Area With a Parameter-Dependent Intersection

Introduction

A definite integral for area depends on more than the formulas for the curves. The curves must intersect in a way that bounds a region, and the limits of integration must match those intersections. When a parameter moves one of the curves, both the endpoints and the existence of the bounded region can change.

We will examine a simple family that makes these changes visible: the parabola $y=x^2$ and the horizontal line $y=p$, where $p$ is a real parameter. The key is to analyze the intersections before writing an area integral.

Central principle: First determine the intersection points and the parameter values for which they enclose a region. Then integrate upper curve minus lower curve between the intersection points.

How the Intersections Depend on $p$

An intersection has the same $y$-value on both curves, so its $x$-coordinate must satisfy $x^2=p$. The number and location of real solutions depend on the sign of $p$.

x^2=p \quad\Longrightarrow\quad \begin{cases} x=-\sqrt{p},\ \sqrt{p}, & p>0,\\ x=0, & p=0,\\ \text{no real solution}, & p<0. \end{cases}

For $p>0$, the two curves meet at $x=-\sqrt p$ and $x=\sqrt p$. Between these points, $x^2\le p$, so the horizontal line is above the parabola. The two intersection points therefore bound a region.

At $p=0$, the curves are $y=x^2$ and $y=0$. They touch at $x=0$, but they do not enclose a positive-area region: there is only one intersection point, and the interval between intersection points has zero length. The degenerate area is zero.

For $p<0$, the equation $x^2=p$ has no real solution. The parabola satisfies $x^2\ge 0$ for every real $x$, while the line lies below the $x$-axis. These two curves do not intersect and do not enclose a bounded region.

Definition: For this family, let $A(p)$ denote the area of the bounded region enclosed by $y=x^2$ and $y=p$ when $p>0$. We assign the degenerate value $A(0)=0$. For $p<0$, there is no enclosed region between these curves, so this bounded-area function is not defined there.

Setting Up the Parameter-Dependent Integral

For $p>0$, the left and right endpoints are $-\sqrt p$ and $\sqrt p$. On this interval, the upper curve is $y=p$ and the lower curve is $y=x^2$. The area-between-curves rule gives the integral of upper minus lower over the interval.

A(p)=\int_{-\sqrt p}^{\sqrt p}(p-x^2)\,dx, \qquad p>0.

Each part of this expression comes from the geometry: the parameter determines both the height of the horizontal line and the two limits of integration. In particular, the limits are not fixed numbers as $p$ changes.

Worked Example: Finding the Area When $p=9$

Set $p=9$. The intersection equation is $x^2=9$, so $x=-3$ or $x=3$. Between these intersections, $9$ is above $x^2$. Thus

A(9)=\int_{-3}^{3}(9-x^2)\,dx.

An antiderivative of $9-x^2$ is $9x-\frac{x^3}{3}$, since differentiating gives $9-x^2$. Evaluate it at both endpoints:

\begin{aligned} A(9) &=\left[9x-\frac{x^3}{3}\right]_{-3}^{3}\\ &=\left(9(3)-\frac{3^3}{3}\right) -\left(9(-3)-\frac{(-3)^3}{3}\right)\\ &=(27-9)-(-27+9)\\ &=18-(-18)\\ &=36. \end{aligned}

Check the setup: the endpoints satisfy $x^2=9$, and the integrand $9-x^2$ is nonnegative for $-3\le x\le3$. Also, the antiderivative differentiates back to the integrand. The enclosed area is $36$ square units.

Deriving the Area Formula

The definite integral can be evaluated for any positive $p$. Since $p$ is constant with respect to $x$, an antiderivative of $p-x^2$ is $px-\frac{x^3}{3}$. Substituting the moving endpoints gives the formula.

\begin{aligned} A(p) &=\int_{-\sqrt p}^{\sqrt p}(p-x^2)\,dx\\ &=\left[px-\frac{x^3}{3}\right]_{-\sqrt p}^{\sqrt p}\\ &=\left(p\sqrt p-\frac{(\sqrt p)^3}{3}\right) -\left(-p\sqrt p-\frac{(-\sqrt p)^3}{3}\right). \end{aligned}

Because $p>0$, we have $p\sqrt p=p^{3/2}$, $(\sqrt p)^3=p^{3/2}$, and $(-\sqrt p)^3=-p^{3/2}$. Therefore, the endpoint values simplify as follows:

\begin{aligned} A(p) &=\left(p^{3/2}-\frac{p^{3/2}}{3}\right) -\left(-p^{3/2}+\frac{p^{3/2}}{3}\right)\\ &=\frac{2}{3}p^{3/2}-\left(-\frac{2}{3}p^{3/2}\right)\\ &=\frac{4}{3}p^{3/2}. \end{aligned}

Result: For the curves $y=x^2$ and $y=p$, the bounded area is $$ A(p)=\frac{4}{3}p^{3/2}\quad\text{for }p>0, $$ with degenerate value $A(0)=0$. For $p<0$, these curves have no bounded region.

This calculation also explains why the endpoints and the integrand must be considered together. Raising $p$ raises the horizontal line and moves each intersection farther from zero, changing both the vertical gap and the width of the region.

Why the Formula Matches the Geometry

The interval has width $\sqrt p-(-\sqrt p)=2\sqrt p$, while the maximum vertical gap occurs at $x=0$ and equals $p$. These observations help interpret the result, although the region is curved rather than rectangular. The integral accounts for the changing gap at every $x$ in the interval.

The formula also shows how the area scales: multiplying $p$ by a positive factor $c$ multiplies the area by $c^{3/2}$, because

A(cp)=\frac{4}{3}(cp)^{3/2} =c^{3/2}\frac{4}{3}p^{3/2} =c^{3/2}A(p), \qquad p>0,\ c>0.

This scaling follows from the formula for this particular pair of curves. It is not a rule that every parameter-dependent area must follow; the intersection equation and the resulting integral determine the relationship in each problem.

Worked Examples at Other Parameter Values

Worked Example: Finding the Area When $p=\frac14$

The intersection equation is $x^2=\frac14$, so the intersections occur at $x=-\frac12$ and $x=\frac12$. On this interval, the line $y=\frac14$ is above $y=x^2$. Set up the area integral:

A\left(\frac14\right) =\int_{-1/2}^{1/2}\left(\frac14-x^2\right)\,dx.

An antiderivative is $\frac{x}{4}-\frac{x^3}{3}$, because its derivative is $\frac14-x^2$. Evaluate at the upper endpoint and then the lower endpoint:

\begin{aligned} A\left(\frac14\right) &=\left[\frac{x}{4}-\frac{x^3}{3}\right]_{-1/2}^{1/2}\\ &=\left(\frac{1/2}{4}-\frac{(1/2)^3}{3}\right) -\left(\frac{-1/2}{4}-\frac{(-1/2)^3}{3}\right)\\ &=\left(\frac18-\frac{1}{24}\right) -\left(-\frac18+\frac{1}{24}\right)\\ &=\frac{1}{12}-\left(-\frac{1}{12}\right)\\ &=\frac16. \end{aligned}

Checking with the derived formula gives $\frac43(\frac14)^{3/2}=\frac43\cdot\frac18=\frac16$, in agreement with the direct evaluation. The intersections and integrand also confirm that this is a positive bounded area.

Worked Example: Checking the Boundary Cases $p=0$ and $p=-1$

When $p=0$, solve $x^2=0$. The only intersection is $x=0$. The curves are $y=x^2$ and $y=0$; they touch at that point, and there is no interval of positive width between intersection points. The degenerate integral is

\int_{0}^{0}(0-x^2)\,dx=0.

Thus $p=0$ gives zero degenerate area, not a positive-area enclosed region. The curves do not coincide: one is the parabola $y=x^2$, and the other is the $x$-axis.

For $p=-1$, the intersection equation would be $x^2=-1$. Since $x^2\ge0$ for every real $x$, there is no real solution. Indeed, $x^2$ is always at least $0$, whereas the line $y=-1$ is below the $x$-axis. There are no intersection limits and no bounded region between these curves.

These checks show why the formula for a positive parameter must not be used as if it described an enclosed region for every real $p$.

A Common Sign Mistake

A frequent error is to integrate lower curve minus upper curve and report the resulting negative value as an area. For $p=4$, the intersections are $x=-2$ and $x=2$. Reversing the order of the curves gives

\int_{-2}^{2}(x^2-4)\,dx =\left[\frac{x^3}{3}-4x\right]_{-2}^{2} =-\frac{32}{3}.

This is a signed integral, not the geometric area: throughout the interval, $x^2-4\le0$. The upper curve is $y=4$ and the lower curve is $y=x^2$, so the correct calculation reverses the integrand:

\int_{-2}^{2}(4-x^2)\,dx =\left[4x-\frac{x^3}{3}\right]_{-2}^{2} =\frac{32}{3}.

The correction is to determine which curve is above on the whole interval and use upper minus lower. If the curves exchange order inside the interval, the integral must instead be split at the exchange point; here they do not exchange order between their two intersections.

A Reliable Setup Process

When a parameter changes the intersections of two curves, keep the geometry and the algebra connected. The following sequence makes the main decisions explicit.

Equate the curve formulas.
Find the equation whose solutions give the intersection coordinates, and determine how its real solutions depend on the parameter.

Identify the bounded case.
Check that the intersections actually enclose a region. A single touching point or no real intersection does not give a positive-area bounded region.

Order the curves.
Determine which curve is upper and which is lower between the intersections.

Write the integral with moving limits.
Use upper minus lower and put the parameter-dependent intersection coordinates at the limits.

Evaluate and check.
Substitute both endpoints carefully, then verify that the result is nonnegative and that the antiderivative differentiates to the integrand.

Check Your Understanding

Use the curves $y=x^2$ and $y=p$ unless a question specifies otherwise.

For $p=16$, find both intersection coordinates and write the area integral, without evaluating it.
For $p=0$, what are the curves and how many intersection points do they have? Why is the degenerate area zero?
Explain why the curves have no real intersection when $p<0$.
For $p>0$, which curve is upper between the intersections, and what is the resulting area formula?
For $p=1$, use the formula to find the bounded area and check that the integral has nonnegative integrand between its limits.

Area With a Parameter-Dependent Intersection

What You'll Learn

Introduction

How the Intersections Depend on \(p\)

Setting Up the Parameter-Dependent Integral

Worked Example: Finding the Area When \(p=9\)

Deriving the Area Formula

Why the Formula Matches the Geometry

Worked Examples at Other Parameter Values

Worked Example: Finding the Area When \(p=\frac14\)

Worked Example: Checking the Boundary Cases \(p=0\) and \(p=-1\)

A Common Sign Mistake

A Reliable Setup Process

Check Your Understanding

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Area With a Parameter-Dependent Intersection

What You'll Learn

Introduction

How the Intersections Depend on \(p\)

Setting Up the Parameter-Dependent Integral

Worked Example: Finding the Area When \(p=9\)

Deriving the Area Formula

Why the Formula Matches the Geometry

Worked Examples at Other Parameter Values

Worked Example: Finding the Area When \(p=\frac14\)

Worked Example: Checking the Boundary Cases \(p=0\) and \(p=-1\)

A Common Sign Mistake

A Reliable Setup Process

Check Your Understanding

Test yourself: Calculus I final exam practice

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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