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Chapter 14: Problem 49

In Exercises \(49-52,\) use a CAS integration utility to evaluate the triple integral of the given function over the specified solid region. \(F(x, y, z)=x^{2} y^{2} z\) over the solid cylinder bounded by \(x^{2}+y^{2}=1\) and the planes \(z=0\) and \(z=1\).

Short Answer

Expert verified
The integral evaluates to \( \frac{\pi}{48} \).

Step by step solution

01

Understand the Region of Integration

The solid region is a cylinder described by the equation \( x^2 + y^2 = 1 \) with its height extending from \( z = 0 \) to \( z = 1 \). The limits for \( z \) are from 0 to 1, and for the cylindrical base, \( x^2 + y^2 \leq 1 \).
02

Set Up the Integral

The triple integral of \( F(x, y, z) = x^2 y^2 z \) over the cylinder is set up as follows: \(\int_{z=0}^{1} \int_{y=-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{x=-1}^{1} x^2 y^2 z \; dx \, dy \, dz\). This accounts for the boundaries of the solid cylinder. But since the base is circular, we can use polar coordinates for \( x \) and \( y \).
03

Convert to Cylindrical Coordinates

We substitute \( x = r \cos(\theta) \), \( y = r \sin(\theta) \), and \( dx \, dy \) by \( rdrd\theta \) due to the Jacobian determinant. The integral becomes \(\int_{z=0}^{1} \int_{\theta=0}^{2\pi} \int_{r=0}^{1} (r^2 \cos^2(\theta) \cdot r^2 \sin^2(\theta) \cdot z) \cdot r \, dr \, d\theta \, dz\).
04

Simplify the Integral Expression

This integral simplifies to \( \int_{z=0}^{1} \int_{\theta=0}^{2\pi} \int_{r=0}^{1} r^5 z \cos^2(\theta) \sin^2(\theta) \, dr \, d\theta \, dz\). The expression can be further simplified by evaluating each part sequentially starting with \( r \).
05

Integrate with Respect to \( r \)

The integral of \( r^5 \) over \( r \, \) is \(\int_{r=0}^{1} r^5 \, dr = \frac{r^6}{6} \bigg|_{0}^{1} = \frac{1}{6}.\).
06

Integrate with Respect to \( z \)

Now integrate with respect to \( z \): \(\int_{z=0}^{1} z \, dz = \frac{z^2}{2} \bigg|_{0}^{1} = \frac{1}{2}.\)
07

Integrate with Respect to \( \theta \)

Finally, integrate with respect to \( \theta \): \(\int_{\theta=0}^{2\pi} \cos^2(\theta) \sin^2(\theta) \, d\theta.\) Use the identity \( \cos^2(\theta) \sin^2(\theta) = \frac{1}{4} \sin^2(2\theta) \), giving \(\frac{1}{4} \int_{0}^{2\pi} \sin^2(2\theta) \, d\theta = \frac{1}{8} \left[\theta - \frac{1}{2}\sin(4\theta)\right]_{0}^{2\pi} = \frac{\pi}{4}.\).
08

Compute the Final Result

Multiplying the results of all parts together: \(\frac{1}{6} \times \frac{1}{2} \times \frac{\pi}{4} = \frac{\pi}{48}.\) Thus the value of the triple integral is \( \frac{\pi}{48}.\)

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Key Concepts

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

Cylindrical Coordinates
Cylindrical coordinates provide a practical way to describe three-dimensional space, particularly for problems involving cylinders or circular symmetry. This system is an extension of polar coordinates in two-dimensional space by adding a third dimension, the height or axial coordinate.
  • Instead of using Cartesian coordinates \(x, y, z\), we express space using the coordinates \(r, \theta, z\).
  • Here, \(r\) represents the radial distance from the origin to the projection of the point in the \(xy\)-plane.
  • The angle \(\theta\) is measured counterclockwise from the positive \((x)\)-axis.
  • The coordinate \(z\) represents the height, just as in Cartesian coordinates.
Using cylindrical coordinates makes it easier to set up integrals over regions with circular bases, such as the cylinder in our original exercise.
Polar Coordinates
Polar coordinates are a two-dimensional coordinate system where each point on a plane is determined by a distance and an angle. As the fundamental base for cylindrical coordinates, understanding polar coordinates is essential.
  • In polar coordinates, a point \( (x, y) \) in Cartesian space can be expressed as \( (r, \theta) \).
  • The variable \( r \) is the distance from the origin to the point, while \( \theta \) is the angle between the positive \( x \)-axis and the line connecting the origin to the point.
This transformation can be denoted mathematically as \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). These equations are pivotal in converting Cartesian integrals to polar or cylindrical coordinates.
Jacobian Determinant
When transforming coordinates, as from Cartesian to cylindrical, we must ensure that the volume element is accurately represented. This is where the Jacobian determinant becomes crucial.
  • The Jacobian determinant helps adjust the volume element from one coordinate system to another.
  • For the conversion from Cartesian to cylindrical coordinates, the Jacobian is \( r \).
  • It reflects the area scaling factor used in integration.
Thus, when integrating in cylindrical coordinates, the differential area \(dx \, dy\) is transformed into \(r \, dr \, d\theta\). This adjustment ensures that we capture the true size of each infinitesimal area in our integration results.
Cylindrical Integration
Cylindrical integration is straightforward once the transformation from rectangular to cylindrical coordinates is complete. The power of cylindrical integration lies in its ability to simplify problems with symmetrical, circular boundaries.
  • Replace Cartesian coordinates \(x, y\) with cylindrical expressions \(r \cos(\theta), r \sin(\theta)\).
  • Convert the differential area \(dx \, dy\) to \(r \, dr \, d\theta\) using the Jacobian.
  • Integrals are then evaluated over the simpler radial and angular limits.
  • This transformation often results in less complex integrals, which are easier to evaluate.
For our exercise, this process took the complex triple integral and transformed it into a series of manageable single-variable integrals.
Multivariable Calculus
Multivariable calculus generalizes the concepts of single-variable calculus to functions of two or more variables. It is the mathematical study underlying our triple integral problem.
  • This branch of calculus explores limits, continuity, differentiation, and integration in multi-dimensional settings.
  • It offers tools like partial derivatives, multiple integrals, and coordinate transformations.
  • For functions of more than one variable, we often need to consider changes in one variable while keeping others constant.
In our exercise, using multivariable calculus allowed us to evaluate the triple integral over the defined solid region, calculating the accumulated continuous sum of the function values throughout the specified volume.

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Most popular questions from this chapter

In Exercises \(37-40,\) find the average value of \(F(x, y, z)\) over the given region. \(F(x, y, z)=x y z\) over the cube in the first octant bounded by the coordinate planes and the planes \(x=2, y=2,\) and \(z=2\).

Noncircular cylinder \(\quad\) A solid right (noncircular) cylinder has its base \(R\) in the \(x y\) -plane and is bounded above by the paraboloid \(z=x^{2}+y^{2} .\) The cylinder's volume is $$V=\int_{0}^{1} \int_{0}^{y}\left(x^{2}+y^{2}\right) d x d y+\int_{1}^{2} \int_{0}^{2-y}\left(x^{2}+y^{2}\right) d x d y$$ Sketch the base region \(R\) and express the cylinder's volume as a single iterated integral with the order of integration reversed. Then evaluate the integral to find the volume.

Let \(D\) be the region in the first octant that is bounded below by the cone \(\phi=\pi / 4\) and above by the sphere \(\rho=3 .\) Express the volume of \(D\) as an iterated triple integral in (a) cylindrical and (b) spherical coordinates. Then (c) find \(V\).

What domain \(D\) in space maximizes the value of the integral $$\iiint_{D}\left(1-x^{2}-y^{2}-z^{2}\right) d V ?$$ Give reasons for your answer.

Improper double integrals can often be computed similarly to improper integrals of one variable. The first iteration of the following improper integrals is conducted just as if they were proper integrals. One then evaluates an improper integral of a single variable by taking appropriate limits, as in Section \(8.7 .\) Evaluate the improper integrals as iterated integrals. $$\int_{-1}^{1} \int_{-1 / \sqrt{1-x^{2}}}^{1 / \sqrt{1-x^{2}}}(2 y+1) d y d x$$
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