Semicircles as Cross Sections (College Board AP® Calculus AB)

Study Guide

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Written by: Roger B

Reviewed by: Dan Finlay

Volumes with cross sections as semicircles

How can I find the volume of a solid with a semicircular cross section?

Use the basic concept
- 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
  - $Volume = \int_{a}^{b} A (x) d x$
You may need to create the cross sectional area function $A (x)$ based on information provided
- For example $A (x)$ may depend on the values of another function (or functions) given to you in the question
Remember that the area of a circle is $π \times {radius}^{2}$
So the area of a semicircle with radius $r$ is
- $Area = \frac{1}{2} \times π r^{2}$

Worked Example

Let $R$ be the triangular region with vertices $(0, 0)$ , $(0, 2)$ and $(4, 0)$ , as shown in the figure below.

A graph of a shaded region labeled R, where R is the triangle with vertices (0, 0), (0, 2) and (4, 0)

Region $R$ 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.

Answer:

Use $Volume = \int_{a}^{b} A (x) d x$

To define $A (x)$ , first find the equation of the line through points $(0, 2)$ and $(4, 0)$

$gradient = \frac{0 - 2}{4 - 0} = - \frac{1}{2}$

$\begin{array}{rcl} y - 0 & = & - \frac{1}{2} (x - 4) \\ y & = & 2 - \frac{1}{2} x \end{array}$

That's the diameter of each semicircle; to find the radius divide by two

$radius = \frac{(2 - \frac{1}{2} x)}{2} = 1 - \frac{1}{4} x$

At each $x$ the cross-sectional area is $\frac{1}{2} π r^{2}$

$\begin{array}{rcl} A (x) & = & \frac{1}{2} π {(1 - \frac{1}{4} x)}^{2} \\ = & \frac{1}{2} π (1 - \frac{1}{2} x + \frac{1}{16} x^{2}) \\ = & \frac{π}{32} (16 - 8 x + x^{2}) \end{array}$

Now the volume integral can be used

$\begin{array}{rcl} Volume & = & \int_{0}^{4} \frac{π}{32} (16 - 8 x + x^{2}) d x \\ = & \frac{π}{32} \int_{0}^{4} (16 - 8 x + x^{2}) d x \\ = & \frac{π}{32} {[16 x - 4 x^{2} + \frac{1}{3} x^{3}]}_{0}^{4} \\ = & \frac{π}{32} ((16 (4) - 4 {(4)}^{2} + \frac{1}{3} {(4)}^{3}) - (0)) \\ = & \frac{π}{32} (64 - 64 + \frac{64}{3}) \\ = & \frac{2 π}{3} \\ = & 2.094395 . . . . . . \end{array}$

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

2.094 units³ (to 3 decimal places)

Last updated: 22 August 2024

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