Wemazi

Topic 8.5

Find the vertical limits of the region

Builds on writing each boundary as x in terms of y. 6 minutes.

Where the strip shrinks to a point

The teal region lies between the line \(x = y\) on the left and the curve \(x = 6y - y^2\) on the right. The orange strip lies across the region at the height \(y\).

Drag the strip up and down. At the lowest and at the highest height, it shrinks to a point, because the line and the curve meet there.

The region starts at \(y = 0\) and ends at \(y = 5\). These are the limits for horizontal strips, and they are y-values, not x-values.

The lowest and the highest height

The figure shows any region between a left boundary and a right boundary. The dashed orange lines mark the lowest height \(c\) and the highest height \(d\) of the region.

Press the first two buttons to move the right boundary. The dashed line at \(d\) moves too, because \(d\) stays at the point where the two boundaries meet.

Press the third button to add a flat top edge. The region now ends at that edge, so \(d\) is its height. Each limit is a y-value, from a given edge or from a point where two boundaries meet.

Notation

  1. The limits are the lowest and the highest y-values of the region. Call them \(c\) and \(d\).

    \[\int_{\cy{c}}^{\cy{d}}\]

  2. If a horizontal edge \(y = k\) bounds the region, then \(k\) is a limit. The x-axis is the line \(y = 0\).

    \[c = 0 \quad \text{for the x-axis}\]

  3. Where two boundaries meet, their x-values are equal. Set the right formula equal to the left formula, and solve for \(y\).

    \[\begin{aligned} &x_R(y) = x_L(y) \\ &\Longrightarrow\; 6y - y^2 = y \\ &\Longrightarrow\; y(5 - y) = 0 \\ &\Longrightarrow\; y = 0 \text{ or } y = 5 \end{aligned}\]

  4. Keep only the solutions on the region, because a meeting point beyond a given edge is not a limit.

    The lowest or the highest point can instead be a vertex, where a boundary turns, so look for one on every boundary.

When it fails

Press the first button to show the triangle with corners \((0, 0)\), \((6, 2)\) and \((2, 4)\). Its right boundary is the line \(x = 3y\) below the corner at \(y = 2\), and the line \(x = 10 - 2y\) above it.

One integral needs one formula for the right boundary. If you use \(x = 3y\) for the whole range, the red region is not the triangle. Split the range at the corner, with one integral from 0 to 2 and another from 2 to 4.

Example 1: A bottom edge and a rejected solution

A region in the first quadrant lies between \(x = y^2 + 1\) on the left and \(x = 7 - y\) on the right, above the x-axis. Find the limits.

  1. The x-axis bounds the region from below, so the lower limit is \(y = 0\).

  2. The curves meet where their x-values are equal. Set the formulas equal and solve for \(y\).

    \[\begin{aligned} &y^2 + 1 = 7 - y \\ &\Longrightarrow\; y^2 + y - 6 = 0 \\ &\Longrightarrow\; (y + 3)(y - 2) = 0 \end{aligned}\]

  3. The solutions are \(y = -3\) and \(y = 2\). The region lies above the x-axis, so reject \(y = -3\).

  4. The upper limit is \(y = 2\). Both formulas give \(x = 5\) there, so the top of the region is the point \((5, 2)\).

    \[\cy{0 \le y \le 2}\]

Example 2: A vertex below the meeting points

The parabola \(y = x^2 - 4\) and the line \(y = x - 2\) bound a region. Find the lowest and the highest y-value of the region.

  1. Find where the boundaries meet. Set the two formulas equal.

    \[\begin{aligned} &x^2 - 4 = x - 2 \\ &\Longrightarrow\; x^2 - x - 2 = 0 \\ &\Longrightarrow\; (x + 1)(x - 2) = 0 \end{aligned}\]

  2. The solutions are \(x = -1\) and \(x = 2\). The line gives their heights, so the meeting points are \((-1, -3)\) and \((2, 0)\).

    \[\begin{aligned} &y = -1 - 2 = -3, \\ &y = 2 - 2 = 0 \end{aligned}\]

  3. A tempting wrong move: Use the heights of the meeting points as the limits.

    \[-3 \le y \le 0\]

    The parabola turns at its vertex \((0, -4)\), below \(y = -3\). The part of the region below \(y = -3\) is not shaded, so these limits miss it.

  4. The lowest point of the whole region is the vertex \((0, -4)\), and the highest is \((2, 0)\).

    \[\cy{-4 \le y \le 0}\]

  5. The left boundary changes from the parabola to the line at \(y = -3\). This height splits an integral into two parts, but it is not a limit.

Practice

The parabola \(y = 4x - x^2\) and the line \(y = x\) bound a region. Find the lowest and the highest y-value of the region.

Find the highest point of each boundary. A parabola that opens downward turns at its highest point, the vertex.

Solution

  1. Set the formulas equal. The solutions are \(x = 0\) and \(x = 3\), so the meeting points are \((0, 0)\) and \((3, 3)\).

    \[4x - x^2 = x \;\Longrightarrow\; x(3 - x) = 0\]

  2. The parabola crosses the x-axis at 0 and 4, so its vertex is midway, at \(x = 2\), where \(y = 4\). This is higher than both meeting points.

    \[4(2) - 2^2 = 4\]

  3. The lowest point is \((0, 0)\), and the highest point is the vertex. The limits are 0 and 4.

    \[0 \le y \le 4\]

Questions

Questions 1 and 2 use the region between the curve \(y = 9 - x^2\) and the line \(y = x + 3\), to the right of the y-axis.

  1. What is the lowest y-value of the region?

    • 3
    • 0
    • 5
    • 9
    Answer

    3

  2. At what height does the right boundary change from the line to the curve?

    • 5
    • 2
    • 3
    • 9
    Answer

    5

  3. The curves \(x = y^2\) and \(x = 2y + 8\) bound a region. Which limits do horizontal strips use?

    • \(-2 \le y \le 4\)
    • \(4 \le y \le 16\)
    • \(-4 \le y \le 2\)
    • \(0 \le y \le 4\)
    Answer

    \(-2 \le y \le 4\)

Summary

You find the lowest and the highest y-value of the region, from a given edge or from the height where two boundaries meet.

This works because every horizontal strip of the region lies between these two heights, and a strip shrinks to a point where the boundaries meet.

You use these limits before you integrate with respect to y, and you look for a vertex beyond the meeting points.

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