Express the boundaries in terms of y

Builds on finding area between curves with vertical strips. 6 minutes.

A strip across a region

The teal region lies between the parabola \(y = x^2\) and the line \(y = 3x\), and the orange strip is a thin horizontal slice of it.

Drag the strip up and down. Each end stays on one boundary, so the length of the strip is the difference of the x-values of its ends.

At the height \(y = 4\), the line is at \(x = \tfrac{4}{3}\) and the parabola is at \(x = 2\).

To find these x-values at any height, solve each equation for \(x\). The line gives \(x = \tfrac{y}{3}\), and the parabola gives \(x = \sqrt{y}\).

The parabola also allows \(x = -\sqrt{y}\), but that point lies left of the y-axis, outside the region.

Any region between two boundaries

This region has a left boundary and a right boundary. Drag the strip up and down.

At each height, the left end of the strip is at \(x = g(y)\) and the right end is at \(x = f(y)\), so the strip has length \(f(y) - g(y)\).

Press the button to reshape the curves. The strip still ends on both boundaries, because each boundary has one x-value at each height.

Notation

Call the x-values of the two ends of the strip \(x_L\) and \(x_R\), where the left end has the smaller x-value.
\[\cy{x_L}\ \text{and}\ \cy{x_R}\]
To write a boundary in terms of \(y\), solve its equation for \(x\). The line gives the left end.
\[y = 3x \;\Longrightarrow\; \cy{x_L = \tfrac{y}{3}}\]
Solving \(y = x^2\) gives two values of \(x\), because both \(\sqrt{y}\) and \(-\sqrt{y}\) square to \(y\). Each value is a branch, which is one half of the parabola.
\[y = x^2 \;\Longrightarrow\; x = \sqrt{y}\ \text{or}\ x = -\sqrt{y}\]
Choose the branch that the region borders. The region lies right of the y-axis, so \(x\) is positive.
\[\cy{x_R = \sqrt{y}}\]
As a check, the point \((2, 4)\) lies on the parabola, and \(\sqrt{4} = 2\).

When it fails

The region inside the parabola \(y = x^2\), below \(y = 4\), touches the parabola on both sides of the y-axis. At \(y = 1\), the ends of the strip are at \(x = -1\) and \(x = 1\).

The left boundary is the branch \(x = -\sqrt{y}\), and the right boundary is \(x = \sqrt{y}\). If you use \(\sqrt{y}\) for both ends, the strip shrinks to length 0, because both ends are at the same point.

Example 1: A logarithm

A region lies between the y-axis and the curve \(y = \ln x\), from the height \(y = 0\) to the height \(y = 2\). Write each boundary as \(x\) in terms of \(y\).

The left boundary is the y-axis, which is the line \(x = 0\).
\[\cy{x_L = 0}\]
The right boundary is the curve. Because \(e^{\ln x} = x\), raise \(e\) to the power of each side to undo the logarithm.
\[\begin{aligned} &y = \ln x \;\Longrightarrow\; e^{y} = x \\ &\Longrightarrow\; \cy{x_R = e^{y}} \end{aligned}\]
As a check, the curve passes through \((e, 1)\), and \(e^{1} = e\).
For every \(y\), \(e^{y} > 0\), so the curve lies right of the y-axis and is the right boundary.

Example 2: The left half of a parabola

A region lies between the y-axis, the x-axis and the curve \(y = (x - 4)^2\), left of the vertex \((4, 0)\), which is the lowest point of the parabola. Write the right boundary as \(x\) in terms of \(y\).

The strip starts on the y-axis, so \(x_L = 0\), and its right end lies on the curve.
\[\cy{x_L = 0}\]
Take the square root of both sides. There are two signs, because \(x - 4\) can be positive or negative.
\[x - 4 = \pm\sqrt{y}\]
A tempting wrong move: Use the plus sign, because a square root symbol means a positive number.
\[x_R = 4 + \sqrt{y}\]
At \(y = 1\), this gives \(x = 5\), on the other half of the parabola. The strip and the shading turn red, because the region in the statement lies left of the vertex.
The region lies left of the vertex, so \(x < 4\) and \(x - 4\) is negative. Use the minus sign. At \(y = 1\), this gives \(x = 3\), and \((3 - 4)^2 = 1\).
\[x - 4 = -\sqrt{y} \;\Longrightarrow\; \cy{x_R = 4 - \sqrt{y}}\]

Practice

The curve \(y = x^3\) and the line \(y = 4x\) bound a region in the first quadrant. Write both boundaries as \(x\) in terms of \(y\). Say which one is the left boundary.

Compare the two x-values at one height, such as \(y = 1\).

Solution

To solve the line for \(x\), divide both sides by 4.
\[y = 4x \;\Longrightarrow\; x = \tfrac{y}{4}\]
To solve the curve \(y = x^3\) for \(x\), take the cube root. A cube root has one real value, so there is no second branch.
\[y = x^3 \;\Longrightarrow\; x = \sqrt[3]{y}\]
At \(y = 1\), the line is at \(x = \tfrac{1}{4}\) and the curve is at \(x = 1\). The line has the smaller x-value, so it is the left boundary.
\[x_L = \tfrac{y}{4}, \qquad x_R = \sqrt[3]{y}\]

Questions

Questions 1 and 2 use the disk bounded by the circle \(x^2 + y^2 = 25\).

At \(y = 4\), the right end of the strip is at \(x = 3\). Which formula gives the right end at every height?
- \(x = \sqrt{25 - y^2}\)
- \(x = -\sqrt{25 - y^2}\)
- \(x = 25 - y^2\)
- \(x = 5 - y\)
Answer
\(x = \sqrt{25 - y^2}\)
At \(y = -3\), what are the x-values of the two ends of the strip?
- \(-4\) and \(4\)
- \(4\) and \(4\)
- \(-3\) and \(3\)
- \(0\) and \(4\)
Answer
\(-4\) and \(4\)
A region lies left of the vertex of \(y = (x + 2)^2\), where \(x \le -2\). Which equation gives this part of the curve?
- \(x = -2 - \sqrt{y}\)
- \(x = -2 + \sqrt{y}\)
- \(x = 2 - \sqrt{y}\)
- \(x = y^2 - 2\)
Answer
\(x = -2 - \sqrt{y}\)

Summary

Each end of a horizontal strip lies on a boundary, so solving each boundary equation for \(x\) gives the ends at any height.

If solving gives two branches, keep the one the region borders, because each branch is one half of the curve.

You do this before you integrate with horizontal strips, because the integral adds the strip lengths \(x_R - x_L\).

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