Calculate area using horizontal slices
Strips that add up to the area
The teal region lies between the y-axis and the curve \(x = 12 - 3y^2\), from \(y = -2\) to \(y = 2\). Horizontal strips cover it.
The length of a strip is an x-distance, and its thickness is a y-distance, \(\Delta y\). The length times the thickness is the area of the strip.
Press the button to use more and thinner strips. With 2 strips, the areas add to 36. With 4 strips, they add to 33, and with 8 strips, they add to 32.25.
The sums approach 32, which is the integral of the length with respect to y.
Length in x, thickness in y
One strip lies across a region that runs from the height c to the height d. The orange segment is the length of the strip, measured along the x-axis, and dy is its thickness, measured along the y-axis.
Move the slider to change the height. Press the button to change the region. The length changes with both, but it stays an x-distance, and the thickness stays a y-distance.
The integrand is the length, and the limits c and d are heights on the y-axis.
Notation
The length of a strip is \(x_R - x_L\), the right boundary minus the left. Find an antiderivative \(F\) of the length with respect to \(y\). Then \(F(d) - F(c)\) is the area.
\[\begin{aligned} \cb{A} &= \int_{\cy{c}}^{\cy{d}} \cy{\left(x_R - x_L\right)}\,dy \\ &= F(d) - F(c) \end{aligned}\]The length is in x-units and the thickness in y-units, so if both are meters, the area is in square meters.
From \(c\) to \(d\), the right boundary must stay at or to the right of the left boundary, and both must be continuous.
Test the answer with a second calculation, such as a known area formula or vertical strips.
Use horizontal strips if the boundaries are easier to write as \(x\) in terms of \(y\). Use them also if vertical strips need two integrals.
Example 1: A triangle
The axes and the line \(x = 6 - 2y\) bound a region. Lengths are in centimeters. Find the area.
The bottom of the region is the x-axis, at \(y = 0\). The top is where the line meets the y-axis, so solve \(6 - 2y = 0\) for that height.
\[\begin{aligned} &6 - 2y = 0 \;\Longrightarrow\; y = 3, \\ &0 \le y \le 3 \end{aligned}\]The left boundary is the y-axis, so the length of a strip is \((6 - 2y) - 0\).
\[\cb{A} = \int_{0}^{3} \cy{(6 - 2y)}\,dy\]An antiderivative of \(6 - 2y\) is \(6y - y^2\). Substitute 3 and 0, and subtract.
\[\Big[6y - y^2\Big]_0^3 = (18 - 9) - 0 = 9\]The area is 9 square centimeters. As a check, the triangle formula gives half of the base 6 times the height 3, which is 9.
Example 2: A parabola and a line
Find the area between the curves \(x = y^2 - 3\) and \(x = 2y\).
Set the x-values equal to find where the curves meet. The factors give \(y = -1\) and \(y = 3\).
\[\begin{aligned} &y^2 - 3 = 2y \\ &\Longrightarrow\; (y - 3)(y + 1) = 0 \end{aligned}\]At \(y = 0\), the line has \(x = 0\) and the parabola has \(x = -3\), so the line is the right boundary. The length is the difference.
\[\begin{aligned} x_R - x_L &= 2y - (y^2 - 3) \\ &= -y^2 + 2y + 3 \end{aligned}\]A tempting wrong move: The curves meet at \((-2, -1)\) and \((6, 3)\). Use the x-values \(-2\) and \(6\) as the limits.
\[\int_{-2}^{6}\left(-y^2 + 2y + 3\right)dy\]The variable is \(y\), so the limits must be heights, and the region lies between \(y = -1\) and \(y = 3\). Outside those heights the line is left of the parabola, so the length is negative, \(-5\) at \(y = 4\), and the shading turns red. An integral from \(-2\) to \(6\) would count the red parts as negative.
Use the y-values \(-1\) and \(3\) as the limits, because they are the heights where the strips begin and end.
\[\begin{aligned} \cb{A} &= \int_{-1}^{3}\left(-y^2 + 2y + 3\right)dy \\ &= \Big[-\tfrac{y^3}{3} + y^2 + 3y\Big]_{-1}^{3} \end{aligned}\]Substitute 3 and \(-1\), and subtract.
\[9 - \left(-\tfrac{5}{3}\right) = \cb{\tfrac{32}{3}}\]With vertical strips, this region needs two integrals, because the lower boundary changes at \(x = -2\). Horizontal strips need one.
Practice
Find the area between the curves \(x = y^2 - 4\) and \(x = 4 - y^2\).
The curves meet where a strip has length 0. Read those heights from the picture. Then subtract the left formula from the right formula.
Solution
The curves meet where the x-values are equal. This gives the limits.
\[\begin{aligned} &y^2 - 4 = 4 - y^2 \\ &\Longrightarrow\; y^2 = 4 \\ &\Longrightarrow\; y = \pm 2 \end{aligned}\]At \(y = 0\), the right boundary has \(x = 4\) and the left boundary has \(x = -4\). The length is the difference.
\[\begin{aligned} x_R - x_L &= (4 - y^2) - (y^2 - 4) \\ &= 8 - 2y^2 \end{aligned}\]Integrate from \(-2\) to \(2\). The antiderivative is \(\tfrac{32}{3}\) at \(y = 2\) and \(-\tfrac{32}{3}\) at \(y = -2\), so the area is \(\tfrac{64}{3}\).
\[\begin{aligned} &\int_{-2}^{2}\left(8 - 2y^2\right)dy \\ &= \Big[8y - \tfrac{2y^3}{3}\Big]_{-2}^{2} \\ &= \cb{\tfrac{64}{3}} \end{aligned}\]
Questions
Questions 1 and 2 use the region between the y-axis and the curve \(x = y^2\), from \(y = 0\) to \(y = 3\).
What is the length of the strip at \(y = 2\)?
- 4
- 2
- 6
- 9
Answer
4
The area is \(\displaystyle\int_0^3 y^2\,dy\). What do the limits 0 and 3 measure?
- The lowest and the highest height of the region
- The shortest and the longest strip
- The left end and the right end of the top strip
- The thickness of the first and the last strip
Answer
The lowest and the highest height of the region
The curves \(x = y^2\) and \(x = 2y\) meet at \(y = 0\) and \(y = 2\), and the line is the right boundary. What is the area between them?
- \(\frac{4}{3}\)
- 4
- \(\frac{8}{3}\)
- \(-\frac{4}{3}\)
Answer
\(\frac{4}{3}\)
Summary
You integrate the strip length with respect to y from the lowest height of the region to the highest.
This works because the length is an x-distance, the thickness is a y-distance, and the sum of their products is the area.
You use it when horizontal strips need one integral, and you test the answer against a second calculation.
This is one of four skills in topic 8.5 of Which & Why: AP Calculus.