Double Integrals over General Regions

Main idea

When the region isn't a rectangle, each slice has a different length, so the inner limits become functions: $y$ runs from the bottom curve to the top curve at each $x$ . Switching the order of integration can turn an impossible integral into an easy one.

\iint_D f\,dA = \int_a^b\!\int_{{\htmlData{sym=0}{g_1(x)}}}^{g_2(x)} f(x, y)\,dy\,dx \quad\text{or}\quad \int_c^d\!\int_{h_1(y)}^{h_2(y)} f(x, y)\,dx\,dy

What the symbols mean

$g_1(x)$
Bottom boundary
The inner limits can depend on the outer variable, never the other way around.

Explanation

A general region is simply any flat shape that isn't a rectangle: the space between two curves, a triangle, a lens. The formula $\int_a^b\!\int_{g_1(x)}^{g_2(x)} f\,dy\,dx$ looks like the rectangle version with something extra, and that's exactly what it is. The only change is that the inner limits $g_1(x)$ (the bottom edge) and $g_2(x)$ (the top edge) move as you slide across, because a vertical strip through the region is longer in some places than others. Same two integrals as before; the strips just change length.

Setting up: sketch the region. That's not optional; almost every wrong set-up comes from skipping the sketch. For vertical strips, the outer limits are the extreme $x$ -values and the inner limits run from the bottom curve to the top curve. For horizontal strips, swap roles: $x$ runs from the left curve to the right curve, and the outer limits are the extreme $y$ -values. This is the same choice you made for areas between curves in Calculus 1, integrating in $x$ or in $y$ .

A common mistake. The inner limits can depend on the outer variable, never the other way around. After the inner integral is done, only the outer variable should be left; if your outer limits contain $y$ while the outer integral is in $x$ , the answer can't be a number. Think about why: the outer integral is what sweeps across the whole region, so its limits have to be fixed numbers.

Reversing the order can turn an impossible integral into an easy one. $\int_0^1\!\int_x^1 e^{y^2}\,dy\,dx$ can't be done as written ( $e^{y^2}$ has no elementary antiderivative), but the region is the triangle $0 \le x \le y \le 1$ , so

\int_0^1\!\int_0^y e^{y^2}\,dx\,dy = \int_0^1 ye^{y^2}\,dy = \frac{e - 1}{2}.

Notice what happened: integrating in $x$ first, $e^{y^2}$ is a constant, so the inner integral just multiplies it by the strip length $y$ . That extra $y$ is exactly what the substitution $u = y^2$ needs. Same region, same tiny boxes, grouped the other way.

See it

The lens between y = x² and y = √x, with a vertical strip sweeping across it. As you can see, the strip is short near both ends, where the curves meet at (0, 0) and (1, 1), and longest near x ≈ 0.63. Its ends always sit on the two curves, which is why the inner limits are x² and √x instead of fixed numbers.

Worked examples

Example 1

Evaluate $\iint_D xy\,dA$ where $D$ is between $y = x^2$ and $y = x$ .

1
Setting $x^2 = x$ gives $x = 0$ and $x = 1$ , where the curves meet. Between them, $y = x$ is the top curve (at $x = \frac12$ it's $\frac12$ , above $\frac14$ ).

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Evaluate $\iint_D xy\,dA$ over the region between $y = x^2$ and $y = x$ .

Problem 2+5 XP

Evaluate $\displaystyle\int_0^1\!\int_0^x(x + y)\,dy\,dx$ .

Problem 3+5 XP

Use $\iint_D 1\,dA$ to find the area between $y = x^2$ and $y = 2x$ .

Problem 4+5 XP

Evaluate $\displaystyle\int_0^1\!\int_x^1 e^{y^2}\,dy\,dx$ by reversing the order.

Problem 5+5 XP

Evaluate $\displaystyle\int_0^2\!\int_0^{4 - x^2}x\,dy\,dx$ .

Problem 6+5 XP

A metal plate fills the region under $y = 4 - x^2$ and above the $x$ -axis (meters). Its density is $\rho = y$ kg/m², so it's heavier toward the top. Find its mass.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Evaluate $\displaystyle\int_0^1\!\int_{\sqrt y}^{1}\sqrt{1 + x^3}\,dx\,dy$ by reversing the order of integration.

Where it's used

Real regions are never rectangles: the area of a lake from a survey, the load on an irregular floor slab, the probability that a point lands in a curved region.

Iterated Integrals Double Integrals in Polar Coordinates