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Writing Integrals Over Type II Regions

Translate horizontal slices into the correct order of integration and write double integrals over Type II regions.

Intermediate 10 min read

What You'll Learn

  • Define Type II regions with lower and upper \(y\)-limits and left and right \(x\)-boundaries.
  • Match the inner \(x\)-limits to a horizontal slice and the outer \(y\)-limits to the full vertical range.
  • Find outer limits from boundary intersections or extreme \(y\)-values.
  • Check whether the proposed left and right boundaries stay in the correct order.
  • Set up and verify iterated integrals over triangular and curved regions.

Horizontal Slices and Type II Regions

A double integral over a nonrectangular region requires limits that describe the region accurately. One useful description starts with horizontal slices: fix a height \(y\), then identify the leftmost and rightmost \(x\)-coordinates in the region at that height. These slices lead to a Type II iterated integral, in which \(x\) is integrated first and \(y\) second.

The key is to match each limit to its role. The outer limits give the full range of heights in the region. For each fixed height in that range, the inner limits give the horizontal interval covered by the region.

Definition: A Type II region has a description of the form \(D=\{(x,y):c\le y\le d,\ g_1(y)\le x\le g_2(y)\}\), where \(g_1\) and \(g_2\) are continuous on \([c,d]\) and \(g_1(y)\le g_2(y)\) throughout that interval. Here, \(g_1(y)\) and \(g_2(y)\) are the left and right boundaries of a horizontal slice.

If \(f\) is continuous on a bounded Type II region \(D\), its double integral can be written as the iterated integral

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

The differential \(dx\) appears first because \(x\) varies along each horizontal slice. The differential \(dy\) appears second because the slices are taken from the bottom of the region to the top. The continuity and boundary conditions in the definition ensure that these limits describe one interval in \(x\) for every \(y\) from \(c\) to \(d\). If a region does not have one such interval at every height, this single Type II description does not apply to the whole region.

Why the Limits Match the Region

The bounds in the definition are a direct translation of the region's inequalities. The condition \(c\le y\le d\) says which horizontal levels occur. At any one of those levels, \(g_1(y)\le x\le g_2(y)\) says that \(x\) begins at the left boundary and ends at the right boundary.

This translation works in both directions. If \((x,y)\) belongs to the described region, then its height satisfies \(c\le y\le d\), and its horizontal position satisfies \(g_1(y)\le x\le g_2(y)\). Conversely, any pair \((x,y)\) satisfying those inequalities lies between the two boundaries at a height in the stated range. Thus the inequalities describe exactly the intended region, not merely a subset of it.

For a continuous integrand, the standard iterated-integral formula above then uses those inequalities as the limits of integration. The geometric task in this tutorial is to determine the bounds correctly before evaluating the integral.

1
Find the vertical range.
Identify the lowest and highest \(y\)-values in the region. These become the outer limits \(c\) and \(d\).
2
Take a horizontal slice.
At a general height \(y\), determine the left and right \(x\)-coordinates. Write them as functions \(g_1(y)\) and \(g_2(y)\).
3
Check the boundary order.
Verify that \(g_1(y)\le g_2(y)\) throughout the full outer interval, including its endpoints.
4
Write the integral in slice order.
Place \(g_1(y)\) and \(g_2(y)\) as the inner \(x\)-limits, with \(c\) and \(d\) as the outer \(y\)-limits.

Worked Examples

Worked Example: A Triangle Between a Diagonal and a Vertical Line

Let \(D\) be the triangle described by \(0\le y\le x\le 2\). Write and evaluate the Type II integral of \(f(x,y)=x+2y\) over \(D\).

For a horizontal slice, first hold \(y\) fixed. The inequality \(0\le y\le x\le 2\) then gives \(y\le x\le 2\). Since \(y\) ranges from \(0\) to \(2\), the region has the Type II description \(0\le y\le 2\), \(y\le x\le 2\). Therefore the setup is

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

Integrate first with respect to \(x\), treating \(y\) as constant:

$$ \int_y^2 (x+2y)\,dx = \left[\frac{x^2}{2}+2yx\right]_{x=y}^{x=2} = \left(2+4y\right)-\left(\frac{y^2}{2}+2y^2\right) = 2+4y-\frac{5y^2}{2}. $$

Now integrate over \(0\le y\le 2\):

$$ \int_0^2\left(2+4y-\frac{5y^2}{2}\right)\,dy = \left[2y+2y^2-\frac{5y^3}{6}\right]_0^2 = 4+8-\frac{20}{3} = \frac{16}{3}. $$

Check. The inner antiderivative differentiates to \(x+2y\), since \(\frac{\partial}{\partial x}(\frac{x^2}{2}+2yx)=x+2y\). The outer antiderivative differentiates to \(2+4y-\frac{5y^2}{2}\), the computed inner integral. The bounds also match the triangle: at \(y=0\), the slice runs from \(x=0\) to \(x=2\), and at \(y=2\), it collapses to \(x=2\).

Worked Example: A Region Between a Parabola and a Horizontal Line

Let \(D\) be the region between \(x=y^2\) and \(x=4\). Set up and evaluate the integral of \(f(x,y)=x\) over \(D\).

The two boundary curves meet where \(y^2=4\). Solving gives \(y=-2\) and \(y=2\), so these are the lowest and highest heights. At each height between them, \(x=y^2\) is the left boundary and \(x=4\) is the right boundary: \(y^2\le x\le 4\). The Type II setup is

$$ \iint_D x\,dA = \int_{-2}^{2}\int_{y^2}^{4} x\,dx\,dy. $$

The inner integral is

$$ \int_{y^2}^{4}x\,dx = \left[\frac{x^2}{2}\right]_{y^2}^{4} = \frac{16}{2}-\frac{(y^2)^2}{2} = 8-\frac{y^4}{2}. $$

Therefore,

$$ \int_{-2}^{2}\left(8-\frac{y^4}{2}\right)\,dy = \left[8y-\frac{y^5}{10}\right]_{-2}^{2} = \left(16-\frac{32}{10}\right) - \left(-16+\frac{32}{10}\right) = \frac{128}{5}. $$

Check. Differentiating \(8-\frac{y^4}{2}\) gives \(-2y^3\), and differentiating \(8y-\frac{y^5}{10}\) gives \(8-\frac{y^4}{2}\). The boundary check also confirms the order: \(y^2\le 4\) for every \(y\) in \([-2,2]\), and equality holds at both endpoints.

Worked Example: Two Curved Boundaries That Meet at the Ends

Suppose \(D\) is bounded by \(x=y^2\) and \(x=2-y\). Write and evaluate the Type II integral of \(1\) over \(D\), which gives the area of the region.

Find the heights where the boundaries meet by solving

$$ y^2=2-y \quad\Longleftrightarrow\quad y^2+y-2=0 \quad\Longleftrightarrow\quad (y+2)(y-1)=0. $$

Thus \(y=-2\) and \(y=1\) are the outer limits. On this interval, the difference between the proposed right and left boundaries is

$$ (2-y)-y^2 = -(y+2)(y-1) \ge 0 \qquad (-2\le y\le 1). $$

So \(y^2\) is the left boundary and \(2-y\) is the right boundary throughout the region. The area integral is

$$ \iint_D 1\,dA = \int_{-2}^{1}\int_{y^2}^{2-y}1\,dx\,dy. $$

Evaluating the inner integral gives

$$ \int_{y^2}^{2-y}1\,dx = (2-y)-y^2 = 2-y-y^2. $$

Then

$$ \int_{-2}^{1}(2-y-y^2)\,dy = \left[2y-\frac{y^2}{2}-\frac{y^3}{3}\right]_{-2}^{1} = \frac{7}{6}-\left(-\frac{10}{3}\right) = \frac{9}{2}. $$

Check. The derivative of \(2y-\frac{y^2}{2}-\frac{y^3}{3}\) is \(2-y-y^2\), exactly the inner integral. At \(y=-2\) and \(y=1\), the two \(x\)-boundaries agree, as required at the ends of this range. Between those heights, the factored difference is nonnegative, so the slice limits never reverse.

A Common Mistake: Reversing the Slice Boundaries

For the triangle \(0\le y\le x\le 2\), a horizontal slice at height \(y\) begins at \(x=y\), not at \(x=0\). A common incorrect setup is

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

That setup describes the other side of the diagonal: it has \(0\le x\le y\le 2\). For instance, the point \((1,0)\) belongs to the original triangle because \(0\le 1\), but it fails the incorrect condition \(x\le y\). Meanwhile, \((0,1)\) satisfies the incorrect condition but is not in the original region.

The correction is to read the original inequalities at a fixed \(y\). They give \(y\le x\le 2\), so the correct integral is \(\int_0^2\int_y^2 (x+2y)\,dx\,dy\). Checking a test point or a slice endpoint can catch a reversed bound before any integration is done.

Using the Setup Carefully

Type II limits are useful when a region is naturally bounded on the left and right by curves given as \(x\)-functions of \(y\). They also provide a precise way to describe how a region extends through its vertical range. Once the limits are correct, the integrand can represent area, accumulated mass, or another quantity modeled by a double integral.

Before proceeding from a sketch to an integral, check three features: the outer limits reach the bottom and top of the region; the inner limits name the left and right boundaries; and the left boundary does not exceed the right boundary at any allowed height. In particular, finding intersections supplies candidate outer limits, while checking the boundary order confirms that those limits actually work.

Check Your Understanding

For each question, focus first on how the inequalities describe a horizontal slice.

  1. Write a Type II description for the region \(0\le y\le 3\), \(y\le x\le 3\).
  2. The boundaries are \(x=y^2\) and \(x=9\). Find the outer \(y\)-limits and state which boundary is left and which is right.
  3. For \(D=\{(x,y): -1\le y\le 2,\ y-1\le x\le y^2+1\}\), what condition must you check before using this as a Type II description over the entire interval?
  4. Set up, but do not evaluate, the integral of a continuous function \(f(x,y)\) over the region \(0\le y\le 1\), \(y^2\le x\le 2-y\).
  5. In a Type II iterated integral, which variable is integrated first, and what geometric direction does that correspond to?

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