How to Set Up a Double Integral Over an Inequality-Defined Region

A double integral over an inequality-defined region is evaluated by translating the inequalities into limits of integration. The main task is geometric: determine which points satisfy every inequality, then describe that set with bounds that cover it exactly once.

Turn the inequalities into bounds

A double integral over a region $D$ is written as

$$\iint_D f(x,y)\,dA.$$

To evaluate it as an iterated integral, describe the region using one variable as the outer variable and the other as the inner variable. For example, if each vertical line through the region meets it in one interval, write

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

Here $a\le x\le b$ is the region’s horizontal projection, and for each such $x$, the coordinate $y$ runs from the lower boundary $g_1(x)$ to the upper boundary $g_2(x)$. This is called a vertically simple description. A horizontally simple description instead has the form

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

The inequalities define the set; the order of integration determines how you describe it. The same region can often be written either way.

A reliable setup method

1. List the constraints. For example, $x\ge 0$ restricts the region to the right of the $y$-axis, while $y\le 1-x$ places it below a line.
2. Find the boundary curves. Replace the relevant inequality by equality to identify a boundary, such as $y=1-x$.
3. Sketch and shade. A test point can help decide which side of a boundary satisfies an inequality. For several inequalities, keep only the part satisfying all of them.
4. Choose an integration order. Imagine vertical or horizontal slices. If the boundary of each slice can be named with one lower and one upper curve, the region needs only one iterated integral in that order.
5. Find the projection limits. These are the full range of the outer variable. If the slice boundaries change partway across the region, split it into pieces or try the other order.

For a curved boundary, solve its equation for the inner variable. For instance, $x^2+y^2\le 1$ gives $-\sqrt{1-x^2}\le y\le\sqrt{1-x^2}$ when using vertical slices. A different coordinate system may be simpler for some regions; the disk $x^2+y^2\le 1$, for example, is naturally described in polar coordinates.

Worked example: a triangular region

Evaluate

$$\iint_D (x+2y)\,dA, \qquad D=\{(x,y):x\ge 0,\ y\ge 0,\ x+y\le 1\}.$$

The inequalities $x\ge0$ and $y\ge0$ restrict the region to the first quadrant. The remaining inequality is $y\le1-x$, so the region lies below the line $y=1-x$. Its vertices are $(0,0)$, $(1,0)$, and $(0,1)$.

Using vertical slices, $x$ ranges from $0$ to $1$. For each fixed $x$, $y$ ranges from the $x$-axis to the line $y=1-x$. Therefore,

$$\begin{aligned} \iint_D (x+2y)\,dA &=\int_0^1\int_0^{1-x}(x+2y)\,dy\,dx\\ &=\int_0^1\left[xy+y^2\right]_{0}^{1-x}\,dx\\ &=\int_0^1\bigl(x(1-x)+(1-x)^2\bigr)\,dx\\ &=\int_0^1(1-x)\,dx\\ &=\frac12. \end{aligned}$$

The simplification in the fourth line follows because $x(1-x)+(1-x)^2=(1-x)(x+1-x)=1-x$.

You can check the setup using horizontal slices instead. Now $y$ runs from $0$ to $1$, and for each fixed $y$, $x$ runs from $0$ to $1-y$:

$$\begin{aligned} \iint_D (x+2y)\,dA &=\int_0^1\int_0^{1-y}(x+2y)\,dx\,dy\\ &=\int_0^1\left(\frac{(1-y)^2}{2}+2y(1-y)\right)\,dy\\ &=\frac12. \end{aligned}$$

Both orders describe the same region, so both must give the same value.

When one set of bounds is not enough

A region may have slices whose endpoints change at an interior point. For example, a vertical slice might be bounded by one curve for $x<1$ and a different curve for $x>1$. In that case, split the outer interval at $x=1$ and add the resulting integrals. Alternatively, reverse the order if horizontal slices give a single pair of boundary curves.

The bounds must represent the region, not just the equation of one of its edges. Check both the outer projection and the inner limits against your sketch. Also check whether the inequalities make the region bounded; if not, an integration limit may be infinite, and the integral must be treated as an improper integral.

Common mistakes

• Using only one inequality. The region is the intersection of all the sets specified by the inequalities. Ignoring one constraint can include points outside the intended region.
• Reversing a boundary. If the slice runs from $y=1-x$ down to $y=0$, the limits are backwards; the lower limit must be the smaller value on that slice.
• Confusing inner and outer limits. In $dy\,dx$, the $y$-limits may depend on $x$, while the $x$-limits are constants. In $dx\,dy$, the roles switch.
• Forgetting to split the region. If the top or bottom boundary changes, one pair of formulas may omit part of the region or count part twice.
• Changing the integrand but not the area element. In Cartesian coordinates use $dA=dy\,dx$ or $dx\,dy$. In polar coordinates use $dA=r\,dr\,d\theta$, including the factor $r$.

A useful final check is to test a point well inside the proposed bounds and confirm that it satisfies the original inequalities. Then compare the bounds with the sketch to make sure every slice has the intended endpoints.