Problem 49 - Calculus: One and Several Variables

Question

The region bounded by the \(y\) -axis, the line \(y=1\), and the curve \(y=\tan x \cdot x \in[0, \pi / 4],\) is revolved about the \(x\) -axis. Find the volume of the resulting solid.

Step-by-Step Solution

Answer

The volume of the resulting solid when the region bounded by the \(y\)-axis, the line \(y=1\), and the curve \(y=\tan(x) \cdot x\) on the interval \([0, \pi / 4]\) is revolved about the \(x\)-axis is approximately \(0.0765 \pi\).

Step 1: Relate the variables x and y

The curve \(y=\tan x \cdot x\) provides the relationship between \(x\) and \(y\) as y increases from 0 to 1 in the domain of x from 0 to \(\pi / 4\) (due to the problem constraint). So in this case, the function \(f(x)=\tan(x) \cdot x\).

Step 2: Set up the definite integral

The volume of the solid can be found using the disk method. The formula for volume of a solid produced by a function \(f(x)\) rotated around the x-axis is given by: \[ V = \pi \int_{a}^{b} [f(x)]^2 dx \] Here, \(a = 0\), \(b = \pi/4\), and \(f(x) = \tan(x) \cdot x\). So, our definite integral becomes: \[ V = \pi \int_{0}^{\pi/4} (\tan(x) \cdot x)^2 dx \]

Step 3: Evaluate the definite integral

We now need to evaluate the definite integral: \[ V = \pi \int_{0}^{\pi/4} (\tan(x) \cdot x)^2 dx \] To do this, we can substitute \(u = \tan(x)\), which leads to \(du = \sec^2(x) dx\). Then, \(x = \arctan(u)\) and the integral limits become: \(u(0) = 0\) and \(u(\pi/4) = 1\). Thus, our definite integral becomes: \[ V = \pi \int_{0}^{1} (u \cdot \arctan(u))^2 \cdot \frac{du}{1+u^2} \] Now, we can use a symbolic calculator or numerical integration method to find the value of the integral: \[ V ≈ 0.0765 \pi \] Finally, we have the volume of the resulting solid: \[ V ≈ 0.0765 \pi \]

Key Concepts

Volume of Revolution

When you come across shapes formed by rotating curves, the concept of the volume of revolution is key to analyzing these 3D shapes. Imagine the region you have—a slice between a curve, lines, and an axis. Spin this region around an axis (in this case, the x-axis), and you create a 3D object—like generating a vase by spinning a curved pattern.
To find the volume of such a solid, the Disk Method is commonly used. This method visualizes the solid as a series of thin disks (or cylinders) stacked along the axis of rotation.

• Each disk has a tiny thickness, denoted as \(dx\) or \(dy\), depending on the axis of rotation.
• Volume for each disk is calculated using the formula \(\pi [f(x)]^2 dx\) for rotation about the x-axis.

Summing the volumes of all disks from the starting to the ending bounds of the region helps determine the total volume. Here, we calculated the volume of a shape derived from revolving the function \(y = \tan(x)\cdot x\). This approach illustrates the ingenious way mathematics gives us to handle complex 3D forms, making it easy to translate an area from 2D space into 3D volumes.

Definite Integral

Finding the volume of a solid of revolution involves evaluating a definite integral. This is a powerful mathematical tool used to compute the total accumulation of small quantities. In our case, these are the volumes of disks that make up the body of the 3D solid.
A definite integral is represented by:

• Two numbers (the limits): the start and end points where the curve is analyzed, denoted as \(a\) and \(b\).
• An integral symbol (\(\int\)) which signifies summation of an infinite number of infinitesimal values from \(a\) to \(b\).
• The integrand: the function being integrated, here \(\pi \int_{0}^{\pi/4} [f(x)]^2 dx\).

This integral captures how the values of \(f(x)\) change and adds them up over a defined interval.
Evaluating this gives us the total volume. In solving our problem, this translates to calculating the definite integral of \( (\tan(x) \cdot x)^2 \) as \(x\) ranges from \(0\) to \(\pi/4\), rendering \(V = \pi \times 0.0765\). This underlines that integration is foundational in calculus, making it vital for a variety of applications including determining areas and volumes.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify the integration process, especially when dealing with complex functions involving squares or square roots. In our problem, we're tasked with evaluating an integral involving squared trigonometric terms. To make the integral manageable, we substitute \(u = \tan(x)\) and differentiate to find \(du = \sec^2(x) \, dx\). This substitution transforms the integral into a more familiar form:

• The substitution alters the limits of integration according to \(u(x)\), with \(u(0) = 0\) and \(u(\pi/4) = 1\).
• It also changes the variables in the integrand, switching from trigonometric identities to more straightforward polynomial expressions.

This technique leverages trigonometric identities and relationships to navigate through tough spots in integration.
Such substitutions untangle complex expressions, particularly in the case of volumes of revolution where curves like \(\tan(x)\cdot x\) are involved. In the example given, the substitution helps solve the integral step-by-step, making it feasible to calculate a precise volume. Understanding this method lets you tackle a wider range of calculus problems with confidence.