Problem 5 - Thomas Calculus

Question

The base of a solid is the region between the curve \(y=2 \sqrt{\sin x}\) and the interval \([0, \pi]\) on the \(x\) -axis. The cross-sections perpendicular to the \(x\) -axis are a. equilateral triangles with bases running from the \(x\) -axis to the curve as shown in the accompanying figure. b. squares with bases running from the \(x\) -axis to the curve.

Step-by-Step Solution

Answer

a) Volume = \(2\sqrt{3}\); b) Volume = 8.

Step 1: Set up the integral for equilateral triangles

The area of an equilateral triangle with side length \(s\) is \( \frac{\sqrt{3}}{4} s^2 \). For cross-sections perpendicular to the x-axis, the base from the x-axis to the curve is \(s = 2 \sqrt{\sin x}\). Substitute into the area formula to get the cross-sectional area: \[ A(x) = \frac{\sqrt{3}}{4} (2\sqrt{\sin x})^2 = \sqrt{3} \sin x \].

Step 2: Integrate to find the volume for equilateral triangles

The volume of the solid is the integral of the cross-sectional area from \(x = 0\) to \(x = \pi\): \[ V = \int_{0}^{\pi} \sqrt{3} \sin x \, dx \]. Use the integral \( \int \sin x \, dx = -\cos x + C \) to solve: \[ V = \sqrt{3}(-\cos x) \bigg|_{0}^{\pi} = \sqrt{3}(-(-1) - (-1)) = 2\sqrt{3} \].

Step 3: Set up the integral for squares

The area of a square with side length \(s\) is \(s^2\). For cross-sections perpendicular to the x-axis, the side from the x-axis to the curve is \(s = 2 \sqrt{\sin x}\). Substitute into the area formula: \[ A(x) = (2\sqrt{\sin x})^2 = 4 \sin x \].

Step 4: Integrate to find the volume for squares

The volume of the solid is the integral of the cross-sectional area from \(x = 0\) to \(x = \pi\): \[ V = \int_{0}^{\pi} 4 \sin x \, dx \]. Using the integral \( \int \sin x \, dx = -\cos x + C \), compute: \[ V = 4(-\cos x) \bigg|_{0}^{\pi} = 4(-(-1) - (-1)) = 8 \].

Key Concepts

Solids of Revolution

In Calculus, solids of revolution are a fascinating concept. They involve generating a three-dimensional shape by rotating a two-dimensional area around an axis.
This creates a visually stunning solid that varies depending on the curve and the rotation method used.
Here's how it works: Imagine you have a curve on the xy-plane. To create a solid of revolution, you spin this curve around an axis - typically the x-axis or y-axis.
Doing this sweeps out a 3D shape. Think of how a potter spins clay on a wheel to form a vase.
In practical applications, solids of revolution are vital. Engineers and designers often use them to craft objects with symmetry around an axis.
Common examples include generating shapes like bowls, vases, or even components of machines.
When dealing with solids of revolution, students often encounter techniques using integrations for finding the volume of these structures. The most common methods are the disk method and the shell method, each with its unique approach to calculation based on the type of solid revolved.

Cross-Sectional Area

The concept of cross-sectional area is key when understanding the construction of solids. In our exercise, we see how cross-sectional areas help to calculate volumes of specific shapes like equilateral triangles and squares.

A cross-section is essentially a slice of the solid, which, when repeated across its length, builds up the entire volume. It is just like slicing a loaf of bread - each slice represents a cross-section.
In our example, the base of these cross-sections stretches from the x-axis up to the curve defined by the function \( y = 2 \sqrt{\sin x} \).
For equilateral triangles, the formula for calculating the area is \( \frac{\sqrt{3}}{4} s^2 \), where \( s \) is the length of the side.
However, when dealing with squares, their cross-sectional area is simply \( s^2 \). This highlights the distinctive way cross-sectional areas can be applied based on the geometric properties of the shape.
Understanding this concept helps you view solids as a sum of infinite small areas, organized one after another across the object’s length.

Definite Integrals

Definite integrals are fundamental in Calculus, giving us a way to calculate areas under curves, which applies directly to finding the volumes of solids. They allow for the accumulation of these areas over a specific interval on the axis.
The process involves setting up an integral where the limits correspond to the boundaries of the object you're studying. In our case, this is from \( x = 0 \) to \( x = \pi \).
This bounds the region over which we want to calculate the volume.

• For equilateral triangle cross-sections, the definite integral is formulated as \( \int_{0}^{\pi} \sqrt{3} \sin x \, dx \).
• For square cross-sections, it's \( \int_{0}^{\pi} 4 \sin x \, dx \).

The definite integral incorporates the cross-sectional area formula and the limits to compute a total volume.
Applying integration rules, such as \( \int \sin x \, dx = -\cos x + C \), aids in solving these integrals to find precise volume values.
Definite integrals thus transform a complex problem of finding volumes into a manageable calculation, leveraging the principles of Calculus.