8.8 - Volumes with Cross Sections: Triangles and Semicircles

Review

V = \int_{a}^{b} A(x)\,dx

where

A(x)

is the area of one cross-section at position

x

. This lesson covers triangles and semicircles.

8.8 - Volumes with Cross Sections: Triangles and Semicircles

Key observation: the same base region produces a totally different solid depending on the cross-section shape.

Equilateral Triangle Cross-Sections

An equilateral triangle has three equal sides.

\boxed{A = \frac{\sqrt{3}}{4}s^2}

where

s

is the length of one side (any side, since they’re all the same).

This formula is not provided on the AP exam (there’s no formula sheet), so you have to memorize it.

The side length equals the distance between the boundary curves:

s = f(x) - g(x)

So:

\boxed{V = \int_{a}^{b} \frac{\sqrt{3}}{4}[f(x) - g(x)]^2\,dx}

Example 1: Equilateral Triangles Perpendicular to the x-axis

Base bounded by

y = \sqrt{x}

and

y = x^2

(from

x = 0

x = 1

), with equilateral triangle cross-sections perpendicular to the

x

-axis.

Step 1: side length

s = \sqrt{x} - x^2

Step 2: set up

V = \int_{0}^{1} \frac{\sqrt{3}}{4}(\sqrt{x} - x^2)^2\,dx = \frac{\sqrt{3}}{4}\int_{0}^{1} (\sqrt{x} - x^2)^2\,dx

Step 3: expand

(\sqrt{x} - x^2)^2 = x - 2x^{5/2} + x^4

Step 4: evaluate

V = \frac{\sqrt{3}}{4}\left[\frac{x^2}{2} - \frac{4x^{7/2}}{7} + \frac{x^5}{5}\right]_{0}^{1} = \frac{\sqrt{3}}{4}\left[\frac{1}{2} - \frac{4}{7} + \frac{1}{5}\right] = \frac{\sqrt{3}}{4} \cdot \frac{9}{70} = \frac{9\sqrt{3}}{280}

Volume:

\frac{9\sqrt{3}}{280}

cubic units

Isosceles Right Triangle Cross-Sections

An isosceles right triangle has two equal legs, a 90° angle between them, and a 45-45-90 configuration.

Important: for these problems, the equal legs (not the hypotenuse) sit on the bounded region.

\boxed{A = \frac{1}{2}s^2}

where

s

is one of the equal legs. The reasoning is simple: an isosceles right triangle is exactly half of a square with side length

s

Since

s = f(x) - g(x)

\boxed{V = \int_{a}^{b} \frac{1}{2}[f(x) - g(x)]^2\,dx}

Example 2: Isosceles Right Triangles

Same base (

y = \sqrt{x}

and

y = x^2

), with isosceles right triangle cross-sections perpendicular to the

x

-axis.

V = \int_{0}^{1} \frac{1}{2}(\sqrt{x} - x^2)^2\,dx = \frac{1}{2}\int_{0}^{1} (x - 2x^{5/2} + x^4)\,dx

= \frac{1}{2}\left[\frac{x^2}{2} - \frac{4x^{7/2}}{7} + \frac{x^5}{5}\right]_{0}^{1} = \frac{1}{2}\left[\frac{1}{2} - \frac{4}{7} + \frac{1}{5}\right] = \frac{1}{2} \cdot \frac{9}{70} = \frac{9}{140}

Volume:

\frac{9}{140}

cubic units

Semicircular Cross-Sections

A semicircle is half a circle, so:

\boxed{A = \frac{1}{2}\pi r^2}

Here’s where people mess up. The distance between the boundary curves gives you the diameter, not the radius.

\text{Diameter} = f(x) - g(x) \qquad\qquad \text{Radius} = \frac{1}{2}[f(x) - g(x)]

So:

\boxed{V = \int_{a}^{b} \frac{1}{2}\pi \left[\frac{f(x) - g(x)}{2}\right]^2\,dx} \quad = \quad \int_{a}^{b} \frac{\pi}{8}[f(x) - g(x)]^2\,dx

Where the

\frac{\pi}{8}

comes from:

\frac{1}{2}\pi \cdot \frac{1}{4}[f(x) - g(x)]^2 = \frac{\pi}{8}[f(x) - g(x)]^2

Example 3: Semicircles, Step by Step

Same base (

y = \sqrt{x}

and

y = x^2

), with semicircular cross-sections perpendicular to the

x

-axis.

Step 1: basic semicircle area

A = \frac{1}{2}\pi r^2

Step 2: identify the radius. The diameter is

\sqrt{x} - x^2

, so:

r = \frac{1}{2}(\sqrt{x} - x^2)

Step 3: substitute

A = \frac{1}{2}\pi \left[\frac{\sqrt{x} - x^2}{2}\right]^2

Step 4: simplify

A = \frac{1}{2}\pi \cdot \frac{(\sqrt{x} - x^2)^2}{4} = \frac{\pi}{8}(\sqrt{x} - x^2)^2

Step 5: set up the volume integral

V = \int_{0}^{1} \frac{\pi}{8}(\sqrt{x} - x^2)^2\,dx = \frac{\pi}{8}\int_{0}^{1} (\sqrt{x} - x^2)^2\,dx

Step 6: evaluate

V = \frac{\pi}{8}\left[\frac{x^2}{2} - \frac{4x^{7/2}}{7} + \frac{x^5}{5}\right]_{0}^{1} = \frac{\pi}{8} \cdot \frac{9}{70} = \frac{9\pi}{560}

Volume:

\frac{9\pi}{560}

cubic units

Why Semicircles Are Tricky

Error 1: using the diameter as the radius

Wrong:

V = \int_{0}^{1} \frac{\pi}{2}(\sqrt{x} - x^2)^2\,dx

This treats the full distance between the curves as the radius.

Error 2: forgetting to square the radius

Wrong:

V = \int_{0}^{1} \frac{\pi}{2} \cdot \frac{\sqrt{x} - x^2}{2}\,dx

You need

r^2

, not just

r

Correct:

V = \int_{0}^{1} \frac{\pi}{2}\left[\frac{\sqrt{x} - x^2}{2}\right]^2\,dx = \int_{0}^{1} \frac{\pi}{8}(\sqrt{x} - x^2)^2\,dx

Where Each Fraction Comes From

V = \int_{a}^{b} \frac{\pi}{8}[f(x) - g(x)]^2\,dx

The first $\frac{1}{2}$ : a semicircle is half a circle.

\text{Circle} = \pi r^2 \quad \Rightarrow \quad \text{Semicircle} = \frac{1}{2}\pi r^2

The second $\frac{1}{2}$ : the radius is half the diameter.

r = \frac{d}{2} = \frac{f(x) - g(x)}{2}

Combined:

\frac{1}{2}\pi \left[\frac{f(x) - g(x)}{2}\right]^2 = \frac{1}{2}\pi \cdot \frac{[f(x) - g(x)]^2}{4} = \frac{\pi}{8}[f(x) - g(x)]^2

Summary Table: Cross-Section Formulas

Shape	Area Formula	Volume Integral
Square	$A = s^2$	$\int_{a}^{b} [f(x) - g(x)]^2\,dx$
Equilateral triangle	$A = \frac{\sqrt{3}}{4}s^2$	$\int_{a}^{b} \frac{\sqrt{3}}{4}[f(x) - g(x)]^2\,dx$
Isosceles right triangle	$A = \frac{1}{2}s^2$	$\int_{a}^{b} \frac{1}{2}[f(x) - g(x)]^2\,dx$
Semicircle	$A = \frac{\pi}{8}d^2$	$\int_{a}^{b} \frac{\pi}{8}[f(x) - g(x)]^2\,dx$

In every case,

s

d

is the distance between the boundary curves,

f(x) - g(x)

Other Cross-Sections

Squares, rectangles, triangles, and semicircles are the usual shapes, but the method works for any cross-section you can find the area of. Write the area of one slice as a function of

x

(or

y

), then integrate it.

Isosceles right triangle with the hypotenuse in the base: now the distance between the curves is the hypotenuse $s$ , not a leg. The height to the hypotenuse is $\frac{s}{2}$ , so $A = \frac{1}{2} \cdot s \cdot \frac{s}{2} = \frac{1}{4}s^2$ . Read the problem carefully to see which side sits in the base.

Full circles: $A = \pi r^2$ . If you know the diameter $d$ , then $A = \pi\left(\frac{d}{2}\right)^2 = \frac{\pi}{4}d^2$ , twice the semicircle area.

Area given directly: if the problem hands you the cross-sectional area $A(x)$ , skip the geometry and integrate it.

Example 4: Hypotenuse in the Base

Same base (

y = \sqrt{x}

and

y = x^2

), with isosceles right triangle cross-sections perpendicular to the

x

-axis whose hypotenuse lies in the base.

V = \int_{0}^{1} \frac{1}{4}(\sqrt{x} - x^2)^2\,dx = \frac{1}{4} \cdot \frac{9}{70} = \frac{9}{280}

Volume:

\frac{9}{280}

cubic units. That’s half of Example 2’s

\frac{9}{140}

, because each triangle has half the area.

Example 5: Cross-Sectional Area Given as a Function

A solid lies along the

x

-axis from

x = 0

x = 4

, measured in centimeters. Its cross-section perpendicular to the

x

-axis at

x

has area

A(x) = 3\sqrt{x}

square centimeters. Find its volume.

V = \int_{0}^{4} 3\sqrt{x}\,dx = \left[2x^{3/2}\right]_{0}^{4} = 2(8) - 0 = 16

Volume:

16

cubic centimeters

With Respect to y

All of these work the same way for cross-sections perpendicular to the

y

-axis:

Convert equations to $x = f(y)$ form

Use $y$ -boundaries as the limits

The distance becomes $x_{\text{right}} - x_{\text{left}}$

Integrate with $dy$ instead of $dx$

For example, if the base is bounded by

x = y^2

(left) and

x = \sqrt{y}

(right) from

y = 0

y = 1

, with semicircular cross-sections:

V = \int_{0}^{1} \frac{\pi}{8}(\sqrt{y} - y^2)^2\,dy

Calculator Strategy

Set up the integral with the correct constant and squared expression

Use MATH → 9:fnInt( to evaluate

Keep $\pi$ symbolic (don’t approximate it as 3.14) until the very end

Example entry:


(π/8) * fnInt((√(X) - X²)², X, 0, 1)

Problem-Solving Checklist

Before integrating, verify: - Shape identified? (which triangle type, or semicircle) - Correct area formula, especially the constant out front? - Distance formula right? (top minus bottom, or right minus left) - Semicircle radius handled? (did you halve the diameter) - Everything squared? - Perpendicular to which axis? (determines

dx

dy

)

Common AP Exam Patterns

Very common: semicircular cross-sections perpendicular to the $x$ -axis (squares from 8.7 are the other big one)

Also common: equilateral triangles perpendicular to the $x$ -axis

Less common: isosceles right triangles, or anything with the $y$ -axis

Master semicircles first. They show up often and have the trickiest setup.

Key Takeaways

Equilateral triangles: $A = \frac{\sqrt{3}}{4}s^2$ (not given on the exam)

Isosceles right triangles: $A = \frac{1}{2}s^2$ (half a square)

Semicircles: $A = \frac{\pi}{8}d^2$ , where $d$ is the diameter

Critical for semicircles: radius = $\frac{1}{2}$ (diameter), so remember to halve the distance

Distance between curves: $s = f(x) - g(x)$ gives the side length or the diameter

Always square it. All these formulas involve $[f(x) - g(x)]^2$

Constants to remember: equilateral $\frac{\sqrt{3}}{4}$ , isosceles right $\frac{1}{2}$ , semicircle $\frac{\pi}{8}$

Perpendicular to the $y$ -axis: same formulas, just use $dy$ and convert equations

Calculator friendly: set up correctly, then use fnInt(

Other shapes: same method, $V = \int A(x)\,dx$ . A hypotenuse in the base gives $\frac{1}{4}s^2$ , and a full circle gives $\frac{\pi}{4}d^2$

Practice semicircles. They’re common and the most error-prone

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