Problem 18 - Calculus

Question

What is the volume of the loaf of bread which comes from rotating \(y=\sin x(0 \leqslant x \leqslant \pi)\) around the \(x\) axis?

Step-by-Step Solution

Answer

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

Step 1: Understand the problem

We are asked to find the volume of a solid generated by rotating the curve \( y = \sin x \) from \( x = 0 \) to \( x = \pi \) about the \( x \)-axis.

Step 2: Recall the formula for volume of revolution

The volume \( V \) of a solid of revolution formed by rotating a curve \( y = f(x) \) from \( x = a \) to \( x = b \) about the \( x \)-axis is given by the integral formula: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] In our case, \( f(x) = \sin x \), \( a = 0 \), and \( b = \pi \).

Step 3: Set up the integral

Substitute \( f(x) = \sin x \), \( a = 0 \), and \( b = \pi \) into the volume formula:\[ V = \pi \int_{0}^{\pi} (\sin x)^2 \, dx \]

Step 4: Simplify the integral

Use the trigonometric identity for \( \sin^2 x \): \[ \sin^2 x = \frac{1 - \cos(2x)}{2} \] Substitute this into the integral:\[ V = \pi \int_{0}^{\pi} \frac{1 - \cos(2x)}{2} \, dx \] This becomes:\[ V = \frac{\pi}{2} \int_{0}^{\pi} (1 - \cos(2x)) \, dx \]

Step 5: Evaluate the integral

Split the integral into two parts:\[ V = \frac{\pi}{2} \left(\int_{0}^{\pi} 1 \, dx - \int_{0}^{\pi} \cos(2x) \, dx\right) \] The first integral is:\[ \int_{0}^{\pi} 1 \, dx = x \Big|_{0}^{\pi} = \pi \] The second integral requires a substitution \( u = 2x \), \( du = 2 \, dx \):\[ \int_{0}^{\pi} \cos(2x) \, dx = \frac{1}{2} \int_{0}^{2\pi} \cos u \, du = \frac{1}{2} [\sin u]_{0}^{2\pi} = 0 \] Therefore:\[ V = \frac{\pi}{2} (\pi - 0) = \frac{\pi^2}{2} \]

Step 6: Conclude the solution

The volume of the solid formed by rotating \( y = \sin x \) from \( x = 0 \) to \( \pi \) around the \( x \)-axis is \( \frac{\pi^2}{2} \).

Key Concepts

Integral Calculus

Integral calculus is a fascinating branch of mathematics. It allows us to find quantities such as areas, volumes, and more.
In this exercise, we used integral calculus to find the volume of a three-dimensional shape. This shape is formed by rotating a curve around an axis. The basic idea is that while differentiation gives us rates and infinitely small changes, integration gives us the sum of infinite small quantities.
Key concepts in integral calculus include:

• Indefinite Integrals: Represent the family of all antiderivatives of a function.
• Definite Integrals: Give specific values that represent cumulative sum over a range.

In the problem, definite integrals helped calculate the volume by summing up infinitesimal disks, which together form the loaf of bread.

Trigonometric Identities

Trigonometric identities are tools that help simplify and solve problems involving trigonometric functions. These identities are crucial in mathematics, especially in calculus, because they provide alternative forms and simplifications of complex expressions.
In the context of this exercise, we made use of the identity for the square of the sine function:

• o \(\sin^2 x = \frac{1 - \cos(2x)}{2}\)

This identity played an essential role in transforming the integral into a simpler form.
Without these identities, solving such problems would be much more complicated and time-consuming. Understanding and applying trigonometric identities effectively enables more efficient problem-solving in integral calculus.

Definite Integrals

Definite integrals concern the evaluation of an integral over a specific interval and provide exact values. This contrasts with indefinite integrals, which provide a general form of antiderivatives.
In problems like calculating a volume of revolution, definite integrals serve as the backbone. They help us compute the actual size or volume based on certain bounds.

• Start by setting up the integral's limits of integration. Here, it was from 0 to \( \pi \).
• Calculate the integral precisely within these limits.

For this problem, the process of applying definite integrals involved integrating the expression derived from using trigonometric identities. Steps followed included breaking down the integral into simpler parts and utilizing key substitution techniques.
The final evaluation provided the solid volume, offering a clear, tangible representation of a mathematical concept.