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Integrals over General Regions · Tutorial 539 of 999

Regions Between A Line And A Parabola

Find the intersections and slice bounds for regions between a line and a parabola, then express a double integral in either order.

Intermediate 9 min read

What You'll Learn

  • Solve the quadratic equation that locates a line–parabola region’s endpoints.
  • Determine which boundary curve is above or below by testing an interior point.
  • Write vertical-slice bounds for a region between two curves.
  • Convert suitable regions to horizontal-slice bounds using inverse relations.
  • Set up and check double-integral expressions in both orders.

From Two Curves to a Region

A line and a parabola can enclose a finite region when they intersect at two distinct points. To integrate over that region, first find the intersections, then determine which curve forms each edge. A vertical slice measures the region from its lower curve to its upper curve; a horizontal slice measures it from its left edge to its right edge.

The algebra finds candidate endpoints, but it does not by itself tell us which curve is above or to the left. That requires checking the region between the intersections. Once the boundaries are identified, they become the variable limits of an iterated integral.

Definition: A region between two curves is the set of points lying between their graphs over an interval whose endpoints are intersection points. For a vertical slice, the bounds are written as lower and upper \(y\)-values. For a horizontal slice, they are written as left and right \(x\)-values.

Finding the Intersections and Vertical Bounds

Suppose the boundaries are \(y=p(x)\), a parabola, and \(y=\ell(x)\), a line. Their intersection points satisfy \(p(x)=\ell(x)\). Since one equation is quadratic and the other is linear, this equality is a quadratic equation. If it has two distinct real roots \(a<b\), those roots are the horizontal endpoints of the enclosed region.

Between \(a\) and \(b\), compare the two curve values. If \(p(x)\leq \ell(x)\) throughout the interval, then the region has vertical description \(a\leq x\leq b\), \(p(x)\leq y\leq\ell(x)\). If the inequality is reversed, the lower and upper curves switch roles. A test value strictly between the endpoints determines which ordering applies, provided the curves do not intersect again inside the interval.

Result: If \(p\) and \(\ell\) are continuous, intersect at \(x=a\) and \(x=b\), and have no other intersection for \(a<x<b\), then \(p(x)-\ell(x)\) has a constant sign on \((a,b)\). Thus one interior test value identifies the lower curve and upper curve throughout the interval.

To describe the same region horizontally, the vertical boundary equations must be expressed as \(x\)-values in terms of \(y\). This may require solving the line and parabola equations for \(x\), and choosing the appropriate branch of a parabola. The resulting left and right bounds must be checked across the full range of \(y\). In the examples below, each horizontal slice is a single interval.

Worked Example: A Line Above a Parabola

Worked Example: A Line Above a Parabola

Find the intersections and describe the region bounded by \(y=x^2\) and \(y=2x\) in both slice directions. Then set up the double integral of a continuous function \(f(x,y)\) over the region.

At an intersection, \(x^2=2x\), so

$$ x^2-2x=0,\qquad x(x-2)=0. $$

Thus \(x=0\) or \(x=2\). Substitution into \(y=2x\) gives the intersection points \((0,0)\) and \((2,4)\). To determine the vertical ordering, test \(x=1\): the parabola has value \(1^2=1\), while the line has value \(2(1)=2\). The parabola is below the line between the intersections.

The vertical description is therefore

$$ D=\{(x,y):0\leq x\leq 2,\ x^2\leq y\leq 2x\}. $$

For horizontal slices, solve each equation for \(x\). The line gives \(x=y/2\). The parabola gives \(x=\pm\sqrt y\); this region lies to the right of the \(y\)-axis, so its relevant branch is \(x=\sqrt y\). The \(y\)-values range from \(0\) to \(4\), and the line is the left boundary while the parabola is the right boundary. Hence

$$ D=\{(x,y):0\leq y\leq 4,\ y/2\leq x\leq\sqrt y\}. $$

The two iterated-integral descriptions are

$$ \iint_D f(x,y)\,dA = \int_0^2\int_{x^2}^{2x} f(x,y)\,dy\,dx = \int_0^4\int_{y/2}^{\sqrt y} f(x,y)\,dx\,dy. $$

As a boundary check, at \(y=1\) the horizontal bounds are \(x=1/2\) and \(x=1\); these lie on the line and the right branch of the parabola, respectively. At the endpoint \(y=4\), both bounds give \(x=2\), the upper intersection.

Why the Integral Bounds Work

The bounds follow directly from describing which points belong to the region. For the first example, a fixed \(x\) between \(0\) and \(2\) gives a vertical segment that starts on \(y=x^2\) and ends on \(y=2x\). Thus \(x^2\leq y\leq2x\). Alternatively, a fixed \(y\) between \(0\) and \(4\) gives a horizontal segment that starts on \(x=y/2\) and ends on \(x=\sqrt y\). Thus \(y/2\leq x\leq\sqrt y\).

For a continuous function \(f\) on this bounded region, the double integral can be computed by integrating along these slices and then across the range of slices. Each description covers the same points, so both iterated integrals represent the same double integral. This is the practical basis for choosing either slice direction when both descriptions use one interval per slice.

1
Find the endpoints.
Set the line equation equal to the parabola equation and solve for the intersection coordinates.
2
Order the vertical boundaries.
Test an interior \(x\)-value to decide which curve has the smaller \(y\)-value and which has the larger one.
3
Describe horizontal slices.
Solve for \(x\) in terms of \(y\), select the branches that bound the region, and identify the full range of \(y\).
4
Check the limits.
Confirm that the bounds meet at the intersection endpoints and that a test slice stays inside the region.

Worked Example: Area in Both Orders

Worked Example: Area in Both Orders

Find the area enclosed by \(y=x^2\) and \(y=x\), using both vertical and horizontal slices.

Set the equations equal to find the intersections:

$$ x^2=x,\qquad x^2-x=0,\qquad x(x-1)=0. $$

The roots are \(x=0\) and \(x=1\), giving intersection points \((0,0)\) and \((1,1)\). At the interior test value \(x=1/2\), the parabola is \(1/4\) and the line is \(1/2\). Thus the parabola is below the line. The vertical-slice area integral is

$$ A=\int_0^1\int_{x^2}^{x}1\,dy\,dx =\int_0^1(x-x^2)\,dx =\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 =\frac12-\frac13 =\frac16. $$

For horizontal slices, \(y=x\) gives \(x=y\), while \(y=x^2\) gives \(x=\pm\sqrt y\). The enclosed region uses the nonnegative branch \(x=\sqrt y\). The horizontal range is \(0\leq y\leq1\), with \(y\leq x\leq\sqrt y\). Therefore

$$ A=\int_0^1\int_y^{\sqrt y}1\,dx\,dy =\int_0^1(\sqrt y-y)\,dy =\left[\frac{2}{3}y^{3/2}-\frac{y^2}{2}\right]_0^1 =\frac23-\frac12 =\frac16. $$

Both slice directions give area \(1/6\). The agreement checks that the bounds describe the same region. The horizontal bounds also meet at both endpoints: at \(y=0\), both are \(0\), and at \(y=1\), both are \(1\).

Worked Example: A Sideways Parabola

Worked Example: A Sideways Parabola

Describe the region between \(x=y^2\) and \(x=2y\) in both slice directions, and compute its area.

The intersections satisfy \(y^2=2y\), so

$$ y^2-2y=0,\qquad y(y-2)=0. $$

Thus \(y=0\) or \(y=2\), and substitution into \(x=2y\) gives \((0,0)\) and \((4,2)\). At \(y=1\), the parabola gives \(x=1\) and the line gives \(x=2\). The parabola is the left boundary and the line is the right boundary. Hence the horizontal description is

$$ 0\leq y\leq2,\qquad y^2\leq x\leq2y. $$

For vertical slices, solve for \(y\). The line gives \(y=x/2\); the parabola gives \(y=\pm\sqrt x\). The region lies above the \(x\)-axis, so the relevant parabola branch is \(y=\sqrt x\). Its \(x\)-range is \(0\leq x\leq4\). The vertical description is \(x/2\leq y\leq\sqrt x\), and the area is

$$ A=\int_0^4\int_{x/2}^{\sqrt x}1\,dy\,dx =\int_0^4\left(\sqrt x-\frac{x}{2}\right)dx =\left[\frac{2}{3}x^{3/2}-\frac{x^2}{4}\right]_0^4 =\frac{16}{3}-4 =\frac43. $$

Using the horizontal description gives an independent check:

$$ A=\int_0^2\int_{y^2}^{2y}1\,dx\,dy =\int_0^2(2y-y^2)\,dy =\left[y^2-\frac{y^3}{3}\right]_0^2 =4-\frac83 =\frac43. $$

The bounds meet at the intersections: at \(x=0\), \(x/2=\sqrt x=0\), and at \(x=4\), both vertical bounds equal \(2\). In the horizontal description, both bounds equal \(0\) at \(y=0\) and \(4\) at \(y=2\).

A Common Mistake: Guessing Which Curve Is Above

Solving for the intersections does not determine which curve is the upper boundary. For \(y=x^2\) and \(y=x\), a mistaken argument might say, “The parabola is above the line because it is quadratic,” and write \(\int_0^1(x^2-x)\,dx\). But at \(x=1/2\), \(x^2=1/4\) while \(x=1/2\), so that ordering is reversed. The proposed integral is negative:

$$ \int_0^1(x^2-x)\,dx =\frac13-\frac12 =-\frac16. $$

An area cannot be negative. The correction is to test an interior value, identify \(x^2\) as the lower curve and \(x\) as the upper curve, and write \(\int_0^1(x-x^2)\,dx=1/6\). A reliable slice description comes from checking the inequalities, not from the names or visual shapes of the curves.

Setting Up a Double Integral

For a continuous integrand \(f(x,y)\), replace the \(1\) in an area integral by \(f(x,y)\). If the vertical description is \(a\leq x\leq b\), \(p(x)\leq y\leq\ell(x)\), then

$$ \iint_D f(x,y)\,dA = \int_a^b\int_{p(x)}^{\ell(x)}f(x,y)\,dy\,dx. $$

If the horizontal description is \(c\leq y\leq d\), \(L(y)\leq x\leq R(y)\), then the corresponding setup is

$$ \iint_D f(x,y)\,dA = \int_c^d\int_{L(y)}^{R(y)}f(x,y)\,dx\,dy. $$

Each pair of bounds must describe the same region: the outer limits give the full range of slices, and the inner limits give the endpoints of each slice. Finding and checking those limits is the essential work in setting up the integral. It also provides a useful geometric check before any integration begins.

Check Your Understanding

Use intersections and test slices to answer these questions.

  1. For \(y=x^2\) and \(y=3x\), find both intersection points and determine which curve is above between them.
  2. For the region between \(y=x^2\) and \(y=x\), why is the right horizontal boundary \(x=\sqrt y\), rather than \(x=-\sqrt y\)?
  3. Write the vertical-slice bounds for the region between \(y=x^2\) and \(y=2x\).
  4. For \(x=y^2\) and \(x=2y\), state the horizontal-slice bounds and the range of \(y\).
  5. What interior test can help determine the upper and lower curves after finding the intersection coordinates?

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