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Magazine Chapter 19: Problem 29
Calculate the flux integral. \(\int_{S}\left(\sin x \vec{i}+\left(y^{2}+z^{2}\right) \vec{j}+y^{2} \vec{k}\right) \cdot d \vec{A}\) where \(S\) is a disk of radius \(\pi\) in the plane \(x=3 \pi / 2,\) oriented in the positive \(x\) -direction.
Short Answer
Expert verified
The flux integral is \(-\pi^3\).
Step by step solution
01
Understand the Problem
We need to calculate a flux integral over a surface \(S\), which is a disk in the plane \(x = \frac{3\pi}{2}\) with radius \(\pi\). The vector field is \(\vec{F} = \sin x \vec{i} + (y^2 + z^2) \vec{j} + y^2 \vec{k}\). The surface is oriented in the positive \(x\)-direction.
02
Describe the Surface
The surface \(S\) is a disk of radius \(\pi\) centered at \(x = \frac{3\pi}{2}\) in the \(yz\)-plane. In terms of \(y\) and \(z\), it's described as \(y^2 + z^2 \leq \pi^2\).
03
Surface Normal Vector
Since the plane is \(x = \frac{3\pi}{2}\) and is oriented in the positive \(x\)-direction, the outward normal vector \(\vec{n}\) is \(\vec{i}\). This means \(d\vec{A} = \vec{i} \, dy \, dz\).
04
Evaluate the Dot Product
Compute the dot product \(\vec{F} \cdot \vec{n}\):\[ \vec{F} \cdot \vec{n} = (\sin x \vec{i} + (y^2 + z^2) \vec{j} + y^2 \vec{k}) \cdot \vec{i} = \sin x \]Because only the \(\vec{i}\) component of \(\vec{F}\) contributes to the dot product.
05
Set up the Double Integral
With \(\vec{F} \cdot d\vec{A} = \sin\left(\frac{3\pi}{2}\right) \ dy \, dz\), we now set up the integral\[ \int_{S} \sin\left(\frac{3\pi}{2}\right) \ dy \, dz \].Since \(\sin\left(\frac{3\pi}{2}\right) = -1\), the integral becomes\[ \int_{\{y^2 + z^2 \leq \pi^2\}} -1 \ dy \, dz \].
06
Convert to Polar Coordinates
Change the double integral to polar coordinates, with \(y = r \cos\theta\) and \(z = r \sin\theta\). The limits for \(r\) are from 0 to \(\pi\), and for \(\theta\) from 0 to \(2\pi\).The double integral is now\[ \int_{0}^{2\pi} \int_{0}^{\pi} -1 \cdot r \, dr \, d\theta \].
07
Solve the Integral
First, solve the inner integral over \(r\):\[ \int_{0}^{\pi} -r \, dr = \left[-\frac{r^2}{2}\right]_{0}^{\pi} = -\frac{\pi^2}{2} \].Then integrate over \(\theta\):\[ \int_{0}^{2\pi} -\frac{\pi^2}{2} \, d\theta = -\frac{\pi^2}{2} \cdot \theta \Big|_{0}^{2\pi} = -\pi^3 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Disk Surface
A disk surface refers to a flat, circular region bordered by a circumference. In three-dimensional space, it can be located in a plane, such as the xy-plane, or any other plane like the yz-plane depending on the orientation. For this problem,
- The disk is positioned in the yz-plane, where every point on the disk maintains a constant x-coordinate of \( x = \frac{3\pi}{2} \).
- Its radius, \( \pi \), dictates how far it extends in the yz-plane, defining the points within \( y^2 + z^2 \leq \pi^2 \).
Polar Coordinates
Polar coordinates provide an alternative to Cartesian coordinates for describing points in a plane, which can be particularly useful in integration problems involving circular regions. In polar coordinates,
- Each point is defined by a distance \( r \) from a reference point (often the origin) and an angle \( \theta \) from a reference direction (often the positive x-axis).
- The conversion formulas from Cartesian to polar coordinates are \( y = r \cos(\theta) \) and \( z = r \sin(\theta) \).
- Converting to polar coordinates simplifies the integration process, taking advantage of the circular region's symmetry.
- The limits for the radial coordinate, \( r \), range from 0 to the disk's radius, \( \pi \), while \( \theta \) sweeps from 0 to \( 2\pi \), covering the entire circle.
Vector Field
A vector field assigns a vector to each point in a specified region of space, effectively describing the velocity of a fluid flow, the intensity and direction of force fields, or other dynamic systems. Here, the given vector field is
- \( \vec{F} = \sin(x) \vec{i} + (y^2 + z^2) \vec{j} + y^2 \vec{k} \).
- The \( \vec{i} \) component, \( \sin(x) \), links the field to the x-position, affecting the flow regardless of changes in y or z.
- For \( \vec{j} \), \( y^2 + z^2 \) ensures the field magnitude is a function of position in the yz-plane.
- The \( y^2 \vec{k} \) component provides a vertical influence varying with y alone.
Surface Normal Vector
The surface normal vector is essential in defining the direction in which we measure flux through a given surface. For the disk in this problem:
- The plane is at a constant \( x = \frac{3\pi}{2} \) and oriented positively along the x-axis. This implies that the outward normal vector, \( \vec{n} \), points directly in the positive x-direction as given by \( \vec{i} \).
- Thus, the differential area vector \( d\vec{A} \) is described as \( \vec{i} \, dy \, dz \), signifying each area element is oriented in this positive direction.
