5.2 Double Integrals over General Regions

Big idea

Rectangles were a warm-up. The regions that actually show up are bounded by curves: the piece between a parabola and a line, a triangle, the inside of a loop. The definition of the double integral needs no repair for these, because you can always extend the integrand by zero outside the region and integrate over a rectangle containing it. What needs repair is the evaluation, since the limits are no longer four constants.

The fix is to let the inner limits be functions. Fix a value of $x$ and slide along the vertical line through it. If that line enters the region at one curve and leaves at another, the inner integral in $y$ runs between those two curves, and its limits are expressions in $x$. Then sweep $x$ across the full horizontal extent of the region, which is a pair of numbers. Regions described this way are called type I. Turning the picture ninety degrees gives type II, where horizontal lines do the slicing and the inner variable is $x$.

The outer limits are always constants. That is not a convention, it is forced: after the inner integral is evaluated and its limits substituted, the inner variable is gone, so the outer integral is an ordinary single-variable integral of a function of one variable, and a number is what it must produce. If your outer limits still contain a variable, the setup is wrong, and no amount of correct integration afterward will fix it.

Many regions are both type I and type II, which gives you two valid setups for the same integral. They are equal, but they are rarely equally pleasant. One order may split into two integrals while the other needs one; one order may lead to an antiderivative that does not exist in elementary terms while the other is immediate. Reversing the order of integration is the technique built on that asymmetry, and it is the part of this section you will use most often later.

Decoder

If the region is bounded above and below by continuous curves $y = g_2(x)$ and $y = g_1(x)$ on an interval $a \le x \le b$, then the double integral over that region is the iterated integral in $y$ from $g_1(x)$ to $g_2(x)$, then in $x$ from $a$ to $b$.

The load-bearing words are “above and below”. The statement requires that each vertical line through the region cross it once going in and once coming out, so that the set of $y$ values on that line is a single interval. A region shaped like a crescent or split into two blobs fails this, and the repair is to cut it into pieces that each pass the test and add the results.

Notice the asymmetry between the two sets of limits. The inner limits are curves and may contain $x$; the outer limits are numbers and may contain nothing. Reading the setup outward, the variable you integrate first disappears first. A limit that mentions the variable currently being integrated is meaningless, so $\int_0^y$ as an outer integral in $y$ signals a bad setup immediately.

The common mistake in reversing an order is to keep the old limits and trade the differentials. Limits are not attached to positions in the expression; they describe a region, and the region has to be redescribed from scratch. Solving the old boundary equations for the other variable helps only once you know which curve the new slices enter on and which they leave on, and a sketch is what tells you that.

Definitions and results

Type I region. A set of the form $D = \{(x,y) : a \le x \le b,\ g_1(x) \le y \le g_2(x)\}$ with $g_1$ and $g_2$ continuous. Vertical slices are single segments.

Type II region. A set of the form $D = \{(x,y) : c \le y \le d,\ h_1(y) \le x \le h_2(y)\}$. Horizontal slices are single segments.

Iterated integral over a type I region.

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

Iterated integral over a type II region.

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

The outer limits are constants. In both formulas the outer limits are numbers. They are the projection of the region onto the outer variable’s axis: the shadow of $D$ on the $x$-axis for type I, on the $y$-axis for type II.

Area. Taking $f = 1$ gives the area of $D$. On a type I region this reduces to $\int_a^b \big( g_2(x) - g_1(x) \big)\, dx$, the area between two curves from single-variable calculus.

Decomposition. If $D$ is the union of two regions that overlap only along a boundary, the integrals over the pieces add. This is how you handle a region that is neither type I nor type II as a whole.

Bounds on an integral. If $m \le f \le M$ on $D$, then $m \cdot A(D) \le \iint_D f\, dA \le M \cdot A(D)$, where $A(D)$ is the area. This is the cheapest check available on a finished answer.

Reversing the order. Both setups over the same region give the same number. To reverse, sketch the region from the given limits, then redescribe it with the roles of the variables exchanged.

Worked examples

The same region read two ways

Let $D$ be the region between the parabola $y = x^2$ and the line $y = 2x$. They meet where $x^2 = 2x$, at $x = 0$ and $x = 2$, and between those values the line is above the parabola. Evaluate $\iint_D 6x\, dA$.

As a type I region, $0 \le x \le 2$ and $x^2 \le y \le 2x$:

$$ \int_0^2 \int_{x^2}^{2x} 6x\, dy\, dx = \int_0^2 6x \big( 2x - x^2 \big)\, dx = \int_0^2 \big( 12x^2 - 6x^3 \big)\, dx = 32 - 24 = 8 $$

As a type II region, $y$ runs from $0$ to $4$, and for fixed $y$ the horizontal line enters at the line $x = y/2$ and leaves at the parabola $x = \sqrt{y}$:

$$ \int_0^4 \int_{y/2}^{\sqrt{y}} 6x\, dx\, dy = \int_0^4 \left[ 3x^2 \right]_{y/2}^{\sqrt{y}} dy = \int_0^4 \left( 3y - \frac{3y^2}{4} \right) dy = 24 - 16 = 8 $$

Two setups, one answer. The region has area $\int_0^2 (2x - x^2)\, dx = \tfrac{4}{3}$ and the integrand $6x$ runs from $0$ to $12$ on it, so $8$ passes the bounding check.

Choosing the order that avoids splitting

Find the area of the region bounded by the parabola $x = y^2$ and the line $x = y + 2$. Setting $y^2 = y + 2$ gives $y^2 - y - 2 = 0$, so $y = -1$ and $y = 2$, and between those values the line lies to the right of the parabola.

Read it as a type II region: $-1 \le y \le 2$ and $y^2 \le x \le y + 2$.

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

A type I reading needs two integrals. For $0 \le x \le 1$ the vertical line enters and leaves on the parabola, so $-\sqrt{x} \le y \le \sqrt{x}$, while for $1 \le x \le 4$ the bottom boundary switches to the line and $x - 2 \le y \le \sqrt{x}$. Same area, twice the work.

Reversing the order to make the integral possible

Evaluate

$$ \int_0^1 \int_x^1 e^{y^2}\, dy\, dx $$

The inner integral cannot be done: $e^{y^2}$ has no elementary antiderivative. So redescribe the region. The limits say $0 \le x \le 1$ and $x \le y \le 1$, which is the triangle with corners $(0,0)$, $(0,1)$ and $(1,1)$, the part of the unit square above the line $y = x$.

Slice it horizontally instead. Now $y$ runs from $0$ to $1$, and for fixed $y$ the horizontal segment inside the triangle runs from $x = 0$ to $x = y$:

$$ \int_0^1 \int_0^y e^{y^2}\, dx\, dy = \int_0^1 y\, e^{y^2}\, dy = \left[ \frac{e^{y^2}}{2} \right]_0^1 = \frac{e - 1}{2} $$

The reversal supplied the factor of $y$ that the substitution $u = y^2$ needs. Check the size: the triangle has area $\tfrac{1}{2}$ and $e^{y^2}$ ranges from $1$ to $e$ on it, so the answer must lie between $0.5$ and about $1.36$. It is about $0.86$.

A triangle sliced vertically

Evaluate $\iint_D 2y\, dA$ where $D$ is the triangle with vertices $(0,0)$, $(2,0)$ and $(0,4)$. The slanted edge joins $(2,0)$ to $(0,4)$, so it is the line $y = 4 - 2x$. Taking $D$ as type I, $0 \le x \le 2$ and $0 \le y \le 4 - 2x$:

$$ \int_0^2 \int_0^{4-2x} 2y\, dy\, dx = \int_0^2 \big( 4 - 2x \big)^2 dx $$

Substituting $u = 4 - 2x$ turns this into $\tfrac{1}{2}\int_0^4 u^2\, du = \tfrac{32}{3}$. The triangle has area $4$ and $2y$ runs from $0$ to $8$ on it, so a value near $10.7$ is reasonable, and positive as it must be.

Practice

Sketch first, every time. Decide whether vertical or horizontal slices cut the region into single segments, write the inner limits as functions of the outer variable, and confirm the outer limits are numbers before integrating anything. Some of these ask you to rewrite a given iterated integral in the other order.

Quiz

Five items on type I and type II setups, areas of regions between curves, and reversing the order of integration.