Why Slicing Produces Iterated Integrals

From a Planar Region to Slices

A double integral adds contributions from an entire region. A useful way to understand that addition is to divide the region into thin strips. Each strip is made up of short slices, and the contributions along a slice can be added first. Adding those slice totals across the region produces an iterated integral.

The order of this addition is not just a computational choice. For a continuous function on a suitably described region, the iterated integral gives the same value as the double integral because both are limits of approximations to the same total.

Key idea: A vertical slice holds \(x\) fixed and adds contributions as \(y\) varies. Integrating those slice totals across the \(x\)-interval gives the double integral. Horizontal slices give the corresponding formula with the roles of \(x\) and \(y\) exchanged.

Simple Regions and the Slicing Formulas

A Type I region has vertical slices that each form one interval. In precise terms, suppose \(a\leq b\), and \(g_1\) and \(g_2\) are continuous functions on \([a,b]\) with \(g_1(x)\leq g_2(x)\) throughout. The Type I region they describe is

D=\{(x,y):a\leq x\leq b,\ g_1(x)\leq y\leq g_2(x)\}.

A Type II region has horizontal slices that each form one interval. If \(c\leq d\), and \(h_1\) and \(h_2\) are continuous on \([c,d]\) with \(h_1(y)\leq h_2(y)\), then

D=\{(x,y):c\leq y\leq d,\ h_1(y)\leq x\leq h_2(y)\}.

Here, \(f\) is assumed continuous on the region \(D\). Continuity ensures that the integral exists and that the values of \(f\) do not change abruptly as the slices move across the region.

Theorem: For a continuous function \(f\) on a Type I region \(D\),

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

For a Type II region \(D\), the corresponding formula is

\iint_D f(x,y)\,dA = \int_c^d\left(\int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\right)dy.

In each formula, the inner integral adds the function values along one slice. The outer integral adds the resulting slice totals across the region. If \(f\) is nonnegative, this total can be interpreted as the volume below the surface \(z=f(x,y)\) and above \(D\). If \(f\) takes both positive and negative values, the integral is a signed total.

Why the Vertical-Slice Formula Works

Consider a Type I region. Divide its horizontal extent \([a,b]\) into short intervals. Over a narrow interval, the region is approximately a vertical strip. Choose a sample \(x_i^*\) in each interval. The contribution of that strip is approximately its width multiplied by the total of \(f\) along the vertical slice at \(x_i^*\).

The slice total at a fixed \(x\) is the function

F(x)=\int_{g_1(x)}^{g_2(x)}f(x,y)\,dy.

Thus the strip approximation is \(\Delta x_i F(x_i^*)\). Adding these approximations across all strips gives a Riemann sum for the integral of \(F\) over \([a,b]\). As the widths of the strips approach zero, the approximation approaches the double integral: the boundary heights and the values of \(f\) vary continuously, so replacing a thin strip by its representative slice produces an error that tends to zero. The Riemann sums therefore give the same limit whether we view them as approximating the double integral or as approximating the integral of \(F\).

Fix a horizontal position.
At \(x\), the region runs from \(y=g_1(x)\) to \(y=g_2(x)\).

Add along that slice.
The contribution per unit change in \(x\) is \(F(x)=\int_{g_1(x)}^{g_2(x)}f(x,y)\,dy\).

Add across the region.
Integrating \(F(x)\) from \(a\) to \(b\) sums the slice contributions and yields the double integral.

The horizontal-slice formula follows by the same reasoning, now dividing the region into thin horizontal strips. Its inner integral adds along a horizontal slice, and its outer integral adds those totals over the \(y\)-interval.

Worked Examples

Worked Example: Integrating Over a Triangular Region

Let \(D=\{(x,y):0\leq x\leq 1,\ 0\leq y\leq x\}\) and \(f(x,y)=x+y\). The region is Type I: for each \(x\), a vertical slice runs from \(y=0\) to \(y=x\). The slicing formula gives

\iint_D(x+y)\,dA = \int_0^1\left(\int_0^x(x+y)\,dy\right)dx.

In the inner integral, \(x\) is held fixed. An antiderivative with respect to \(y\) is \(xy+\frac{y^2}{2}\), since its \(y\)-derivative is \(x+y\). Evaluating at the slice endpoints gives

\int_0^x(x+y)\,dy = \left[xy+\frac{y^2}{2}\right]_0^x = \left(x^2+\frac{x^2}{2}\right)-0 = \frac{3x^2}{2}.

Now integrate the slice totals across the \(x\)-interval:

\int_0^1\frac{3x^2}{2}\,dx = \left[\frac{x^3}{2}\right]_0^1 = \frac12-0 = \frac12.

For a check, differentiating the inner antiderivative \(xy+\frac{y^2}{2}\) with respect to \(y\) recovers \(x+y\), and differentiating the outer antiderivative \(\frac{x^3}{2}\) with respect to \(x\) recovers \(\frac{3x^2}{2}\). Both integrations and their limits are consistent with the slice formula.

Worked Example: Using Horizontal Slices

Let \(D=\{(x,y):0\leq y\leq 2,\ 0\leq x\leq y\}\), and let \(f(x,y)=1+x\). Each horizontal slice starts at \(x=0\) and ends at \(x=y\), so the Type II formula is

\iint_D(1+x)\,dA = \int_0^2\left(\int_0^y(1+x)\,dx\right)dy.

With \(y\) fixed, an antiderivative of \(1+x\) with respect to \(x\) is \(x+\frac{x^2}{2}\). Therefore, the slice total is

\int_0^y(1+x)\,dx = \left[x+\frac{x^2}{2}\right]_0^y = \left(y+\frac{y^2}{2}\right)-0 = y+\frac{y^2}{2}.

Integrating that total over \(0\leq y\leq 2\) gives

\int_0^2\left(y+\frac{y^2}{2}\right)dy = \left[\frac{y^2}{2}+\frac{y^3}{6}\right]_0^2 = \left(2+\frac{8}{6}\right)-0 = \frac{10}{3}.

As a check, the derivative of \(x+\frac{x^2}{2}\) with respect to \(x\) is \(1+x\), and the derivative of \(\frac{y^2}{2}+\frac{y^3}{6}\) with respect to \(y\) is \(y+\frac{y^2}{2}\). Evaluating at the stated endpoints gives the computed value \(\frac{10}{3}\).

Worked Example: A Region Between a Curve and a Line

Consider \(D=\{(x,y):-1\leq x\leq 1,\ x^2\leq y\leq 1\}\) and \(f(x,y)=x^2+y\). A vertical slice begins at the curve \(y=x^2\) and ends at the line \(y=1\). Thus

\iint_D(x^2+y)\,dA = \int_{-1}^1\left(\int_{x^2}^1(x^2+y)\,dy\right)dx.

For fixed \(x\), an antiderivative with respect to \(y\) is \(x^2y+\frac{y^2}{2}\). Evaluating at \(y=1\) and \(y=x^2\) gives

\begin{aligned} \int_{x^2}^1(x^2+y)\,dy &= \left[x^2y+\frac{y^2}{2}\right]_{x^2}^1\\ &= \left(x^2+\frac12\right) - \left(x^4+\frac{x^4}{2}\right)\\ &= \frac12+x^2-\frac32x^4. \end{aligned}

Now integrate with respect to \(x\):

\begin{aligned} \int_{-1}^1\left(\frac12+x^2-\frac32x^4\right)dx &= \left[\frac{x}{2}+\frac{x^3}{3}-\frac{3x^5}{10}\right]_{-1}^1\\ &= \left(\frac12+\frac13-\frac{3}{10}\right) - \left(-\frac12-\frac13+\frac{3}{10}\right)\\ &= \frac{8}{15}-\left(-\frac{8}{15}\right) = \frac{16}{15}. \end{aligned}

For a check, the derivative of \(x^2y+\frac{y^2}{2}\) with respect to \(y\) is \(x^2+y\). The derivative of \(\frac{x}{2}+\frac{x^3}{3}-\frac{3x^5}{10}\) is \(\frac12+x^2-\frac32x^4\), the slice total found above. Subtracting the evaluated outer endpoints gives \(\frac{16}{15}\).

A Common Bounds Mistake

The bounds of the inner integral must describe the endpoints of each slice, not merely the overall smallest and largest coordinates of the region. For the triangle \(0\leq x\leq 1,\ 0\leq y\leq x\), the vertical slice at \(x\) ends at \(y=x\), not at \(y=1\) for every \(x\). Using constant bounds instead would calculate the integral over a square:

\int_0^1\int_0^1 1\,dy\,dx=1.

But the triangle's area, found using its actual vertical slices, is

\int_0^1\int_0^x 1\,dy\,dx = \int_0^1x\,dx = \frac12.

The wrong calculation does not fail because slicing is inaccurate. It fails because the constant inner bounds describe a different region. Always make sure the slice endpoints vary with the outer variable when the boundary does.

Why the Formula Matters

The slicing formulas connect the geometric shape of a region to the calculation of its integral. They also explain why integration is performed in two stages: first summing the function along a slice, then summing those slice totals across the region. The continuous-boundary and continuity conditions matter because they ensure that the region and the function change smoothly enough for the strip approximations to converge to the double integral.

When choosing a formula, the essential question is whether the region can be described by one interval on each slice in the chosen direction. If it can, the corresponding single iterated integral applies. If the required slice description fails, a single pair of inner bounds will not represent the whole region in that direction.

Check Your Understanding

Use the slicing interpretation and formulas to answer these questions.

For a Type I region, which variable is held fixed while the inner integral is evaluated?
In the vertical-slice formula, what does the inner integral represent?
Why does continuity help justify replacing thin strips by representative slices?
For \(0\leq x\leq 2\) and \(0\leq y\leq x\), write the vertical-slice formula for the integral of \(1\) over the region.
What region would be described by \(\int_0^1\int_0^1 1\,dy\,dx\), and why is it not the triangle \(0\leq y\leq x\leq 1\)?