1. The double integral
The single-variable integral $\int_a^b f(x)\,dx$ is defined by chopping $[a,b]$ into $n$ tiny intervals, multiplying each interval's width by a function value, and taking the limit as the chopping gets fine. Geometrically: signed area between the curve $y = f(x)$ and the $x$-axis.
For a function of two variables $z = f(x, y)$, the analogue is to chop a region $R$ in the $xy$-plane into tiny rectangles, multiply each rectangle's area by a function value, sum, and take the limit. The result is the double integral.
For $f$ continuous on a bounded region $R \subset \mathbb{R}^2$, partition $R$ into $n$ tiny pieces with areas $\Delta A_i$, pick a sample point $(x_i^*, y_i^*)$ in each, and form the Riemann sum
$$ S_n = \sum_{i=1}^{n} f(x_i^*, y_i^*)\,\Delta A_i. $$The double integral is the limit as the pieces shrink to zero size:
$$ \iint_R f(x, y)\,dA \;=\; \lim_{\max \Delta A_i \to 0}\; S_n. $$The picture is the same one you've seen for single integrals — just one dimension richer. Each term $f(x_i^*, y_i^*)\,\Delta A_i$ is the volume of a thin rectangular column standing on the patch $\Delta A_i$ and rising to the surface $z = f(x, y)$. Adding them gives an approximate volume; the limit gives the exact one.
When $f$ is non-negative, $\iint_R f\,dA$ is the volume of the solid that sits above $R$ and below the surface $z = f(x, y)$. When $f$ takes both signs, you get signed volume: regions where $f < 0$ subtract. Setting $f \equiv 1$ collapses the formula to the area of $R$:
$$ \text{Area}(R) \;=\; \iint_R 1\,dA. $$Don't take "volume under a surface" too literally — it's just the easiest picture. If $f$ is a mass density, $\iint_R f\,dA$ is total mass. If $f$ is a probability density, it's a probability. The double integral is a general "accumulate $f$ over $R$" operation; volume is one interpretation among many.
2. Iterated integrals & Fubini's theorem
The definition tells you what a double integral is, but not how to compute one. The computational tool is the iterated integral: a double integral done as two nested single integrals — first one variable with the other held constant, then the result integrated against the second.
On a rectangle $R = [a, b] \times [c, d]$, the recipe is
$$ \iint_R f(x, y)\,dA \;=\; \int_a^b \!\! \left(\int_c^d f(x, y)\,dy\right)\!dx. $$The inner integral treats $x$ as a constant and integrates over $y$ from $c$ to $d$; the result is a function of $x$ alone, which the outer integral then handles. You could also do it the other way around — integrate $x$ first, then $y$. Fubini's theorem says it doesn't matter.
If $f$ is continuous on the rectangle $R = [a, b] \times [c, d]$, then both iterated integrals exist, are equal, and equal the double integral:
$$ \iint_R f\,dA \;=\; \int_a^b\!\int_c^d f(x,y)\,dy\,dx \;=\; \int_c^d\!\int_a^b f(x,y)\,dx\,dy. $$More generally, the same conclusion holds whenever $f$ is integrable on $R$.
This is more than a notational convenience — it's a license to choose the easier order. Some integrands are tame in $x$ first; some are tame in $y$ first; for many computations the choice is the whole game.
A quick worked example on a rectangle. Take $f(x, y) = x + y$ on $[0, 2] \times [0, 3]$:
$$ \int_0^2 \int_0^3 (x + y)\,dy\,dx \;=\; \int_0^2 \left[xy + \tfrac{y^2}{2}\right]_0^3 dx \;=\; \int_0^2 \left(3x + \tfrac{9}{2}\right) dx \;=\; 6 + 9 \;=\; 15. $$The inner step treats $x$ as a constant; the outer cleans up. Doing it in the other order gives $15$ as well — try it and confirm.
If the integrand factors as $f(x, y) = g(x)\,h(y)$ and the region is a rectangle, the double integral factors too:
$$\iint_{[a,b]\times[c,d]} g(x)\,h(y)\,dA \;=\; \left(\int_a^b g(x)\,dx\right)\!\left(\int_c^d h(y)\,dy\right).$$
Both conditions are required — factoring fails on a non-rectangular region because the inner limits then depend on the outer variable.
3. Integrating over non-rectangular regions
Most interesting regions aren't rectangles. The fix is to describe the region as a "stack" of vertical or horizontal slices and let the inner limits depend on the outer variable.
Type-I region (vertical slices)
A type-I region is one you can describe by fixing $x$ and letting $y$ vary between two functions of $x$:
$$ R = \{(x, y) : a \le x \le b,\; g_1(x) \le y \le g_2(x)\}. $$Picture vertical strips: for each $x \in [a, b]$, $y$ runs from a lower boundary curve $g_1(x)$ to an upper boundary curve $g_2(x)$. The iterated integral becomes
$$ \iint_R f\,dA \;=\; \int_a^b \!\!\int_{g_1(x)}^{g_2(x)} f(x, y)\,dy\,dx. $$The inner integral's limits are functions of $x$; the outer's are constants.
Type-II region (horizontal slices)
A type-II region is the mirror image: fix $y$, let $x$ vary between two functions of $y$:
$$ R = \{(x, y) : c \le y \le d,\; h_1(y) \le x \le h_2(y)\}. $$Now picture horizontal strips. The iterated integral is
$$ \iint_R f\,dA \;=\; \int_c^d \!\!\int_{h_1(y)}^{h_2(y)} f(x, y)\,dx\,dy. $$Many regions are both type-I and type-II — and you'll often pick whichever description gives easier integrals.
Setting up the limits — the recipe
- Sketch the region. Find the boundary curves and where they intersect.
- Decide outer variable. Type-I means $x$ outer; type-II means $y$ outer.
- Find the outer range — the two extreme values of the outer variable. These are constants.
- Find the inner range as functions of the outer variable — the lower and upper boundary curves.
- Integrate inner, then outer.
Worked example. Take $R$ bounded by $y = x$ and $y = x^2$ for $0 \le x \le 1$ — the curves meet at $(0, 0)$ and $(1, 1)$, and on $[0, 1]$ the parabola sits below the line. As a type-I region:
$$ R = \{(x, y) : 0 \le x \le 1,\; x^2 \le y \le x\}. $$Then $\iint_R xy\,dA$ becomes
$$ \int_0^1 \!\!\int_{x^2}^{x} xy\,dy\,dx \;=\; \int_0^1 x\cdot \tfrac{1}{2}(x^2 - x^4)\,dx \;=\; \tfrac{1}{2}\!\int_0^1 (x^3 - x^5)\,dx \;=\; \tfrac{1}{2}\!\left(\tfrac{1}{4} - \tfrac{1}{6}\right) = \tfrac{1}{24}. $$4. Changing the order of integration
Sometimes one order of integration leads to an integral with no elementary antiderivative. The cure: re-describe the same region with the variables swapped, then integrate the other way.
The canonical example is
$$ \int_0^1 \!\!\int_x^1 e^{y^2}\,dy\,dx. $$The inner integral $\int e^{y^2}\,dy$ has no closed form. Stuck — until you redraw the region.
The region is $\{(x, y) : 0 \le x \le 1,\; x \le y \le 1\}$ — the triangle with vertices $(0, 0)$, $(0, 1)$, $(1, 1)$. Described horizontally instead:
$$ R = \{(x, y) : 0 \le y \le 1,\; 0 \le x \le y\}. $$Reversed integral:
$$ \int_0^1\!\!\int_0^y e^{y^2}\,dx\,dy \;=\; \int_0^1 y\,e^{y^2}\,dy \;=\; \tfrac{1}{2}\!\left[e^{y^2}\right]_0^1 \;=\; \tfrac{e - 1}{2}. $$The inner $x$-integral is trivial because $e^{y^2}$ doesn't involve $x$; the outer $y$-integral becomes elementary by the substitution $u = y^2$. The whole calculation runs in three lines once you reorder.
If the inner antiderivative looks impossible — $e^{y^2}$, $\sin(x^2)$, $1/\ln y$ — and the region is symmetric enough to swap, redraw and try the other order. Fubini guarantees you'll get the same answer; you just won't be stuck.
5. Polar coordinates & the Gaussian integral
Rectangular coordinates are the wrong language for circular regions. Trying to integrate over a disk in $(x, y)$ forces inner limits like $\pm\sqrt{R^2 - x^2}$ — algebraically grim. Polar coordinates $(r, \theta)$ are the cure: they match the geometry, and they convert circular boundaries into rectangular ones in the $r$-$\theta$ plane.
Recall $x = r\cos\theta$, $y = r\sin\theta$. The area element is not $dr\,d\theta$ — there's a stretching factor, because a "polar rectangle" $[r, r + dr] \times [\theta, \theta + d\theta]$ has sides $dr$ (radial) and $r\,d\theta$ (arc length):
$$ \boxed{\,dA = r\,dr\,d\theta\,} $$The $r$ is the Jacobian of the change of coordinates, and forgetting it is one of the most common errors in this whole topic.
The disk, made easy
Watch what happens to the area of a disk of radius $R$ in polar form:
$$ \text{Area} \;=\; \iint_D 1\,dA \;=\; \int_0^{2\pi}\!\!\int_0^R r\,dr\,d\theta \;=\; \int_0^{2\pi}\! \tfrac{R^2}{2}\,d\theta \;=\; \pi R^2. $$The formula you've known since school, derived in two lines.
The Gaussian integral
Now the showpiece. The integral
$$ I \;=\; \int_{-\infty}^{\infty} e^{-x^2}\,dx $$has no elementary antiderivative — you cannot evaluate it by one-dimensional methods alone. But it can be evaluated by a trick that's almost unfair: compute $I^2$ as a double integral and switch to polar.
Step 1. Write $I^2$ as a product of two identical integrals, then recombine as a double integral:
$$ I^2 \;=\; \left(\int_{-\infty}^{\infty} e^{-x^2} dx\right)\!\!\left(\int_{-\infty}^{\infty} e^{-y^2} dy\right) \;=\; \iint_{\mathbb{R}^2} e^{-(x^2 + y^2)}\,dA. $$Step 2. Convert to polar. The whole plane becomes $0 \le r < \infty$, $0 \le \theta < 2\pi$, and $x^2 + y^2 = r^2$:
$$ I^2 \;=\; \int_0^{2\pi}\!\!\int_0^{\infty} e^{-r^2}\,r\,dr\,d\theta. $$Step 3. The inner integral surrenders immediately to $u = r^2$, $du = 2r\,dr$:
$$ \int_0^{\infty} r\,e^{-r^2}\,dr \;=\; \tfrac{1}{2}\!\int_0^{\infty} e^{-u}\,du \;=\; \tfrac{1}{2}. $$Step 4. Then $I^2 = 2\pi \cdot \tfrac{1}{2} = \pi$, and taking a square root:
$$ \boxed{\,\int_{-\infty}^{\infty} e^{-x^2}\,dx \;=\; \sqrt{\pi}\,} $$The 1D Gaussian has no elementary antiderivative — but its square, viewed as a 2D integral, has radial symmetry. The radial symmetry is what makes polar coordinates kill it. The trick recurs throughout probability and physics: when a 1D integral looks impossible, ask whether its 2D version has a symmetry you can exploit.
6. Triple integrals
Going from two to three dimensions adds nothing conceptually new. A function $f(x, y, z)$ over a region $E \subset \mathbb{R}^3$ has a triple integral
$$ \iiint_E f(x, y, z)\,dV $$defined as the limit of Riemann sums over tiny boxes, where $\Delta V_i = \Delta x_i\,\Delta y_i\,\Delta z_i$.
The geometric pictures generalize:
- $f \equiv 1$ gives the volume of $E$: $\iiint_E 1\,dV = \text{Vol}(E)$.
- If $f$ is mass density, you get total mass.
- If $f$ is a probability density on three random variables, you get a probability.
Iterated form (with $E$ described as a stack of vertical slabs):
$$ \iiint_E f\,dV \;=\; \int_a^b \!\!\int_{g_1(x)}^{g_2(x)}\!\!\int_{h_1(x,y)}^{h_2(x, y)} f(x, y, z)\,dz\,dy\,dx. $$The innermost limits are functions of both outer variables; the middle's are functions of the outermost; the outermost are constants. There are six possible orderings of $dx\,dy\,dz$ — pick the one that makes the limits cleanest.
Volume of the rectangular box $[0, a] \times [0, b] \times [0, c]$ is the easiest possible example:
$$ V = \int_0^a\!\!\int_0^b\!\!\int_0^c 1\,dz\,dy\,dx = abc. $$The unit tetrahedron with vertices $(0,0,0), (1,0,0), (0,1,0), (0,0,1)$ — bounded by $x + y + z \le 1$ in the first octant — is a one-page calculation that gives volume $1/6$.
7. Cylindrical and spherical coordinates
Just as polar coordinates rescue circular regions in 2D, the natural 3D coordinate systems rescue regions with cylindrical or spherical symmetry. Each has its own volume element — and getting the volume element right is the whole game.
Cylindrical coordinates
Cylindrical coordinates $(r, \theta, z)$ are polar coordinates in the $xy$-plane, with $z$ measured the usual Cartesian way. The conversion is
$$ x = r\cos\theta, \quad y = r\sin\theta, \quad z = z. $$The volume element inherits the polar $r$:
$$ \boxed{\,dV = r\,dz\,dr\,d\theta\,} $$Use cylindrical coordinates whenever the region has an axis of rotational symmetry — cylinders, paraboloids of revolution, anything written as "$r \le \text{something}$ and $z$ between two heights".
The volume of a cylinder of radius $a$ and height $h$ falls out in two lines:
$$ V = \int_0^{2\pi}\!\!\int_0^a\!\!\int_0^h r\,dz\,dr\,d\theta \;=\; 2\pi \cdot \tfrac{a^2}{2} \cdot h \;=\; \pi a^2 h. $$Spherical coordinates
Spherical coordinates $(\rho, \varphi, \theta)$ use one radial distance and two angles:
- $\rho \ge 0$ — distance from the origin.
- $\varphi \in [0, \pi]$ — polar angle from the $+z$-axis ("down from the north pole").
- $\theta \in [0, 2\pi)$ — azimuthal angle in the $xy$-plane, same as in polar/cylindrical.
The conversion is
$$ x = \rho\sin\varphi\cos\theta,\quad y = \rho\sin\varphi\sin\theta,\quad z = \rho\cos\varphi. $$The volume element carries a $\rho^2 \sin\varphi$ — derive it geometrically by noting that a spherical "box" spans $d\rho$ radially, $\rho\,d\varphi$ in the polar direction, and $\rho \sin\varphi\,d\theta$ azimuthally:
$$ \boxed{\,dV = \rho^2 \sin\varphi\,d\rho\,d\varphi\,d\theta\,} $$Use spherical coordinates whenever the region is bounded by spheres or cones from the origin. The sphere of radius $R$ has volume
$$ V = \int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^R \rho^2 \sin\varphi\,d\rho\,d\varphi\,d\theta \;=\; 2\pi \cdot 2 \cdot \tfrac{R^3}{3} \;=\; \tfrac{4}{3}\pi R^3. $$Picking the right coordinates
| Region type | Coordinates | Volume element |
|---|---|---|
| Boxes, slabs | Cartesian $(x, y, z)$ | $dV = dx\,dy\,dz$ |
| Cylinders, paraboloids of revolution | Cylindrical $(r, \theta, z)$ | $dV = r\,dz\,dr\,d\theta$ |
| Spheres, cones from origin | Spherical $(\rho, \varphi, \theta)$ | $dV = \rho^2 \sin\varphi\,d\rho\,d\varphi\,d\theta$ |
Every "extra factor" — the $r$ in polar, the $r$ in cylindrical, the $\rho^2 \sin\varphi$ in spherical — is the absolute value of the determinant of the Jacobian matrix of the coordinate change. The general statement is the change-of-variables theorem: for a smooth invertible map $T$, $\,dV_x = |\det J_T|\,dV_u$. Polar, cylindrical, and spherical are the three named cases everyone memorizes; the underlying principle handles them all.
The general change-of-variables formula
Suppose $T:(u, v) \mapsto (x, y)$ is a smooth, invertible map from a region $S$ in $uv$-space to a region $R$ in $xy$-space. The Jacobian determinant is
$$ \frac{\partial(x, y)}{\partial(u, v)} \;=\; \det\!\begin{pmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{pmatrix}. $$Then the change-of-variables theorem says
$$ \iint_R f(x, y)\,dA \;=\; \iint_S f\bigl(x(u,v), y(u,v)\bigr)\,\left|\frac{\partial(x, y)}{\partial(u, v)}\right|\,du\,dv. $$Polar coordinates are the famous instance: $\,\partial(x, y)/\partial(r, \theta) = r$. For the ellipse $x^2/a^2 + y^2/b^2 \le 1$, the substitution $x = ar\cos\theta$, $y = br\sin\theta$ has Jacobian $ab\,r$, and the area falls out as $\pi ab$.
Applications: mass and center of mass
For a plate occupying region $D$ with density $\sigma(x, y)$:
- Mass: $\displaystyle M = \iint_D \sigma(x, y)\,dA.$
- Moments about the axes: $\displaystyle M_y = \iint_D x\,\sigma\,dA, \quad M_x = \iint_D y\,\sigma\,dA.$
- Center of mass: $\displaystyle \bar{x} = \frac{M_y}{M}, \quad \bar{y} = \frac{M_x}{M}.$
(Note the indexing: $M_y$, the moment about the $y$-axis, uses $x$ as its lever arm. Many students invert this — pay attention.) The same templates work in 3D with $\iiint_E \rho\,dV$ for mass and $\bar{x} = \tfrac{1}{M}\iiint_E x\,\rho\,dV$, etc., for the centroid.