Choose bounds and slicing direction

Slices along the axis

The teal region lies between the parabola \(y = x^2\) and the line \(y = x + 2\). It turns about the x-axis.

A slice sweeps a washer only when it is perpendicular to the axis. The x-axis is horizontal, so each slice is a vertical segment. Drag the slice along the x-axis to see the washer that it sweeps.

Turn on the stack. The washers fill the solid from the left end of the region to the right end. The curves meet at both ends: at \(x = -1\) both are at height 1, and at \(x = 2\) both are at height 4.

The bounds of the volume integral are \(x = -1\) and \(x = 2\). They are \(x\)-values, because each washer sits at one \(x\) along the axis.

From one end to the other

Without numbers, the region lies between two curves that meet at \(x = a\) and \(x = b\). The orange letters mark these two ends.

Stack the washers. They fill the solid from the first slice, at \(a\), to the last slice, at \(b\). Then make them thinner.

However thin they are, the washers stay perpendicular to the axis, and the stack runs from \(a\) to \(b\). These two choices, the slicing direction and the bounds, set up every washer integral.

Notation

1. The slicing variable tells where each slice is along the axis. A horizontal axis gives vertical slices, one at each \(x\), so the slicing variable is \(x\). A vertical axis gives horizontal slices, one at each \(y\), so it is \(y\).

   \[\begin{aligned} \text{horizontal axis} &\to \cy{x} \\ \text{vertical axis} &\to \cy{y} \end{aligned}\]

2. Write both curves as functions of that variable, so the orange radii are \(\cy{R(x)}\) and \(\cy{r(x)}\). For a vertical axis, solve each curve for \(x\) in terms of \(y\).

3. The bounds are the two values of the slicing variable where the region starts and ends. Where two boundary curves meet, set their expressions equal and solve.

   \[f(s) = g(s) \;\Rightarrow\; s = \cy{a} \text{ or } \cy{b}\]

   Here \(s\) stands for the slicing variable, \(x\) or \(y\), and \(f\) and \(g\) are the two curves written in \(s\).

When it fails

The curves meet at the points \((0, 1)\) and \((4, 3)\). The heights 1 and 3 are \(y\)-values, so they are not the bounds.

The washers stack along the x-axis, so the bounds are the \(x\)-values 0 and 4. With bounds 1 and 3, the red washer at \(x = 1\) would be the first and the one at \(x = 3\) the last. The faint washers outside them would be left out.

Example 1: a horizontal axis

The region between \(y = 10 - x^2\) and \(y = x^2 + 2\) turns about the x-axis. Find the slicing variable and the bounds.

1. The axis is horizontal, so the slices are vertical and the slicing variable is \(x\). Both curves are already functions of \(x\).

2. The region ends where the curves meet. Set the two expressions equal and solve for \(x\).

   \[\begin{aligned} 10 - x^2 &= x^2 + 2 \\ x^2 &= 4 \\ x &= -2 \text{ or } x = 2 \end{aligned}\]

3. The bounds are \(-2\) and \(2\), so the washers stack from \(x = -2\) to \(x = 2\) and fill the solid.

   \[\cb{-2 \le x \le 2}\]

Example 2: a vertical axis

The region between \(y = \sqrt{x}\) and \(y = \tfrac{x}{3}\) turns about the y-axis. Find the slicing variable and the bounds.

1. The axis is vertical, so the slices are horizontal and the slicing variable is \(y\).

2. Solve each curve for \(x\), so that each one is a function of \(y\).

   \[\begin{aligned} y = \sqrt{x} &\;\Rightarrow\; x = y^2 \\ y = \tfrac{x}{3} &\;\Rightarrow\; x = 3y \end{aligned}\]

3. A tempting wrong move: The original curves meet where \(\sqrt{x} = \tfrac{x}{3}\), at \(x = 0\) and \(x = 9\), so the bounds are 0 and 9.

   \[\int_0^9 \cdots \, dy\]

   The curves do meet at \(x = 9\), but the washers stack along the y-axis, so the bounds are \(y\)-values. At \(x = 9\) the height is only \(y = 3\), so the region stops there. The red label at \(y = 9\) is far above it.

4. Set the two expressions in \(y\) equal and solve for \(y\). The bounds are 0 and 3.

   \[\begin{aligned} y^2 = 3y \;&\Rightarrow\; y(y - 3) = 0 \\ &\Rightarrow\; \cb{0 \le y \le 3} \end{aligned}\]

Practice

The region between \(y = 3x\) and \(y = x^2 + 2\) turns about the x-axis. Find the slicing variable and the bounds.

Drag the slice to each end of the region. At which \(x\)-values do the two curves have the same height?

Questions

The region between \(y = 7 - x\) and \(y = x^2 + 1\) turns about the x-axis. The curves meet at the two points shown.

1. What are the bounds of integration?

   • \(-3\) and \(2\)
   • 5 and 10
   • 0 and 2
   Answer

   \(-3\) and \(2\)

2. The region turns about the x-axis. Which slices sweep the washers?

   • Vertical slices, one for each \(x\)
   • Horizontal slices, one for each \(y\)
   • Either direction, because the solid is the same
   Answer

   Vertical slices, one for each \(x\)

3. The region between \(y = x^2\) and \(y = 4x\) turns about the y-axis. What are the bounds of integration?

   • 0 and 16
   • 0 and 4
   • 0 and 64
   Answer

   0 and 16

Summary

You cut the region perpendicular to the axis, write the curves in the variable along the axis, and find the bounds where the curves meet.

This works because every washer sits at one value of the slicing variable, so its radii and the bounds are all measured in that variable.

You use it before you write any washer integral.

This is one of four skills in topic 8.11 of Which & Why: AP Calculus.