Set up an integral with respect to y

Adding up horizontal strips

The teal region lies between the curve \(x = y^2 - 1\) on the left and the line \(x = 3\) on the right. Four horizontal strips cover it, and the slider selects the orange one.

Each strip is a rectangle. Its length is the right x-value minus the left x-value, and its thickness is \(\Delta y\), a small step in \(y\).

The strip at the height \(y = 0.5\) runs from \(x = -0.75\) to \(x = 3\), so its length is \(3 - (-0.75) = 3.75\). Its thickness is 1, so its area is 3.75.

Press the button to make the strips thinner. The rectangles fit the region better, and their total moves toward the area of the region, \(\tfrac{32}{3}\).

Any two boundaries

The figure shows a region between a left boundary \(x = g(y)\) and a right boundary \(x = f(y)\). The strip at the height \(y\) has the length \(f(y) - g(y)\) and the thickness \(dy\).

Drag the strip, or reshape the boundaries. The length changes, but it is still \(f(y) - g(y)\), and the area of the strip is still length times thickness.

Add all the strips from the lowest height c to the highest height d. As they become thinner, the sum becomes an integral, whose integrand is an x-distance and whose variable and limits are y-values.

Notation

1. The area of the region is the sum of the strip areas. The integral adds length times thickness over every height from \(c\) to \(d\).

   \[\cb{A} = \int_{\cy{c}}^{\cy{d}} \left(\cy{x_R - x_L}\right) dy\]

   \(x_R\) and \(x_L\) are the x-values on the right and left boundaries, so \(x_R - x_L\) is the length of a strip.

2. Write \(x_R\) as \(f(y)\) and \(x_L\) as \(g(y)\), formulas in \(y\). Then the integrand contains only \(y\), and you can evaluate the integral.

   \[\cb{A} = \int_{c}^{d} \left(\cy{f(y) - g(y)}\right) dy\]

3. The length \(f(y) - g(y)\) is not negative if the right boundary stays on the right from \(c\) to \(d\). If the boundaries do not cross there, they do not change sides, so one test height is enough.

When it fails

If you subtract in the other order, each length is negative. Each strip then counts as negative area, and the strips turn red.

For the region from the opening, between \(x = y^2 - 1\) and \(x = 3\), the integral gives \(-\tfrac{32}{3}\), not \(\tfrac{32}{3}\). An area cannot be negative, so the order is wrong.

Example 1: A parabola and a line

Set up an integral for the area between the curves \(x = 8 - y^2\) and \(x = 2y + 5\). Do not evaluate it.

1. Find the limits. The boundaries meet where their x-values are equal. With every term on one side, \(y^2 + 2y - 3 = 0\) factors.

   \[\begin{aligned} &8 - y^2 = 2y + 5 \\ &\Longrightarrow\; (y + 3)(y - 1) = 0 \end{aligned}\]

2. The limits are \(y = -3\) and \(y = 1\). Decide which curve is the right boundary. At \(y = 0\), the parabola has \(x = 8\) and the line has \(x = 5\), so the parabola is on the right.

3. The length of the strip is the right x-value minus the left x-value.

   \[\begin{aligned} x_R - x_L &= \cy{(8 - y^2) - (2y + 5)} \\ &= \cy{3 - 2y - y^2} \end{aligned}\]

4. Write the integral with the limits. At \(y = -1\), the formula gives \(3 + 2 - 1 = 4\), the length of the orange strip.

   \[\cb{A} = \int_{-3}^{1} \left(3 - 2y - y^2\right) dy\]

Example 2: A cubic written as y

The region lies between the y-axis, the curve \(y = x^3\) and the line \(y = 27\). Set up an integral for its area with horizontal strips.

1. The strip starts on the y-axis and ends on the curve. The lowest point of the region is the origin, where the curve meets the y-axis, so the heights run from 0 to 27.

   \[0 \le y \le 27\]

2. A tempting wrong move: Use the formula of the curve as the length of the strip.

   \[\int_{0}^{27} x^3 \, dy\]

   At \(y = 8\), this formula gives 8, so the red strip runs to \(x = 8\) and the boundary becomes the line \(x = y\). The curve \(y = x^3\) passes through \((2, 8)\), so the true length is 2. On the curve, \(x^3\) is the height \(y\), not a distance from the y-axis.

3. Solve the curve for \(x\). A cube root has one value, so there is no sign to choose: the right end is \(\sqrt[3]{y}\), and the left end is \(x = 0\).

   \[\begin{aligned} &y = x^3 \;\Longrightarrow\; x_R = \sqrt[3]{y}, \\ &x_L = 0 \end{aligned}\]

4. Subtract the left end, 0, and write the integral. At \(y = 8\), the formula gives \(\sqrt[3]{8} = 2\), the length of the orange strip.

   \[\cb{A} = \int_{0}^{27} \cy{\sqrt[3]{y}} \, dy\]

Practice

Set up an integral for the area between the curves \(x = y^2\) and \(x = y + 12\). Do not evaluate it.

Move the strip in the picture. Read which curve is at its right end before you subtract.

Questions

Questions 1 and 2 use the region between the line \(x = y + 4\) and the parabola \(x = y^2 - 2\).

1. What is the length of the strip at \(y = 1\)?

   • 6
   • 5
   • -6
   • 4
   Answer

   6

2. Which integral gives the area of the whole region?

   • \(\displaystyle\int_{-2}^{3}\left[(y + 4) - (y^2 - 2)\right]dy\)
   • \(\displaystyle\int_{-2}^{3}\left[(y^2 - 2) - (y + 4)\right]dy\)
   • \(\displaystyle\int_{2}^{7}\left[(y + 4) - (y^2 - 2)\right]dy\)
   • \(\displaystyle\int_{-2}^{3}\left[(y + 4) - (y^2 - 2)\right]dx\)
   Answer

   \(\displaystyle\int_{-2}^{3}\left[(y + 4) - (y^2 - 2)\right]dy\)

3. The curves \(x = 3y\) and \(x = y^2 + 2\) meet at \(y = 1\) and \(y = 2\). At \(y = 1.5\), \(3y = 4.5\) and \(y^2 + 2 = 4.25\). Which integral gives the area between them?

   • \(\displaystyle\int_{1}^{2}\left(3y - y^2 - 2\right)dy\)
   • \(\displaystyle\int_{1}^{2}\left(y^2 + 2 - 3y\right)dy\)
   • \(\displaystyle\int_{3}^{6}\left(3y - y^2 - 2\right)dy\)
   • \(\displaystyle\int_{1}^{2}\left(3y - y^2 + 2\right)dy\)
   Answer

   \(\displaystyle\int_{1}^{2}\left(3y - y^2 - 2\right)dy\)

Summary

You subtract the left x-value from the right x-value and integrate with respect to y from the lowest height to the highest.

This works because each horizontal strip has area length times thickness, and the integral adds the strips.

You use it when the boundaries are easier to write as x in terms of y.

This is one of four skills in topic 8.5 of Which & Why: AP Calculus.