4-x^{2}}\) and the \(x\) -axis.

Evaluate \(\iint_{s} x^{2} z^{2} d S\) if \(S\) is the top half of the cylinder \(y^{2}+z^{2}=4\) between \(x=0\) and \(x=1\).

The force at a point \((x, y, z)\) in three dimensions is given by \(\mathbf{F}(x, y, z)=y \mathbf{i}+z \mathbf{j}+x \mathbf{k}\). Find the work done by \(\mathbf{F}(x, y, z)\) along the twisted cubic \(x=t, y=t^{2}, z=t^{3}\) from (0.0,0) to (2,4,8)

Use the divergence theorem (18.26) to find \(\iint_{s} F \cdot \mathbf{n} d S\) \(\mathbf{F}=y^{3} e^{z} \mathbf{i}-x y \mathbf{j}+x \arctan y \mathbf{k}\) \(S\) is the surface of the region bounded by the coordinate planes and the plane \(x+y+z=1\).

Evaluate \(\int_{C} x y d s ; C\) is the graph of \(y=x^{4}\) from (-1,1) to (2,16).
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