Setting Bounds on Parameter Domains

Why Parameter-Domain Bounds Matter

A scalar integral over a parametrized surface is evaluated over a region in the parameter plane. Once the surface integrand has been determined, the remaining task is to describe that region accurately using integration bounds. A small error in the bounds can omit part of the surface, include points outside it, or count part of it more than once.

The key idea is to think of the domain as a collection of slices. For vertical slices, fix the first parameter and describe the allowed values of the second. For horizontal slices, fix the second parameter and describe the allowed values of the first. The correct bounds are those whose slices cover the domain exactly.

Definition: A parameter domain \(D\) is a region in the parameter plane. An iterated description of \(D\) gives outer bounds for one parameter and, for each fixed value of that parameter, inner bounds for the other. For example, \(a\le u\le b\) and \(g_1(u)\le v\le g_2(u)\) describes the points between the two curves \(v=g_1(u)\) and \(v=g_2(u)\) over \(a\le u\le b\).

For a continuous function \(G\) on a bounded domain \(D\), an iterated integral can be used to calculate \(\iint_D G(u,v)\,dA\). The bounds describe \(D\); they do not replace the surface-area factor in a surface integral. In a scalar surface integral, that factor is already part of the integrand \(G\).

Start with the Boundary and the Slices

Before writing an integral, identify the boundary curves and their intersection points. Then choose a slicing direction and ask two questions: what values can the outer variable take, and where does each slice enter and leave the region? A quick sketch is often the most reliable way to answer them.

For vertical slices, the usual form is \(a\le u\le b\), with \(v\) between a lower and an upper boundary. For horizontal slices, the roles of \(u\) and \(v\) are reversed. The lower and upper boundaries can change as a slice moves across the domain; if they do, the bounds must be split into separate integrals.

Worked Example: A Triangular Parameter Domain

Let \(D\) be the triangle with vertices \((0,0)\), \((2,0)\), and \((0,1)\). The sloping edge has equation \(v=1-\frac{u}{2}\). The vertical slices run from \(v=0\) to that edge while \(u\) ranges from \(0\) to \(2\). Thus

D=\left\{(u,v):0\le u\le2,\ 0\le v\le1-\frac{u}{2}\right\}.

For the integrand \(G(u,v)=u+2v\), the iterated integral is

\begin{aligned} \iint_D (u+2v)\,dA &=\int_0^2\int_0^{1-u/2}(u+2v)\,dv\,du\\ &=\int_0^2\left[uv+v^2\right]_0^{1-u/2}\,du\\ &=\int_0^2\left(u\left(1-\frac{u}{2}\right)+\left(1-\frac{u}{2}\right)^2\right)\,du\\ &=\int_0^2\left(1-\frac{u^2}{4}\right)\,du\\ &=\left[u-\frac{u^3}{12}\right]_0^2 =2-\frac{8}{12} =\frac{4}{3}. \end{aligned}

To reverse the order, use horizontal slices. Here \(v\) ranges from \(0\) to \(1\); at each height, \(u\) runs from the left edge \(u=0\) to the sloping edge \(u=2-2v\). This gives

\begin{aligned} \iint_D (u+2v)\,dA &=\int_0^1\int_0^{2-2v}(u+2v)\,du\,dv\\ &=\int_0^1\left[\frac{u^2}{2}+2vu\right]_0^{2-2v}\,dv\\ &=\int_0^1\left(2(1-v)^2+4v(1-v)\right)\,dv\\ &=\int_0^1(2-2v^2)\,dv\\ &=\left[2v-\frac{2v^3}{3}\right]_0^1 =\frac{4}{3}. \end{aligned}

The two calculations agree. The triangle’s area is \(1\), and the computed integral is positive, as expected for the nonnegative integrand \(u+2v\) on this region.

Annular Domains

An annulus centered at the origin, with inner radius \(a\) and outer radius \(b\), where \(0<a<b\), consists of points whose distance from the origin is between those radii. In polar coordinates, its bounds are especially direct: the radius runs from \(a\) to \(b\), and the angle makes one complete turn.

a\le r\le b,\qquad 0\le\theta\le2\pi.

In Cartesian coordinates, vertical slices are more complicated because the inner circle removes the middle of some slices. For the annulus with \(a=1\) and \(b=2\), a slice with \(-1<u<1\) consists of two intervals of \(v\): one below the hole and one above it. For \(1<|u|<2\), the slice does not meet the hole and is a single interval. At \(u=\pm1\), the intervals meet at a boundary point, which does not affect the integral.

Worked Example: Integrating Over an Annulus

Let \(D\) be the annulus \(1\le u^2+v^2\le4\). In polar coordinates, \(u=r\cos\theta\), \(v=r\sin\theta\), and \(dA=r\,dr\,d\theta\). To find the area of \(D\), integrate the constant function \(1\):

\begin{aligned} \iint_D 1\,dA &=\int_0^{2\pi}\int_1^2 r\,dr\,d\theta\\ &=\int_0^{2\pi}\left[\frac{r^2}{2}\right]_1^2\,d\theta\\ &=\int_0^{2\pi}\frac{3}{2}\,d\theta =3\pi. \end{aligned}

This is also the area of the radius-2 disk minus the area of the radius-1 disk: \(4\pi-\pi=3\pi\). The Cartesian vertical slices show why simply using the outer circle’s bounds would be wrong. For \(-1\le u\le1\), the allowed \(v\)-intervals are \([-\sqrt{4-u^2},-\sqrt{1-u^2}]\) and \([\sqrt{1-u^2},\sqrt{4-u^2}]\). For \(-2\le u\le-1\) and \(1\le u\le2\), the full slice runs from \(-\sqrt{4-u^2}\) to \(\sqrt{4-u^2}\).

As a check on the description, the inner-circle gap occurs only when \(|u|\le1\), exactly the range in which \(\sqrt{1-u^2}\) is defined. The polar bounds describe the same annulus without this split.

Curved Domains and Changing Boundaries

Curved boundaries can make one order of integration simpler than the other. Reversing the order requires solving boundary equations for the other variable and then checking where the resulting expressions are valid. Sometimes a horizontal or vertical slice changes which curve forms its endpoint. That is the point where the domain must be split.

Worked Example: Reversing the Order Requires a Split

Consider the region \(D\) between \(v=u^2\) and \(v=2-u\), from their intersection at \(u=-2\) to their intersection at \(u=1\). Since the line is above the parabola on this interval, vertical slices give

-2\le u\le1,\qquad u^2\le v\le2-u.

To describe horizontal slices, first find the full range of \(v\). Its minimum is \(0\), at the vertex of the parabola, and its maximum is \(4\), at the intersection \((-2,4)\). The parabola inequality \(u^2\le v\) gives \(-\sqrt v\le u\le\sqrt v\). The line inequality \(v\le2-u\) gives \(u\le2-v\). Together these require \(-\sqrt v\le u\le\min(\sqrt v,2-v)\).

The upper endpoint changes where \(\sqrt v=2-v\), which occurs at \(v=1\) in the range \(0\le v\le4\). Therefore the horizontal description must be split:

\begin{cases} 0\le v\le1,\quad -\sqrt v\le u\le\sqrt v,\\ 1\le v\le4,\quad -\sqrt v\le u\le2-v. \end{cases}

Check the region’s area using the vertical bounds:

\begin{aligned} \iint_D 1\,dA &=\int_{-2}^1\int_{u^2}^{2-u}1\,dv\,du\\ &=\int_{-2}^1(2-u-u^2)\,du\\ &=\left[2u-\frac{u^2}{2}-\frac{u^3}{3}\right]_{-2}^1\\ &=\frac{7}{6}-\left(-\frac{10}{3}\right) =\frac{9}{2}. \end{aligned}

Now use the split horizontal bounds:

\begin{aligned} \iint_D 1\,dA &=\int_0^1\int_{-\sqrt v}^{\sqrt v}1\,du\,dv +\int_1^4\int_{-\sqrt v}^{2-v}1\,du\,dv\\ &=\int_0^1 2\sqrt v\,dv+\int_1^4(2-v+\sqrt v)\,dv\\ &=\frac{4}{3} +\left[2v-\frac{v^2}{2}+\frac{2}{3}v^{3/2}\right]_1^4\\ &=\frac{4}{3}+\left(\frac{16}{3}-\frac{13}{6}\right)\\ &=\frac{8}{6}+\frac{19}{6} =\frac{9}{2}. \end{aligned}

Both orders give the same area. In the reversed description, the split at \(v=1\) is essential: below that height the parabola sets the right endpoint, and above it the line does.

Why Reversing or Splitting Works

For a region that can be described by vertical slices, integrating first with respect to \(v\) accumulates the contribution along each slice, and integrating those slice totals over \(u\) covers the whole region. Horizontal slices do the same in the other direction. If the function is continuous on the bounded region, changing the order is valid when the new bounds describe exactly the same region. A split is simply a way to express that region as pieces with consistent slice boundaries.

For example, in the curved-region calculation, every point in \(D\) satisfies both \(u^2\le v\) and \(u\le2-v\). For \(0\le v\le1\), \(\sqrt v\le2-v\), so the parabola provides the tighter upper limit. For \(1\le v\le4\), \(2-v\le\sqrt v\), so the line provides the tighter upper limit. These inequalities establish why the two pieces cover the region without overlap except at their shared boundary.

Practical rule: Bounds are correct when their slices cover every point of the intended domain and no point outside it. If the slice’s entry or exit boundary changes, split the integral at the change.

A Common Bounds Error

A frequent mistake is to describe an annulus by the outer-circle bounds alone. For \(1\le u^2+v^2\le4\), the bounds \(-2\le u\le2\) and \(-\sqrt{4-u^2}\le v\le\sqrt{4-u^2}\) describe the entire radius-2 disk, not the annulus. Integrating \(1\) with those bounds gives \(4\pi\), rather than the correct area \(3\pi\). The missing condition is the exclusion of the inner disk. Correct the bounds by removing the inner-circle intervals in Cartesian slices, or use the polar radial bound \(1\le r\le2\).

Other checks help catch errors before calculation: verify that boundary curves meet at the stated endpoints, confirm each square root is defined throughout its interval, and test a point from each proposed slice against the original domain inequalities. When reversing an order, compare the resulting domain—not just the algebraic expressions—to the original one.

Check Your Understanding

For each question, focus first on the region described by the bounds.

For the triangle \(0\le u\le3\), \(0\le v\le1-\frac{u}{3}\), what are the bounds when the order is reversed?
Why does a Cartesian vertical-slice description of an annulus require two intervals for some values of the outer variable?
For the region \(0\le u\le2\), \(u^2\le v\le4\), what is the maximum value of \(v\), and which boundary equations should be solved to describe horizontal slices?
What does a split in iterated bounds represent geometrically?
What region is described if the inner-radius restriction is omitted from the bounds for an annulus?