Unit 8 · Topic 8.8 Beta

Volumes with Triangle and Semicircle Cross Sections

Triangle and semicircle cross sections use the same integral as squares, V = ∫ from a to b of A(x) dx , with s = f(x) − g(x) across the base.

Practice 1: Implementing Mathematical Processes

Question set for this topic

Part 1 · Hook

Why this matters

A tent ridge, a Toblerone bar, a half-pipe: each is a solid whose slices are all the same kind of shape, triangles or half circles, growing and shrinking along its length. The method from squares carries over unchanged. Only one thing is new: the area formula for the slice.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The base of a solid is the region under $y = \sqrt{x}$ for $0 \le x \le 4$ , and its cross sections perpendicular to the $x$ -axis are squares. What is $A(x)$ ?

$x$
$\sqrt{x}$
$x^2$
$2\sqrt{x}$

Show the answer

The side is $s = \sqrt{x}$ , so $A(x) = s^2 = x$ .

Correct: $x$ :
$\sqrt{x}$ :
$x^2$ :
$2\sqrt{x}$ :

2. What is the area of a semicircle of radius $r$ ?

$\frac{1}{2}\pi r^2$
$\pi r^2$
$\frac{1}{4}\pi r^2$
$\pi r$

Show the answer

Half of a circle's area $\pi r^2$ .

Correct: $\frac{1}{2}\pi r^2$ :
$\pi r^2$ :
$\frac{1}{4}\pi r^2$ :
$\pi r$ :

Part 4 · See it

See it first

A 3D view of a solid on the region under y = √x from x = 0 to 4, lying flat on the floor. Equilateral triangles stand on the base, each perpendicular to the x-axis with one side across the base, so the triangle at x has side s = √x; one is highlighted in blue. A side panel shows the triangle with side s and height (√3/2)s, its area (√3/4)s², and the volume, the integral from 0 to 4 of (√3/4)(√x)² dx = 2√3, compared with 8 for squares on the same base. — Equilateral triangles standing on the region under $y = \sqrt{x}$ . Each has side $s = \sqrt{x}$ and area $\frac{\sqrt{3}}{4}s^2$ , so the volume is $2\sqrt{3}$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Each slice stands on a segment of the base of length $s = f(x) - g(x)$ the segment is a side of the triangle, or the diameter of the semicircle
Every slice is the same kind of shapeits area is a fixed number times $s^2$
Equilateral triangle: height $\frac{\sqrt{3}}{2}s$ $A$ $= \frac{1}{2} \cdot s \cdot \frac{\sqrt{3}}{2}s$ $= \frac{\sqrt{3}}{4}s^2$
Semicircle: radius $\frac{s}{2}$ $A = \frac{1}{2}\pi\left(\frac{s}{2}\right)^2 = \frac{\pi}{8}s^2$
Volume is $\int_a^b A(x)\,dx$ each volume is its constant times $\int_a^b s^2\,dx$

Part 6 · Key ideas

Key ideas

Triangular and semicircular cross sections: same method as squares, $V$ $= \int_a^b A(x)\,dx$ , with a new area formula.
Equilateral triangle with side $s$ : $A = \frac{\sqrt{3}}{4}s^2$ .
Isosceles right triangle: leg $s$ in the base gives $\frac{1}{2}s^2$ ; hypotenuse $s$ in the base gives $\frac{1}{4}s^2$ .
Semicircle with diameter $s$ in the base: radius $\frac{s}{2}$ , so $A = \frac{\pi}{8}s^2$ .
On one base, every volume is a constant times $\int s^2\,dx$ : squares $1$ , equilateral $0.433$ , semicircles $0.393$ , right isosceles $0.5$ or $0.25$ .

Part 7 · Misconception

A common mistake

The wrong idea: For semicircles, the segment across the base is the radius, so $A = \frac{1}{2}\pi s^2$ .

What actually happens: The segment across the base is the diameter. The radius is $\frac{s}{2}$ , and $A = \frac{1}{2}\pi\left(\frac{s}{2}\right)^2 = \frac{\pi}{8}s^2$ . Using $s$ as the radius makes the volume four times too big.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The base: the region between $y = 2x$ and $y = x^2$

1. The shaded region is the base of a solid whose cross sections perpendicular to the $x$ -axis are equilateral triangles. Given $\int_0^2 (2x - x^2)^2\,dx$ $= \frac{16}{15}$ , find the volume.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$s = 2x - x^2$ and $A(x) = \frac{\sqrt{3}}{4}s^2$ , so $V$ $= \frac{\sqrt{3}}{4}\int_0^2 (2x - x^2)^2\,dx$ $= \frac{\sqrt{3}}{4} \cdot \frac{16}{15}$ $= \frac{4\sqrt{3}}{15}$ .

Answer: $\frac{4\,\sqrt{3}}{15}$ ≈ 0.462

2. A solid on the same base has semicircular cross sections perpendicular to the $x$ -axis, each with its diameter in the base. Find the volume.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$A(x)$ $= \frac{1}{2}\pi\left(\frac{s}{2}\right)^2$ $= \frac{\pi}{8}(2x - x^2)^2$ , so $V = \frac{\pi}{8} \cdot \frac{16}{15} = \frac{2\pi}{15}$ .

Answer: $\frac{2\,\pi}{15}$ ≈ 0.419

3. A solid on the same base has cross sections perpendicular to the $x$ -axis that are isosceles right triangles with one leg in the base. What is its volume?

$\frac{8}{15}$
$\frac{4}{15}$
$\frac{16}{15}$
$\frac{32}{15}$

Show the answer

Both legs have length $s$ , so $A = \frac{1}{2}s \cdot s = \frac{s^2}{2}$ and $V = \frac{1}{2} \cdot \frac{16}{15} = \frac{8}{15}$ .

Correct: $\frac{8}{15}$ : Half of $\int s^2\,dx$ .
$\frac{4}{15}$ : $\frac{s^2}{4}$ is the area when the hypotenuse, not a leg, lies in the base.
$\frac{16}{15}$ : That is the volume with squares; a right triangle with legs $s$ is half a square.
$\frac{32}{15}$ : That doubles the square's area instead of halving it.

4. For the solid with equilateral triangle cross sections, write the area $A(x)$ of the cross section at $x$ .

Type an expression for $A(x) =$ .

Show the answer

$s = 2x - x^2$ and $A$ $= \frac{\sqrt{3}}{4}s^2$ $= \frac{\sqrt{3}}{4}(2x - x^2)^2$ .

Answer: $A(x) =$ $\frac{\sqrt{3}}{4} (2x - x^{2})^{2}$

5. What is the area of an equilateral triangle with side $s$ ?

$\frac{\sqrt{3}}{4}s^2$
$\frac{\sqrt{3}}{2}s^2$
$\frac{1}{2}s^2$
$\frac{\sqrt{3}}{4}s$

Show the answer

Its height is $\frac{\sqrt{3}}{2}s$ (from the Pythagorean theorem with half the base, $\frac{s}{2}$ ), so its area is $\frac{1}{2} \cdot s \cdot \frac{\sqrt{3}}{2}s$ .

Correct: $\frac{\sqrt{3}}{4}s^2$ : Half of base times height.
$\frac{\sqrt{3}}{2}s^2$ : That forgets the $\frac{1}{2}$ in a triangle's area.
$\frac{1}{2}s^2$ : That uses $s$ as the height; the height is shorter, $\frac{\sqrt{3}}{2}s$ .
$\frac{\sqrt{3}}{4}s$ : An area needs $s^2$ .

6. A semicircle has its diameter $s$ in the base. What is its area?

$\frac{\pi}{8}s^2$
$\frac{\pi}{2}s^2$
$\frac{\pi}{4}s^2$
$\pi s^2$

Show the answer

The radius is $\frac{s}{2}$ : $\frac{1}{2}\pi\left(\frac{s}{2}\right)^2 = \frac{\pi}{8}s^2$ .

Correct: $\frac{\pi}{8}s^2$ : Half a circle of radius $\frac{s}{2}$ .
$\frac{\pi}{2}s^2$ : That uses $s$ as the radius.
$\frac{\pi}{4}s^2$ : That is a full circle of diameter $s$ .
$\pi s^2$ : That is a full circle of radius $s$ .

7. The base of a solid is the disk $x^2 + y^2 \le 9$ . Cross sections perpendicular to the $x$ -axis are equilateral triangles. Find the volume.

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$s = 2\sqrt{9 - x^2}$ , so $A$ $= \frac{\sqrt{3}}{4} \cdot 4(9 - x^2)$ $= \sqrt{3}(9 - x^2)$ and $V$ $= \sqrt{3}\int_{-3}^{3} (9 - x^2)\,dx$ $= \sqrt{3} \cdot 36$ $= 36\sqrt{3}$ .

Answer: $36\,\sqrt{3}$ ≈ 62.354

Part 9 · Summary

Summary

Triangle and semicircle cross sections use the same integral as squares,

V

= \int_a^b A(x)\,dx

, with

s = f(x) - g(x)

across the base. Learn the areas: equilateral

\frac{\sqrt{3}}{4}s^2

, isosceles right

\frac{1}{2}s^2

(leg in base) or

\frac{1}{4}s^2

(hypotenuse in base), semicircle

\frac{\pi}{8}s^2

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The base: the region between y=2xy = 2x and y=x2y = x^2

Summary

What comes next

Connections

The base: the region between $y = 2x$ and $y = x^2$