Problem 3 - Thomas Calculus in SI Units

Question

Find the volumes of the solids. The solid lies between planes perpendicular to the \(x\) -axis at \(x=-1\) and \(x=1 .\) The cross-sections perpendicular to the \(x\) -axis between these planes are squares whose bases run from the semicircle \(y=-\sqrt{1-x^{2}}\) to the semicircle \(y=\sqrt{1-x^{2}}\).

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{16}{3}\).

Step 1: Understanding the Problem

First, we need to visualize the cross-section, which is a square. The base of the square is given by the distance between the semicircles described by the equation \(y = \pm \sqrt{1-x^2}\). This forms the diameter of the semicircle.

Step 2: Calculate the Base of the Square

For a given \(x\) value, the base of the square is the vertical distance between the top and bottom semi-circle, which is \(\sqrt{1-x^2} - (-\sqrt{1-x^2}) = 2\sqrt{1-x^2}\).

Step 3: Determine the Area of the Square

Since the base also forms the side of the square, the area \(A\) of the square is given by \(A = (2\sqrt{1-x^2})^2 = 4(1-x^2)\).

Step 4: Setup the Integral for Volume

The volume \(V\) of the solid is found by integrating the area of the square cross-sections along \(x\) from -1 to 1. Thus: \[ V = \int_{-1}^{1} 4(1-x^2) \, dx \]

Step 5: Solve the Integral

Integrate the function: \[ \int_{-1}^{1} 4(1-x^2) \, dx = 4 \left[ \int_{-1}^{1} 1 \, dx - \int_{-1}^{1} x^2 \, dx \right] \]First integral: \(\int_{-1}^{1} 1 \, dx = [x]_{-1}^{1} = (1 - (-1)) = 2\).Second integral: \(\int_{-1}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{1} = \left(\frac{1}{3} - \frac{-1}{3}\right) = \frac{2}{3}\).Substitute back: \[ V = 4(2 - \frac{2}{3}) = 4(\frac{4}{3}) = \frac{16}{3}\]

Step 6: Conclusion

The volume of the solid, which has square cross-sections with bases running from \(y = -\sqrt{1-x^2}\) to \(y = \sqrt{1-x^2}\), is \(\frac{16}{3}\).

Key Concepts

Volume of Solids

When we discuss the volume of solids in calculus, we often think about how shapes extend in three-dimensional space.
In the particular problem of this exercise, we are looking at a solid object defined between two planes. To solve such a problem, we need to imagine the 3D form of the solid. Here, it is confined between the planes at \(x = -1\) and \(x = 1\).
The solid lies between these planes, stretching along the x-axis. The unique quality of this solid is that its cross-sections are squares. Understanding how to find these volumes helps in various mathematical and real-world applications. For instance:

• In engineering, designing components often requires knowing their volumes.
• In computer graphics, rendering realistic scenes relies on calculating volumes.

Grasping this concept is about visualizing the piecing together of many small sectional areas into a complete volume.

Cross-Sectional Area

The cross-sectional area is a crucial concept when finding the volume of objects that aren't straightforward shapes like cubes or spheres.
For this exercise, the cross-sections of the solid are squares.Each cross-section, or slice, is perpendicular to the x-axis and spans from the lower semicircle \(y = -\sqrt{1-x^2}\) to the upper semicircle \(y = \sqrt{1-x^2}\).
This distance provides the side length of each square.The length of the base is calculated as \(2\sqrt{1-x^2}\), meaning the entire side of the square is determined by this expression.
Squaring this side length to find the area gives us the formula \(A = (2\sqrt{1-x^2})^2 = 4(1-x^2)\).By understanding cross-sectional areas, you learn:

• How different parts of a solid contribute to its total volume.
• Why calculus is a powerful tool for measuring irregular shapes.

This insight is fundamental to mastering volume calculation in calculus.

Definite Integrals

Definite integrals play a pivotal role in finding the volume of solids with complex shapes.
In this exercise, we use a definite integral to sum up all the cross-sectional areas along the x-axis from \(x = -1\) to \(x = 1\).The integral \(\int_{-1}^{1} 4(1-x^2) \, dx\) helps us to add up the areas to find the total volume of the solid.
By breaking this into manageable parts:

• First, we calculate \(\int_{-1}^{1} 1 \, dx\), which measures the entire length.
• Then, \(\int_{-1}^{1} x^2 \, dx\) calculates the area influenced by the parabolic shape.

After integrating these parts and substituting back, we end up with the volume \(\frac{16}{3}\).
This example of integration showcases how calculus allows us to handle complex geometric problems. Definite integrals are your tool for connecting individual sections into a meaningful whole, making them an indispensable part of calculus. Understanding them is all about seeing how integration stretches calculus into solving real-world geometry issues.