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Area of regions Use a line integral on the boundary to find the area of the following regions. $$\left\\{(x, y): x^{2}+y^{2} \leq 16\right\\}$$

Short Answer

Expert verified
The area of the region defined by the inequality \(x^2+y^2\leq 16\) using a line integral on the boundary is \(16\pi\) square units.

Step by step solution

01

Define the curve

The boundary of the region in the given inequality is a circle centered at the origin with a radius of 4. We can parametrize this curve as a function of a single parameter, \(t\), as follows: $$\textbf{r}(t) = \langle x(t), y(t) \rangle = \langle 4\cos(t), 4\sin(t) \rangle$$ where \(t\in[0, 2\pi]\).
02

Compute the derivatives of x(t) and y(t)

We need to find the derivatives of \(x(t)\) and \(y(t)\) with respect to \(t\) to compute the line integral. The derivatives are as follows: $$x'(t) = -4\sin(t)$$ $$y'(t) = 4\cos(t)$$
03

Calculate the area using the line integral

Now, using the Green's theorem formula for the area, we can compute the area A as: $$A = \frac{1}{2} \int_{0}^{2\pi} |4\cos(t)(4\cos(t)) - (-4\sin(t))(4\sin(t))|\, dt$$ $$A = \frac{1}{2} \int_{0}^{2\pi} |16\cos^2(t) + 16\sin^2(t)|\, dt$$ Using the Pythagorean identity, \(\sin^2(t) + \cos^2(t) = 1\), we can simplify the integral: $$A = \frac{1}{2} \int_{0}^{2\pi} |16|\, dt$$
04

Evaluate the integral

The integral is now straightforward to evaluate: $$A = \frac{1}{2} \cdot 16 \int_{0}^{2\pi} dt$$ $$A=8 [t]_{0}^{2\pi} = 8(2\pi - 0) = 16\pi$$
05

State the final answer

The area of the region defined by the inequality \(x^2+y^2\leq 16\) using a line integral on the boundary is \(16\pi\) square units.

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Key Concepts

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

Green's theorem
Green's theorem provides a powerful tool to convert a line integral around a simple, closed curve into a double integral over the plane region bounded by the curve. It's fundamental in calculating the area enclosed by a curve or the circulation of a vector field around the curve.

Here's the relation at the heart of Green's theorem: if you have a positively oriented, simple, closed curve C and a continuously differentiable vector field \textbf{F} = \textbf{M} \textbf{i} + \textbf{N} \textbf{j}, the line integral around the curve can be rewritten as a double integral over the region D enclosed by C, like so:
\[\text{Line integral: } \/oint_{C} (M dx + N dy) \]\[\text{Double integral: } = \/int_\/int_{D} (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y})dA\]
In the context of area calculations, we can manipulate the components of the vector field such that the curl (here represented by \(\frac{\textbf{\partial N}}{\textbf{\partial x}} - \frac{\textbf{\partial M}}{\textbf{\partial y}}\)) computes to 1, thus reducing the double integral to simply calculating the area of region D.
Parametric equations
Parametric equations describe a curve in the plane by expressing the coordinates of the points on the curve as functions of a single variable, usually denoted as t (for parameter). Using parametric equations makes it easier to describe and work with complex curves, such as circles and ellipses.

For a circle with a radius r centered at the origin, we can use the following parametric equations:
\[\textbf{r}(t) = \langle r\cos(t), r\sin(t) \rangle\]
This parametric representation is useful because it breaks the position down into x and y components, which can change with respect to the parameter, t. The circular motion is captured by relating the x and y coordinates through the cosine and sine functions, both of which are periodic and bounded.
Pythagorean identity
The Pythagorean identity is an essential tool in trigonometry that states that for any real number t:
\[\sin^2(t) + \cos^2(t) = 1\]
This fundamental relationship comes directly from the Pythagorean theorem, as it relates the squares of the sine and cosine of an angle, which represent the normalized lengths of the legs of a right triangle, to the square of the hypotenuse, which is always 1 in a unit circle. Not only is this identity helpful in simplifying trigonometric expressions, but it also plays a critical role in converting trigonometric forms that seem complicated at first glance into much simpler terms.
Integrating trigonometric functions
Integrating trigonometric functions becomes substantially easier when identities, like the Pythagorean identity, are utilized. Such identities allow you to modify the integrand into a familiar form that's easier to integrate.

For instance, when integrating a function like \(16\cos^2(t) + 16\sin^2(t)\), recognizing that this is just 16 times the Pythagorean identity can reduce the integral to \(16\int dt\), which is straightforward to evaluate. Always be on the lookout for these patterns in trigonometric functions—they can transform a daunting integration task into a few simple steps.

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Most popular questions from this chapter

Maximum surface integral Let \(S\) be the paraboloid \(z=a\left(1-x^{2}-y^{2}\right),\) for \(z \geq 0,\) where \(a>0\) is a real number. Let \(\mathbf{F}=\langle x-y, y+z, z-x\rangle .\) For what value(s) of \(a\) (if any) does \(\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S\) have its maximum value?

Fourier's Law of heat transfer (or heat conduction ) states that the heat flow vector \(\mathbf{F}\) at a point is proportional to the negative gradient of the temperature; that is, \(\mathbf{F}=-k \nabla T,\) which means that heat energy flows from hot regions to cold regions. The constant \(k>0\) is called the conductivity, which has metric units of \(J /(m-s-K)\) A temperature function for a region \(D\) is given. Find the net outward heat flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=-k \iint_{S} \nabla T \cdot \mathbf{n} d S\) across the boundary S of \(D\) In some cases, it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume \(k=1 .\) $$\begin{aligned} &T(x, y, z)=100+x+2 y+z\\\ &D=\\{(x, y, z): 0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1\\} \end{aligned}$$

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. $$T(x, y, z)=100 e^{-x^{2}+y^{2}+z^{2}}$$

Prove the following identities. Assume \(\varphi\) is a differentiable scalar- valued function and \(\mathbf{F}\) and \(\mathbf{G}\) are differentiable vector fields, all defined on a region of \(\mathbb{R}^{3}\). $$\nabla \times(\mathbf{F} \times \mathbf{G})=(\mathbf{G} \cdot \nabla) \mathbf{F}-\mathbf{G}(\nabla \cdot \mathbf{F})-(\mathbf{F} \cdot \nabla) \mathbf{G}+\mathbf{F}(\nabla \cdot \mathbf{G})$$

The rotation of a threedimensional velocity field \(\mathbf{V}=\langle u, v, w\rangle\) is measured by the vorticity \(\omega=\nabla \times \mathbf{V} .\) If \(\omega=\mathbf{0}\) at all points in the domain, the flow is irrotational. a. Which of the following velocity fields is irrotational: \(\mathbf{V}=\langle 2,-3 y, 5 z\rangle\) or \(\mathbf{V}=\langle y, x-z,-y\rangle ?\) b. Recall that for a two-dimensional source-free flow \(\mathbf{V}=\langle u, v, 0\rangle,\) a stream function \(\psi(x, y)\) may be defined such that \(u=\psi_{y}\) and \(v=-\psi_{x} .\) For such a two-dimensional flow, let \(\zeta=\mathbf{k} \cdot \nabla \times \mathbf{V}\) be the \(\mathbf{k}\) -component of the vorticity. Show that \(\nabla^{2} \psi=\nabla \cdot \nabla \psi=-\zeta\) c. Consider the stream function \(\psi(x, y)=\sin x \sin y\) on the square region \(R=\\{(x, y): 0 \leq x \leq \pi, 0 \leq y \leq \pi\\} .\) Find the velocity components \(u\) and \(v\); then sketch the velocity field. d. For the stream function in part (c), find the vorticity function \(\zeta\) as defined in part (b). Plot several level curves of the vorticity function. Where on \(R\) is it a maximum? A minimum?

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