Distinguish arc length from area and perimeter
Three measurements of one region
The curve is \(y = \tfrac23 x^{3/2}\) from \(x = 0\) to \(x = 3\). The curve, the x-axis and the vertical line \(x = 3\) enclose a region.
Press Measure the curve. The curve turns orange. Its arc length, the length of a string laid along it, is \(\tfrac{14}{3} \approx 4.67\) units.
Press Measure the region. The region turns teal, and its area is about 4.16 square units. Area counts squares, so it is not a length.
Press Measure the whole boundary. The boundary is the curve, the base of length 3 and the vertical side of length \(2\sqrt{3} \approx 3.46\). The perimeter is their total, about 11.13 units.
Which pieces does the question include?
The curve is now any smooth curve \(y = f(x)\) from \(x = a\) to \(x = b\), above the x-axis.
Raise the curve with the slider. The arc length stays the same, because raising the curve does not change its slope at any point. The area and the vertical sides grow.
The arc length measures the curve only, and the area measures the region inside. The perimeter adds the curve and all the straight pieces that close the region.
The words of the question decide which of these pieces count.
Notation
The arc length uses the slope \(f'(x)\) and measures the curve only, in units of length.
\[\cy{L} = \int_a^b \sqrt{1 + f'(x)^2}\,dx\]The area uses the height \(f(x)\) and measures the region between the curve and the x-axis, in square units.
\[\cb{A} = \int_a^b f(x)\,dx\]The perimeter is the arc length plus the closing pieces: the left side, the base and the right side.
\[\cy{P} = L + f(a) + (b - a) + f(b)\]Add the closing pieces only when the question asks for the boundary of a region. A closing piece has length 0 when the curve touches the x-axis at that end.
The area formula needs \(f(x) \ge 0\) on \([a, b]\) for the integral to equal the area.
When it fails
Lower the curve by 2, so that part of it lies below the x-axis. The integral counts the red part as negative. For \(y = \tfrac23 x^{3/2} - 2\) it is about \(-1.84\), which cannot be an area.
The arc length is still \(\tfrac{14}{3} \approx 4.67\), because lowering the curve does not change its slope. A length is never negative.
Example 1: a straight line
A region lies under the line \(y = \tfrac34 x\) from \(x = 0\) to \(x = 4\), above the x-axis. Find the arc length, the area and the perimeter.
The slope is \(\tfrac34\) at every point, so the root is the constant \(\tfrac54\).
\[\begin{aligned} L &= \int_0^4 \sqrt{1 + \tfrac{9}{16}}\,dx \\ &= \int_0^4 \tfrac54\,dx = 5 \end{aligned}\]As a test, the line is the hypotenuse of a right triangle with legs 4 and 3, and \(\sqrt{4^2 + 3^2} = 5\).
The area is the integral of the height. The triangle formula \(\tfrac12 \cdot 4 \cdot 3\) gives the same 6.
\[A = \int_0^4 \tfrac34 x\,dx = 6\]For the perimeter, add the base 4 and the vertical side 3. The left side has length 0, because the line starts on the axis.
\[P = 5 + 4 + 3 = 12\]
Example 2: a perimeter
A region lies under \(y = \tfrac13\left(x^2 + 2\right)^{3/2}\) from \(x = 0\) to \(x = 1\), above the x-axis. Find its perimeter.
The perimeter is the whole boundary: the curve, the base and both vertical sides, because the curve does not touch the axis at either end.
Find the arc length. Here \(1 + f'(x)^2\) is a perfect square, so the root is \(x^2 + 1\).
\[\begin{aligned} f'(x) &= x\sqrt{x^2 + 2} \\ 1 + f'(x)^2 &= 1 + x^2(x^2 + 2) \\ &= (x^2 + 1)^2 \\ L &= \int_0^1 (x^2 + 1)\,dx = \tfrac43 \end{aligned}\]A tempting wrong move: The arc length is a length, so the perimeter is \(\tfrac43\).
\[P = \tfrac43\]The orange path along the curve is not a closed loop. The red base and the red vertical sides are also part of the boundary, and the base alone has length 1.
Add the left side \(f(0) = \tfrac13 \cdot 2^{3/2} = \tfrac{2\sqrt2}{3}\), the base 1 and the right side \(f(1) = \tfrac13 \cdot 3^{3/2} = \sqrt3\).
\[P = \tfrac43 + \tfrac{2\sqrt2}{3} + 1 + \sqrt3 \approx 5.008\]
Practice
A region lies under \(y = \sqrt{4 - x^2}\), the upper half of a circle of radius 2, above the x-axis. Find the arc length, the area and the perimeter.
The curve is half a circle of radius 2, and its base is a diameter.
Solution
The arc length is half the circumference of the circle.
\[L = \tfrac12 \cdot 2\pi \cdot 2 = 2\pi\]The area is half the area of the disk. It equals the arc length in number, but it is in square units.
\[A = \tfrac12 \pi \cdot 2^2 = 2\pi\]The perimeter adds the base, which is the diameter 4. Both vertical sides have length 0, because the curve meets the axis at both ends.
\[P = 2\pi + 4\]
Questions
The graph shows the region under \(y = \sqrt{x}\) from \(x = 0\) to \(x = 9\). The labels give the arc length, the area, the base and the vertical side.
What is the perimeter of the region?
- 21.75
- 9.75
- 12
- 39.75
Answer
21.75
A student integrates \(\sqrt{x}\) from 0 to 9 and finds 18. Which quantity is this?
- The area of the region
- The arc length of the curve
- The perimeter of the region
Answer
The area of the region
A region lies under \(y = 2x\) from \(x = 0\) to \(x = 3\), above the x-axis. What is its perimeter?
- \(9 + 3\sqrt5\)
- \(3\sqrt5\)
- 9
- \(6 + 3\sqrt5\)
Answer
\(9 + 3\sqrt5\)
Summary
The arc length is the integral of the root, which uses the slope, and the area is the integral of the height.
A perimeter adds the straight pieces that close the region to the arc length, so the three measures differ.
Add the closing pieces only when the question asks for the perimeter of a region.
This is one of four skills in topic 8.13 of Which & Why: AP Calculus.