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Chapter 15: Problem 28

Find the volumes of the regions. The region in the first octant bounded by the coordinate planes, the plane \(y=1-x,\) and the surface \(z=\cos (\pi x / 2), 0 \leq x \leq 1\)

Short Answer

Expert verified
The volume of the region is \( \frac{4}{\pi^2} \).

Step by step solution

01

Understand the Problem

We need to find the volume of a region in the first octant, meaning all coordinates are non-negative. The region is bounded by the coordinate planes, the plane \( y = 1 - x \), and the surface \( z = \cos(\frac{\pi x}{2}) \). The limits for \( x \) are from 0 to 1.
02

Set Up the Integral

The goal is to set up a triple integral that calculates the volume. The bounds for \( x \) are from 0 to 1. For a specific \( x \), \( y \) ranges from 0 to \( 1-x \), and \( z \) goes from 0 to \( \cos(\frac{\pi x}{2}) \). Therefore, the integral can be expressed as: \[ V = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{\cos(\frac{\pi x}{2})} dz \, dy \, dx \]
03

Integrate with Respect to z

Perform the integration with respect to \( z \): \[ \int_{0}^{\cos(\frac{\pi x}{2})} dz = \left[ z \right]_{0}^{\cos(\frac{\pi x}{2})} = \cos(\frac{\pi x}{2}) \] So, the integral becomes: \[ \int_{0}^{1} \int_{0}^{1-x} \cos(\frac{\pi x}{2}) \, dy \, dx \]
04

Integrate with Respect to y

Next, integrate with respect to \( y \): \[ \int_{0}^{1-x} \cos(\frac{\pi x}{2}) \, dy = \cos(\frac{\pi x}{2}) \cdot \int_{0}^{1-x} dy = \cos(\frac{\pi x}{2}) (1-x) \] which simplifies the volume integral to: \[ \int_{0}^{1} \cos(\frac{\pi x}{2}) (1-x) \, dx \]
05

Integrate with Respect to x

Now integrate with respect to \( x \):\[ \int_{0}^{1} \cos(\frac{\pi x}{2}) (1-x) \, dx \] This requires using integration by parts or substitution. Let \( u = 1-x \) which simplifies to: \[ V = \int_{0}^{1} (1-x) \cos(\frac{\pi x}{2}) \, dx \] This evaluates to \( \frac{4}{\pi^2} \), which is the volume of the region.

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Key Concepts

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

Triple Integral
In mathematics, a triple integral is highly useful for calculating volumes of three-dimensional regions. Think of a triple integral as a natural extension of double integrals. Instead of finding the volume under a surface, triple integrals compute volumes of a spatial region. When setting up a triple integral, it鈥檚 crucial to identify the correct bounds for each variable鈥攗sually denoted as \(x\), \(y\), and \(z\).

The integration occurs in stages, typically from the innermost integral to the outermost. Here, the integral structure helps us systematically account for the full depth, width, and height of the region in question. For example:
  • Integrating with respect to \(z\) gives the height at a given \(x, y\)-point.
  • Integrating with respect to \(y\) aggregates this height over the region's width.
  • Finally, integrating with respect to \(x\) sums over its full length.
Implementing this approach in the problem, first, we tackle \(z\), then \(y\), and ultimately \(x\), leading up to the total volume calculation.
First Octant
The first octant is a specific region in three-dimensional Cartesian coordinates. It consists of all points where \(x, y,\) and \(z\) are non-negative. This is essentially one-eighth of the entire coordinate space.

In many problems, such as the given exercise, focusing on the first octant simplifies boundary conditions because it restricts each variable to only consider values greater than or equal to zero. This significantly helps in keeping the bounds clear and reducing complexity when formulating integrals.

The first octant essentially serves as a boundary-defining feature that ensures none of the coordinates are negative, making it simpler to manage for volume calculations.
Bounded Region
A bounded region refers to a defined space in a mathematical context, which is enclosed by specific surfaces or constraints. In the exercise, the region is bounded by the coordinate planes, the plane \(y = 1-x\), and the surface \(z = \cos(\frac{\pi x}{2})\).

Understanding the boundaries is crucial here:
  • The coordinate planes (\(x=0\), \(y=0\), \(z=0\)) form natural limits along each axis.
  • The plane \(y = 1 - x\) acts as a slanted cap over the region, tracking how \(x\) and \(y\) are inter-dependent.
  • The surface \(z = \cos(\frac{\pi x}{2})\) restricts the height of the region.
Together, these boundaries define a finite three-dimensional "box" within the first octant, crucial for calculating its volume accurately using a triple integral.

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Most popular questions from this chapter

Average distance to a given point inside a disk Let \(P_{0}\) be a point inside a circle of radius \(a\) and let \(h\) denote the distance from \(P_{0}\) to the center of the circle. Let \(d\) denote the distance from an arbitrary point \(P\) to \(P_{0} .\) Find the average value of \(d^{2}\) over the region enclosed by the circle. (Hint: Simplify your work by placing the center of the circle at the origin and \(P_{0}\) on the \(x\) -axis.)

Use a CAS to change the Cartesian integrals into an equivalent polar integral and evaluate the polar integral. Perform the following steps in each exercise. \begin{equation} \begin{array}{l}{\text { a. Plot the Cartesian region of integration in the } x y \text { -plane. }} \\ {\text { b. Change each boundary curve of the Cartesian region in part }} \\ \quad {\text { (a) to its polar representation by solving its Cartesian equation }} \\ \quad {\text { for } r \text { and } \theta .} \\ {\text { c. Using the results in part (b), plot the polar region of integration }} \\ \quad {\text { in the } r \theta \text { -plane. }} \\\ {\text { d. Change the integrand from Cartesian to polar coordinates. }} \\\ \quad {\text { Determine the limits of integration from your plot in part (c) }} \\ \quad {\text { and evaluate the polar integral using the CAS integration utility.}} \end{array}\end{equation} \begin{equation}\int_{0}^{1} \int_{0}^{x / 2} \frac{x}{x^{2}+y^{2}} d y d x\end{equation}

Find the average value of \(F(x, y, z)\) over the given region. \(F(x, y, z)=x^{2}+y^{2}+z^{2}\) over the cube in the first octant bounded by the coordinate planes and the planes \(x=1, y=1\) and \(z=1\)

Finding an upper limit of an iterated integral Solve for \(a :\) $$ \int_{0}^{1} \int_{0}^{4-a-x^{2}} \int_{a}^{4-x^{2}-y} d z d y d x=\frac{4}{15} $$

Centroid of a solid semiellipsoid Assuming the result that the centroid of a solid hemisphere lies on the axis of symmetry three-eighths of the way from the base toward the top, show, by transforming the appropriate integrals, that the center of mass of a solid semiellipsoid \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)+\left(z^{2} / c^{2}\right) \leq 1, \quad z \geq 0\) lies on the \(z\) -axis three-eighths of the way from the base toward the top. (You can do this without evaluating any of the integrals.)
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