Volumes with Cross Sections

Introduction In this guide, we will explore how to find the volume of a solid with semicircular cross sections. This involves integrating the area of the cross-section along the axis of the solid.

#Volume of Solids with Semicircular Cross Sections

#Basic Concept

To find the volume of a solid with a given cross-sectional area:

• If the area of the cross section of a solid is given by $A(x)$ and $A(x)$ is continuous on $[a, b]$, then the volume of the corresponding solid from $x = a$ to $x = b$ is: $$\mathrm{Volume} = \int_{a}^{b} A(x) , dx$$

#Creating the Cross-Sectional Area Function

You may need to create the cross-sectional area function $A(x)$ based on the information provided in the problem. Often, $A(x)$ may depend on the values of another function (or functions) given in the question.

#Area of a Semicircle

The area of a circle is $\pi \times \text{radius}^2$. Therefore, the area of a semicircle with radius $r$ is: $$\mathrm{Area} = \frac{1}{2} \times \pi r^2$$

#Worked Example

Let $R$ be the triangular region with vertices $(0, 0)$, $(0, 2)$, and $(4, 0)$. $R$ is the base of a solid. For the solid, at each $x$, the cross-section perpendicular to the $x$-axis is a semicircle. Find the volume of the solid.

#Solution Steps

1. Find the Equation of the Line:

   • The line passes through points $(0, 2)$ and $(4, 0)$.
   • Slope (gradient) is: $$\text{gradient} = \frac{0-2}{4-0} = -\frac{1}{2}$$
   • Equation of the line: $$y - 0 = -\frac{1}{2}(x - 4) \implies y = 2 - \frac{1}{2}x$$

2. Determine the Radius:

   • The diameter of each semicircle is $2 - \frac{1}{2}x$.
   • Therefore, the radius is: $$\mathrm{radius} = \frac{2 - \frac{1}{2}x}{2} = 1 - \frac{1}{4}x$$

3. Find the Cross-Sectional Area $A(x)$:

   • Area of a semicircle: $$A(x) = \frac{1}{2} \pi \left(1 - \frac{1}{4}x\right)^2$$
   • Simplify: $$A(x) = \frac{1}{2} \pi \left(1 - \frac{1}{2}x + \frac{1}{16}x^2\right) = \frac{\pi}{32} \left(16 - 8x + x^2\right)$$

4. Calculate the Volume:

   • Integrate $A(x)$ from 0 to 4: $$\mathrm{Volume} = \int_{0}^{4} \frac{\pi}{32} \left(16 - 8x + x^2\right) , dx$$
   • Evaluate the integral: $$\mathrm{Volume} = \frac{\pi}{32} \left[ 16x - 4x^2 + \frac{1}{3}x^3 \right]_{0}^{4}$$
   • Substitute and calculate: $$\mathrm{Volume} = \frac{\pi}{32} \left( 64 - 64 + \frac{64}{3} \right) = \frac{\pi}{32} \cdot \frac{64}{3} = \frac{2\pi}{3}$$
   • Numerical value: $$\mathrm{Volume} \approx 2.094 , \text{units}^3$$

The question doesn't specify units, so the units of volume will be $\text{units}^3$.

#Practice Questions

Practice Question

1. Find the volume of a solid with a base region defined by $y = \sqrt{x}$, from $x=0$ to $x=4$, with cross sections perpendicular to the $x$-axis that are semicircles.

Practice Question

2. The base of a solid is a rectangle with length 6 and width 2. Cross sections perpendicular to the length are semicircles. Calculate the volume of the solid.

Practice Question

3. A solid has a base defined by the region $y = x^2$ and $y = 4$. Cross sections perpendicular to the $y$-axis are semicircles. Determine the volume of the solid.

#Glossary

• Cross-Sectional Area ($A(x)$): The area of the shape obtained by cutting through a solid perpendicular to a given axis at position $x$.
• Integral: A mathematical operation that calculates the area under a curve.
• Semicircle: A half of a circle.
• Radius: The distance from the center of a circle to any point on its circumference.
• Diameter: The distance across a circle through its center, equal to twice the radius.

#Summary and Key Takeaways

#Summary

• To find the volume of a solid with semicircular cross sections, use the integral of the area function $A(x)$.
• The area of a semicircle is $\frac{1}{2} \pi r^2$.
• The integral is evaluated over the range of $x$ values that define the solid.

#Key Takeaways

• Understand the formula: $\mathrm{Volume} = \int_{a}^{b} A(x) , dx$
• Identify the radius: Based on the geometry of the problem.
• Set up the integral: Use the correct limits and area function.
• Evaluate the integral: Carefully substitute and simplify to find the volume.