Double integrals: rectangular and general regions

The single-variable integral gave the area under a curve. The double integral gives the volume under a surface: the solid between the surface $z = f(x,y)$ and a region $D$ in the $xy$-plane. The computation technique is surprisingly simple: you take two single-variable integrals nested. The real difficulty is not computing the integral but setting up the limits correctly. The exam on this topic usually has one “evaluate”, one “sketch the region and reverse the order” and one volume or area question.

Golden rules:

1. In the inner integral the outer variable behaves as a constant. Work from inside out.
2. The limits of the outer integral are always numbers. The limits of the inner integral may depend on the outer variable.
3. For a general region, first sketch the region. Find the intersections of the curves, decide which is below and which is above.
4. If the inner integral cannot be computed (no antiderivative), reverse the order of integration.

Rectangular region and Fubini’s theorem

Let $R = [a,b]\times[c,d]$. The double integral is defined as the limit of Riemann sums: divide $R$ into small rectangles, take a value of $f$ in each, multiply by the area, add up, make the pieces smaller. If $f\ge0$, the result is the volume under the surface.

Fubini’s theorem. If $f$ is continuous on $R$,

$\iint_Rf(x,y)\,dA = \int_a^b\!\!\int_c^d f(x,y)\,dy\,dx = \int_c^d\!\!\int_a^b f(x,y)\,dx\,dy.$

On a rectangle the order is free. Choose whichever is easier.

Intuition. $\int_c^df(x,y)\,dy$ is the area of the slice formed when you cut the solid with the plane $x =$ constant at a fixed value of $x$. If you add up (integrate) these areas along $x$, you find the volume. It is like slicing a loaf of bread and adding up the areas of the slices.

Example 1. $\displaystyle\int_0^3\!\!\int_1^2x^2y\,dy\,dx$.

inner (with respect to y, x constant):  ∫_1^2 x²y dy = x² [y²/2]_1^2 = x² (2 - 1/2) = (3/2) x²
outer:                                  ∫_0^3 (3/2) x² dx = (3/2)(9) = 27/2

Since the integrand has the form $g(x)h(y)$, the shortcut: $\left(\int_0^3x^2dx\right)\left(\int_1^2y\,dy\right) = 9\cdot\frac32 = \frac{27}{2}$.

Example 2 (choosing the order). $\displaystyle\iint_Ry\sin(xy)\,dA$, $R = [0,\pi]\times[0,1]$. If you integrate with respect to $y$ first, integration by parts is needed. Integrate with respect to $x$ first: $\displaystyle\int_0^\pi y\sin(xy)\,dx = \left[-\cos(xy)\right]_0^\pi = 1-\cos(\pi y)$. Then $\displaystyle\int_0^1(1-\cos\pi y)\,dy = 1-0 = 1$.

TIP · choose the inner variable. If, next to the integrand in the inner integral, there is a factor that is the derivative with respect to the inner variable (in $y\sin(xy)$, when integrating with respect to $x$, $y$ is the derivative of the inner function), that order is easy. Example 2 and Q3 are based on this idea.

General regions

If the region is not a rectangle, its boundaries are curves. There are two standard forms:

Type I (vertical strips). $D = \{a\le x\le b,\ g_1(x)\le y\le g_2(x)\}$:

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

Type II (horizontal strips). $D = \{c\le y\le d,\ h_1(y)\le x\le h_2(y)\}$:

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

Template for setting limits

1) SKETCH the region. Compute the intersections of the boundary curves.
2) For type I: draw a vertical arrow inside the region (from bottom to top).
     where the arrow enters → lower limit g1(x);  where it exits → upper limit g2(x)
     the smallest and largest values x takes → outer limits a, b
3) For type II: a horizontal arrow (from left to right). Entry h1(y), exit h2(y).
     the smallest and largest values of y → c, d
4) If the arrow changes boundary inside the region (if the entry/exit curve changes)
     split the integral in two or try the other type.

Example 3. $D$: between $y = x^2$ and $y = 2x$. Intersection $x^2 = 2x$, $x = 0$ and $x = 2$. On $[0,2]$, $2x\ge x^2$ (test: at $x = 1$, $2>1$). $\displaystyle\iint_Dx\,dA$:

$\text{Type I:}\ \int_0^2\!\!\int_{x^2}^{2x}x\,dy\,dx = \int_0^2x(2x-x^2)\,dx = \frac{16}{3}-4 = \frac43.$

The same region as type II: $y$ goes from $0$ to $4$; $y = 2x\Rightarrow x = y/2$ (left), $y = x^2\Rightarrow x = \sqrt y$ (right). $\displaystyle\int_0^4\!\!\int_{y/2}^{\sqrt y}x\,dx\,dy$ also gives $\frac43$.

Reversing the order of integration

Some integrands have no antiderivative with respect to one variable: $e^{y^2}$, $\sin(y^3)$, $\cos(x^2)$, $\frac{\sin y}{y}$. If such an integral is given with that variable inside, you reverse the order. The job is to rewrite the region as the other type.

Example 4. $\displaystyle\int_0^1\!\!\int_y^1\cos(x^2)\,dx\,dy$. $\cos(x^2)$ has no antiderivative with respect to $x$.

Given: 0 ≤ y ≤ 1,  y ≤ x ≤ 1     → the triangle with vertices (0,0), (1,0), (1,1)
Sketch. With vertical strips (Type I): 0 ≤ x ≤ 1,  0 ≤ y ≤ x
∫_0^1 ∫_0^x cos(x²) dy dx = ∫_0^1 x cos(x²) dx = [sin(x²)/2]_0^1 = sin(1)/2

The new inner integral $\int_0^xdy = x$ produced a factor, and this factor is exactly what the substitution $u = x^2$ needs. In order-reversal questions this “magic” happens almost every time; if it does not, you have set up the limits wrong.

PITFALL · just swapping the limits. Turning $\int_0^1\int_y^1\ldots dx\,dy$ into $\int_y^1\int_0^1\ldots dy\,dx$ is meaningless: a variable remains in the limit of the outer integral. Reversing the order always means sketching the region and defining it again.

Area, volume and average value

Example 5. The volume under the paraboloid $z = 4-x^2-y^2$ and above the square $[0,1]\times[0,1]$: $\displaystyle\int_0^1\!\!\int_0^1(4-x^2-y^2)\,dy\,dx = 4-\frac13-\frac13 = \frac{10}{3}$.

If the region is not given explicitly in a volume question, the curve where the surface meets the $xy$-plane (or the projection of the intersection of two surfaces) is the boundary of the region. In Q9 the plane gives the line $2x+3y = 6$ at $z = 0$, and the base is a triangle.

Common mistakes