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Writing Integrals Over Type I Regions

Set up double integrals over vertically described regions by matching the inner \(y\)-limits and outer \(x\)-limits to the region’s boundaries.

Intermediate 9 min read

What You'll Learn

  • Recognize a Type I region from its vertical slices.
  • Determine the interval of \(x\)-values covered by the region.
  • Identify the lower and upper \(y\)-boundaries for each slice.
  • Write the corresponding iterated integral in the correct order.
  • Check that boundary intersections and sample slices agree with the proposed limits.

Introduction

A double integral over a rectangle has constant limits in both variables. Over a more general planar region, one or both pairs of limits may depend on the other coordinate. To write the integral correctly, first describe the region in a way that matches the order of integration.

This tutorial focuses on Type I regions, which are described by vertical slices: choose an \(x\)-value, then let \(y\) run from a lower boundary to an upper boundary. The corresponding iterated integral has \(y\) as its inner variable and \(x\) as its outer variable.

Type I Regions and Their Limits

Definition: A bounded planar region \(D\) is Type I if it can be written as $$ D=\{(x,y):a\le x\le b,\; g_1(x)\le y\le g_2(x)\}, $$ where \(g_1\) and \(g_2\) are continuous on \([a,b]\) and \(g_1(x)\le g_2(x)\) throughout that interval. For each fixed \(x\), the vertical slice of \(D\) runs from \(y=g_1(x)\) to \(y=g_2(x)\).

The interval \([a,b]\) is the region’s horizontal projection: it contains the \(x\)-coordinates attained by points in \(D\). The functions \(g_1\) and \(g_2\) give the lower and upper edges of each vertical slice. The curves may meet at an endpoint, where a slice can have zero length.

If \(f\) is continuous on \(D\), the standard iterated-integral formula for a Type I region is

$$ \iint_D f(x,y)\,dA = \int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y)\,dy\,dx. $$

The inner integral is with respect to \(y\), so \(x\) is held fixed while the slice is traversed. The outer integral is with respect to \(x\), sweeping across the region from \(a\) to \(b\). Writing \(dy\,dx\) records that order. Here the goal is to set up the integral; the formula does not require the integral to be evaluated as part of identifying the limits.

A Systematic Setup

Begin with a sketch or a precise description of the region. Determine its leftmost and rightmost \(x\)-coordinates, then examine a vertical slice at a general \(x\) between them. Label the smaller \(y\)-value as the lower boundary and the larger as the upper boundary. This gives the inner limits; the \(x\)-range gives the outer limits.

1
Find the \(x\)-interval.
Project the region onto the \(x\)-axis to identify its left and right endpoints, \(a\) and \(b\).
2
Describe a typical vertical slice.
For a fixed \(x\), identify where the slice enters and leaves the region.
3
Assign lower and upper boundaries.
Write the smaller \(y\)-value as \(g_1(x)\) and the larger as \(g_2(x)\).
4
Write the iterated integral.
Use \(g_1(x)\) and \(g_2(x)\) as the inner \(y\)-limits, and \(a\) and \(b\) as the outer \(x\)-limits.

When the boundaries are given as curves that meet, their intersection \(x\)-coordinates may determine the endpoints of the outer interval. It is not enough to find an intersection, however: check that the proposed lower curve really lies below the proposed upper curve throughout the interval.

Worked Example: A Triangular Region

Worked Example: A Triangular Region

Set up the double integral of \(f(x,y)=x+2y\) over the region bounded by \(x=0\), \(x=3\), \(y=0\), and \(y=x\).

Step 1: Find the \(x\)-interval. The vertical sides are \(x=0\) and \(x=3\), so \(0\le x\le 3\).

Step 2: Find the vertical-slice boundaries. At each fixed \(x\) in this interval, the slice begins at the horizontal boundary \(y=0\) and ends at the slanted boundary \(y=x\). Since \(x\ge0\), the upper value \(x\) is not below \(0\).

Step 3: Write the integral.

$$ \iint_D (x+2y)\,dA = \int_0^3\int_0^x(x+2y)\,dy\,dx. $$

Check: At \(x=0\), the slice has \(0\le y\le0\), as expected at the triangle’s tip. At \(x=2\), it has \(0\le y\le2\), between the axis and the line \(y=x\). At \(x=3\), it runs from \(y=0\) to \(y=3\), the full right edge of the region. The limits describe the stated triangle.

Worked Example: The Region Between Two Curves

Worked Example: The Region Between Two Curves

Set up the double integral of \(f(x,y)=x+y\) over the region between \(y=x^2\) and \(y=2x\).

Step 1: Find the intersections. At an intersection, the two \(y\)-values are equal, so

$$ x^2=2x \quad\Longrightarrow\quad x^2-2x=0 \quad\Longrightarrow\quad x(x-2)=0. $$

Thus the intersection \(x\)-coordinates are \(0\) and \(2\). The corresponding \(y\)-coordinates are \(0\) and \(4\).

Step 2: Confirm which curve is lower. On \(0\le x\le2\), the difference between the candidate upper and lower boundaries is

$$ 2x-x^2=x(2-x)\ge0. $$

Both factors are nonnegative on this interval. Therefore \(y=x^2\) is the lower curve and \(y=2x\) is the upper curve there.

Step 3: Write the integral.

$$ \iint_D (x+y)\,dA = \int_0^2\int_{x^2}^{2x}(x+y)\,dy\,dx. $$

Check: At \(x=1\), the lower and upper values are \(1^2=1\) and \(2(1)=2\). At \(x=0\) and \(x=2\), the two curves meet. The slice therefore runs between the curves throughout \(0\le x\le2\), as required.

This example illustrates why boundary intersections and boundary ordering are separate checks. The intersections identify candidate endpoints, while the inequality \(2x-x^2\ge0\) verifies which curve supplies each limit between those endpoints.

Worked Example: A Circular Region

Worked Example: A Circular Region

Set up the double integral of \(f(x,y)=1\) over the disk \(x^2+y^2\le9\).

Step 1: Find the \(x\)-interval. Since \(y^2\ge0\), every point in the disk satisfies \(x^2\le9\). Thus \(-3\le x\le3\).

Step 2: Solve for the \(y\)-boundaries. For a fixed \(x\) in that interval, rearrange the disk inequality:

$$ x^2+y^2\le9 \quad\Longrightarrow\quad y^2\le9-x^2 \quad\Longrightarrow\quad -\sqrt{9-x^2}\le y\le\sqrt{9-x^2}. $$

The expression under the square root is nonnegative for \(-3\le x\le3\). The negative square root is the lower boundary, and the positive square root is the upper boundary.

Step 3: Write the integral.

$$ \iint_D 1\,dA = \int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}}1\,dy\,dx. $$

Check: At \(x=0\), the bounds are \(-3\le y\le3\), the full vertical diameter. At \(x=3\) or \(x=-3\), both bounds are \(0\), giving a slice of zero length at the edge of the disk. For every intermediate \(x\), the lower limit is no greater than the upper limit. These checks agree with the disk’s vertical slices.

Why the Limits Match the Slice Description

The defining description of a Type I region states that, for each \(x\) from \(a\) to \(b\), its points have \(y\)-coordinates between \(g_1(x)\) and \(g_2(x)\). In other words, fixing \(x\) produces exactly the interval used for the inner integral. Letting \(x\) range over \([a,b]\) then covers the region’s horizontal projection.

Consequently, the integral limits must reflect two different roles: the inner limits trace a single vertical slice, while the outer limits specify which slices are included. This is the standard iterated-integral rule for continuous functions on Type I regions. If the proposed limits do not include all the region’s slices, or if the inner limits reverse the slice’s lower and upper edges, the written integral does not describe the intended region.

Common Pitfall: Reversing the Inner Limits

For the region between \(y=x^2\) and \(y=2x\), a common error is to write

$$ \int_0^2\int_{2x}^{x^2}(x+y)\,dy\,dx. $$

At \(x=1\), this uses the interval \(2\le y\le1\). There are no \(y\)-values satisfying those inequalities, and the limits put the upper curve first and the lower curve second. This is not the intended vertical slice.

The correction is to put the smaller boundary first. Since \(x^2\le2x\) on \([0,2]\), the correct setup is

$$ \int_0^2\int_{x^2}^{2x}(x+y)\,dy\,dx. $$

A reliable check is to choose an interior \(x\)-value and compare the proposed limits with the actual slice. If the lower limit exceeds the upper limit, recheck the boundary order.

Practical Checks Before Finalizing

Before accepting a Type I setup, verify that the outer interval spans the entire \(x\)-projection of the region. Then check that the inner limits are functions of \(x\), that they describe the lower and upper edges, and that the lower limit does not exceed the upper one on the stated interval. Testing a typical interior slice and, where useful, the endpoint slices often exposes a mistaken boundary or an incorrect intersection.

For the three examples, the resulting forms follow the same pattern: constant outer limits with variable inner limits for the triangle; intersection-based outer limits with curve-based inner limits for the region between curves; and square-root boundaries for the disk. In every case, \(dy\) comes before \(dx\) because the limits first describe a vertical slice and then sweep across the \(x\)-interval.

Check Your Understanding

For each question, focus on how the region determines the inner and outer limits.

  1. In the definition of a Type I region, what do the outer limits \(a\) and \(b\) represent?
  2. For \(0\le x\le1\) and \(x\le y\le1\), write the iterated integral of a continuous function \(f(x,y)\).
  3. For the region between \(y=x^2\) and \(y=3x\), find the intersection \(x\)-coordinates and determine which curve is lower between them.
  4. For the disk \(x^2+y^2\le16\), identify the \(x\)-interval and the lower and upper \(y\)-boundaries.
  5. What does it indicate if the proposed lower \(y\)-limit is greater than the upper \(y\)-limit at an interior \(x\)-value?

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