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Chapter 17: Problem 39

Line integrals Use Green's Theorem to evaluate the following line integrals. Assume all curves are oriented counterclockwise. A sketch is helpful. The circulation line integral of \(\mathbf{F}=\left(x^{2}+y^{2}, 4 x+y^{3}\right),\) where \(C\) is the boundary of \(\\{(x, y): 0 \leq y \leq \sin x, 0 \leq x \leq \pi\\}\)

Short Answer

Expert verified
Using Green's Theorem, we can compute the circulation line integral of the given vector field \(\mathbf{F} = (x^2 + y^2, 4x + y^3)\) along the boundary of the region defined by \(0 \leq y \leq \sin x, 0 \leq x \leq \pi\). Following the steps to apply Green's Theorem, we calculate the partial derivatives \(\frac{\partial Q}{\partial x} = 4\) and \(\frac{\partial P}{\partial y} = 2y\). Then, we set up and compute the double integral over the region \(D\): \[\iint_{D} (4 - 2y) \, dA = \int_{0}^{\pi} \int_{0}^{\sin x} (4 - 2y) \, dy \, dx\] After evaluating the double integral, we find that the circulation line integral of the vector field along the curve \(C\) is 8.

Step by step solution

01

Identify P and Q from the vector field

We are given the vector field \(\mathbf{F} = (x^2 + y^2, 4x + y^3)\). From this, we can identify \(P = x^2 + y^2\) and \(Q = 4x + y^3\).
02

Calculate partial derivatives

Next, we need to compute the partial derivatives \(\frac{\partial Q}{\partial x}\) and \(\frac{\partial P}{\partial y}\). For \(\frac{\partial Q}{\partial x}\), we get: \[\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(4x + y^3) = 4\] For \(\frac{\partial P}{\partial y}\), we get: \[\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(x^2 + y^2) = 2y\]
03

Set up and compute the double integral

According to Green's Theorem, the circulation line integral is equal to the double integral of \((\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})\) over the region \(D\). We already have the partial derivatives \(\frac{\partial Q}{\partial x} = 4\) and \(\frac{\partial P}{\partial y} = 2y\). Thus, we need to compute the double integral: \[\iint_{D} (4 - 2y) \, dA\] The given region \(D\) is described by \(0 \leq y \leq \sin x\), \(0 \leq x \leq \pi\), so we set up the double integral as follows: \[\int_{0}^{\pi} \int_{0}^{\sin x} (4 - 2y) \, dy \, dx\] First, we integrate with respect to \(y\): \[\int_{0}^{\pi} [(4y - y^2) \Big|_{0}^{\sin x}] \, dx = \int_{0}^{\pi} (4\sin x - \sin^2 x) \, dx\] Now, we integrate with respect to \(x\): \[\int_{0}^{\pi} (4\sin x - \sin^2 x) \, dx = [-4\cos x + \frac{1}{3}\sin 3x]_{0}^{\pi}\] Evaluating the integral, we get: \[-4\cos \pi + \frac{1}{3}\sin 3\pi - (-4\cos 0 + \frac{1}{3}\sin 0) = (-4)(-1) - 0 - (-4)(1) - 0 = 8\] So, the circulation line integral of the given vector field \(\mathbf{F} = (x^2 + y^2, 4x + y^3)\) along the curve \(C\) is 8.

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Key Concepts

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

Line Integrals
A line integral is a type of integral where the function to be integrated is evaluated along a curve. Imagine you have a curve in a plane or in space, and you want to sum up values of another function along that curve. This is what a line integral does.
In the context of vector fields, line integrals help us measure things like:
  • Work done by a force field in moving an object along a path
  • Flow across a curve
To calculate a line integral, you often need to parameterize your curve. This means figuring out a way to describe all the points along the curve using a single variable. Then, using Green's Theorem, like in this exercise, line integrals can be converted to double integrals over a region if the curve is closed.
Here, the line integral relates to the vector field \(\mathbf{F} = \left(x^2 + y^2, 4x + y^3\right)\) and the region is bounded by a sine curve from \(x = 0\) to \(x = \pi\).
Vector Fields
A vector field is essentially a function that assigns a vector to every point in a space. You can visualize it as a field of arrows, where each arrow represents the direction and magnitude of the vector at that particular point. Vector fields are crucial in physics for representing things like gravitational, electric, or magnetic fields.
In our exercise, the vector field \(\mathbf{F} = (x^2 + y^2, 4x + y^3)\) represents a field in a two-dimensional plane. Each vector in the field depends on its position (determined by x and y values).
When dealing with exercises involving Green's Theorem and line integrals, identifying components of the vector field (here denoted \(P\) and \(Q\)) is essential. These components play a pivotal role when applying Green's Theorem, which essentially finds a relationship between line integrals around closed curves and double integrals over the region inside those curves.
Double Integral
A double integral extends the concept of a single integral to functions of two variables. Instead of finding areas under a curve, a double integral helps you find volumes under surfaces or over regions in the plane.
In mathematical terms, a double integral over a region \(D\) is expressed as \(\iint_{D} f(x, y) \, dA\), where \(dA\) represents an infinitesimal area element.
Applying Green's Theorem in this solution, the line integral of a vector field around a closed curve is transformed into a double integral over the region enclosed by the curve. This transformation makes it easier to compute the desired values in relation to the vector field.
In our specific problem, after computing necessary partial derivatives, we set up a double integral over defined limits: from \(0\) to \(\pi\) for \(x\) and from \(0\) to \(\sin x\) for \(y\). This results in a more straightforward process than directly computing the line integral.

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Most popular questions from this chapter

Suppose a solid object in \(\mathbb{R}^{3}\) has a temperature distribution given by \(T(x, y, z) .\) The heat flow vector field in the object is \(\mathbf{F}=-k \nabla T,\) where the conductivity \(k>0\) is a property of the material. Note that the heat flow vector points in the direction opposite to that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is \(\nabla \cdot \mathbf{F}=-k \nabla \cdot \nabla T=-k \nabla^{2} T(\text {the Laplacian of } T) .\) Compute the heat flow vector field and its divergence for the following temperature distributions. $$T(x, y, z)=100 e^{-\sqrt{x^{2}+y^{2}+z^{2}}}$$

Inverse square fields are special Let \(F\) be a radial ficld \(\mathbf{F}=\mathbf{r} /|\mathbf{r}|^{p},\) where \(p\) is a real number and \(\mathbf{r}=\langle x, y, z\rangle .\) With \(p=3, \mathbf{F}\) is an inverse square field. a. Show that the net flux across a sphere centered at the origin is independent of the radius of the sphere only for \(p=3\) b. Explain the observation in part (a) by finding the flux of \(\mathbf{F}=\mathbf{r} /|\mathbf{r}|^{p}\) across the boundaries of a spherical box \(\left\\{(\rho, \varphi, \theta): a \leq \rho \leq b, \varphi_{1} \leq \varphi \leq \varphi_{2}, \theta_{1} \leq \theta \leq \theta_{2}\right\\}\) for various values of \(p\)

A scalar-valued function \(\varphi\) is harmonic on a region \(D\) if \(\nabla^{2} \varphi=\nabla \cdot \nabla \varphi=0\) at all points of \(D\). Show that if \(\varphi\) is harmonic on a region \(D\) enclosed by a surface \(S\) then \(\iint_{S} \nabla \varphi \cdot \mathbf{n} d S=0\)

Suppose a solid object in \(\mathbb{R}^{3}\) has a temperature distribution given by \(T(x, y, z) .\) The heat flow vector field in the object is \(\mathbf{F}=-k \nabla T,\) where the conductivity \(k>0\) is a property of the material. Note that the heat flow vector points in the direction opposite to that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is \(\nabla \cdot \mathbf{F}=-k \nabla \cdot \nabla T=-k \nabla^{2} T(\text {the Laplacian of } T) .\) Compute the heat flow vector field and its divergence for the following temperature distributions. $$T(x, y, z)=100 e^{-x^{2}+y^{2}+z^{2}}$$

Rain on roofs Let \(z=s(x, y)\) define a surface over a region \(R\) in the \(x y\) -plane, where \(z \geq 0\) on \(R\). Show that the downward flux of the vertical vector field \(\mathbf{F}=\langle 0,0,-1\rangle\) across \(S\) equals the area of \(R .\) Interpret the result physically.
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