Problem 28 - Thomas Calculus

Question

The region in the first octant bounded by the coordinate planes, the plane \(y=1-x,\) and the surface \(z=\cos (\pi x / 2)\) \(0 \leq x \leq 1\) (GRAPH NOT COPY)

Step-by-Step Solution

Answer

The region's volume is determined using a triple integral with specific bounds and transformations.

Step 1: Understand the Problem

We need to find the region in the first octant that is bounded by the planes and surface given by the equations. In the first octant, all coordinates \(x, y, \) and \(z\) are non-negative.

Step 2: Identify the Bounding Surfaces

The region is bounded by the plane \(y = 1 - x\) and the surface \(z = \cos(\pi x / 2)\). The coordinate planes \(x=0, y=0, z=0\) also provide boundaries.

Step 3: Determine Relationships Between Bounds

Since \(0 \leq x \leq 1\), the plane \(y = 1 - x\) implies \(0 \leq y \leq 1 - x\). The surface \(z = \cos(\pi x / 2)\) implies \(0 \leq z \leq \cos(\pi x / 2)\).

Step 4: Set Up the Integral for Volume

We will compute the volume of the region using a triple integral. The limits for \(x\) are \([0, 1]\), for \(y\) are \([0, 1-x]\), and for \(z\) are \([0, \cos(\pi x / 2)]\). The integral is set up as: \[ \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{\cos(\pi x / 2)} dz \, dy \, dx. \]

Step 5: Evaluate the Integral with Respect to \(z\)

Calculating the innermost integral with respect to \(z\), we have: \(\int_{0}^{\cos(\pi x / 2)} dz = \cos(\pi x / 2)\).

Step 6: Evaluate the Integral with Respect to \(y\)

Next, we integrate the result with respect to \(y\): \[ \int_{0}^{1-x} \cos(\pi x / 2) \, dy = (1-x) \cos(\pi x / 2). \]

Step 7: Integrate with Respect to \(x\)

Finally, integrate with respect to \(x\): \[ \int_{0}^{1} (1-x) \cos(\pi x / 2) \, dx. \] Use the substitution \(u = \pi x/2\), \(du = (\pi/2)dx\), to simplify and compute the resulting integral.

Step 8: Simplify and Calculate the Final Answer

The substitution \(u = \pi x/2\) simplifies the integral to \(\int_{0}^{\pi/2} \frac{2}{\pi} (1-\frac{2u}{\pi}) \cos(u) \, du\). Evaluate this integral to find the volume, which provides the final answer.

Key Concepts

Volume Calculation

Calculating the volume of a solid region using triple integration can be a fascinating process. Triple integration involves nested integrals, which allow us to compute the volume under a surface in three-dimensional space. In the given exercise, the region we're dealing with is described by several boundaries, including a plane and a surface. Knowing how to set up these integrals correctly is key to finding the volume. First, we determine the limits of integration, which represent the boundaries of our region in terms of the coordinate axes. In this case, we integrate over the variables from the outermost integral to the innermost: starting with the variable representing width (like in our case, x), then height (y), and finally depth (z). This particular dimension order is chosen based on our given boundaries, but it can vary. Once the integrals are established, we systematically calculate them from the innermost to the outermost, substituting where necessary to simplify the expressions. The end result is the total volume of the region we were interested in.

Bounded Region

A bounded region in three-dimensional space is defined by surfaces and planes that restrict the volume. In simple terms, these boundaries create a 'box' around the region of interest. For this exercise, the region is confined to the first octant, where all coordinate values are non-negative. The specific boundaries include the plane given by the equation \( y = 1 - x \) and the surface described by \( z = \, \cos(\pi x /2) \).
Additionally, we have the coordinate planes \( x = 0 \), \( y = 0 \), and \( z = 0 \), which further restrict the region. So, the bounds form a region with a peculiar shape influenced by trigonometric and linear functions. Being in the first octant implies that x, y, and z cannot be negative. Thus, the interplay of these boundaries determines a finite region in space, whose volume is calculated through the integration process.

Coordinate Planes

Coordinate planes form fundamental limits in the process of calculating bounded regions. The three principal coordinate planes in three-dimensional space are:

• The XY-plane, defined by \(z = 0\)
• The YZ-plane, defined by \(x = 0\)
• The XZ-plane, defined by \(y = 0\)

These planes establish the basic boundaries along each axis and are crucial in defining the extent of our region in one or more directions. In the provided problem, these planes determine a part of the region as they ensure orientation in the first octant. Serving as edges, they effectively help 'clip' the space, keeping the region finite and manageable for integration. As foundational constraints, these planes help frame the problem and limit the volume calculations.

Substitution Method

The substitution method is a powerful technique often used to simplify integrals, especially when dealing with complex or non-linear boundaries. In this exercise, the substitution helps transform the variable in the integral, making it easier to solve. For the final integral with respect to x, we used the substitution \( u = \pi x/2 \), resulting in \( du = (\pi/2)dx \). This substitute effectively translates the bounds of the integral from x-space to u-space, simplifying the integration process.
By using the substitution method, we aim to find a function form or variable that makes the integral easier to evaluate. It turns a challenging integral into a manageable one, often involving trigonometric identities or other algebraic manipulations to simplify terms. The key is to match the integral function closer to its antiderivative form, leading to easier computations and eventually the complete evaluation of the integral efficiently.