Question

Find the volume of the solid generated when the region bounded by \(y=1 / \sqrt{x(3-x)}, y=0, x=1\), and \(x=2\) is revolved about the \(x\) -axis.

Short answer

The volume is \(\frac{2}{3} \pi \ln 2\).

Step-by-step solution

Understanding the Problem

We are tasked with finding the volume of a solid formed by rotating a region around the x-axis. The boundaries of this region are given by the function \(y = \frac{1}{\sqrt{x(3-x)}}\), \(y=0\), \(x=1\), and \(x=2\).

Setting Up the Integral

The volume \(V\) of a solid of revolution generated by rotating a function \(y = f(x)\) from \(x = a\) to \(x = b\) about the x-axis is given by the formula:\[V = \pi \int_{a}^{b} [f(x)]^2 \, dx\]Here, we have \(f(x) = \frac{1}{\sqrt{x(3-x)}}\), \(a = 1\), and \(b = 2\). So, we set up the integral as:\[V = \pi \int_{1}^{2} \left( \frac{1}{\sqrt{x(3-x)}} \right)^2 \, dx\]

Simplifying the Integrand

The integrand \( \left( \frac{1}{\sqrt{x(3-x)}} \right)^2 \) simplifies to \( \frac{1}{x(3-x)} \). Therefore, the integral becomes:\[V = \pi \int_{1}^{2} \frac{1}{x(3-x)} \, dx\]

Decomposing the Fraction

Using partial fraction decomposition, we decompose the integrand:\[\frac{1}{x(3-x)} = \frac{A}{x} + \frac{B}{3-x}\]Solving for \(A\) and \(B\), we find that \(A = \frac{1}{3}\) and \(B = \frac{1}{3}\). Thus, the integral becomes:\[V = \pi \left( \frac{1}{3} \int_{1}^{2} \frac{1}{x} \, dx + \frac{1}{3} \int_{1}^{2} \frac{1}{3-x} \, dx \right)\]

Calculating Each Integral

Calculate each integral separately:\[\frac{1}{3} \int_{1}^{2} \frac{1}{x} \, dx = \frac{1}{3} [\ln|x|]_{1}^{2} = \frac{1}{3}(\ln 2 - \ln 1) = \frac{1}{3} \ln 2\]\[\frac{1}{3} \int_{1}^{2} \frac{1}{3-x} \, dx = \frac{1}{3} [-\ln|3-x|]_{1}^{2} = \frac{1}{3} (-\ln 1 + \ln 2) = \frac{1}{3} \ln 2\]

Combining the Results

Add the results of the two integrals:\[V = \pi \left( \frac{1}{3} \ln 2 + \frac{1}{3} \ln 2 \right) = \frac{2}{3} \pi \ln 2\]

Final Volume Expression

The volume of the solid generated when the region is revolved about the x-axis is:\[ V = \frac{2}{3} \pi \ln 2 \]

Key concepts

Integral Calculus

Integral calculus is a core concept used to solve problems involving areas, volumes, and more. It is centered around the idea of "summing up" small quantities to determine a whole. Take the problem of finding the volume of a solid of revolution, for instance. Think of a pancake made up of many thin disks. The volume of the whole pancake (or solid) is the sum of the volumes of these disks.

To find this volume using integral calculus, we use the definite integral. In this problem, we rotate the function around the x-axis, creating a solid. By integrating the square of the function over a certain interval, we determine the volume. The formula for the volume when revolving around the x-axis is:

• \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \)

Here, \( f(x) \) is the function describing the boundary, and \( a \) and \( b \) are the limits of integration, which are the x-values of the region we're rotating.
This approach allows us to transform the problem of calculating volumes into a manageable calculation of integrals.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify complex fractions into simpler, more manageable parts. It's particularly useful in integral calculus, where integrating over complex rational expressions can be challenging without simplification.

When faced with an integrand like \( \frac{1}{x(3-x)} \), breaking it down into partial fractions is key. This expression can be decomposed into:

• \( \frac{A}{x} + \frac{B}{3-x} \)

Here, \( A \) and \( B \) are constants that we determine by solving a simple algebraic equation resulting from equating the original fraction to the sum of partial fractions after bringing them under the same denominator.

In this problem, solving gives \( A = \frac{1}{3} \) and \( B = \frac{1}{3} \). This decomposition makes it easy to integrate the original complex fraction, because it leads to simple logarithmic integrals.

Definite Integrals

Definite integrals are a fundamental tool in calculus used to calculate exact areas under curves or the accumulated quantity over an interval. They provide the means to evaluate integrals from one specific limit to another, giving an exact numerical result.

In this volume calculation problem, after using partial fraction decomposition, the definite integrals we need are:

• \( \int_{1}^{2} \frac{1}{x} \, dx \)
• \( \int_{1}^{2} \frac{1}{3-x} \, dx \)

Each of these integrals can be evaluated straightforwardly using known antiderivatives, such as logarithmic functions. Thus, we calculate:

• \( \frac{1}{3} [\ln|x|]_{1}^{2} \)
• \( \frac{1}{3} [-\ln|3-x|]_{1}^{2} \)

These definite integrals add up to give the total volume. This precision offered by definite integrals ensures we have an exact measure of the accumulated volume as the area under the curve rotates.