Double integral

On an n × n grid over a region R, each cell whose center lies in R carries a column as tall as f at that center, and the columns’ total volume, $\sum f(x_i, y_j)\,\Delta x\,\Delta y$, is a Riemann sum. As the grid gets finer the sums approach one number, the double integral $\iint_R f\,dA$. Where f is negative a column hangs below the floor and counts negative, so the double integral is a signed volume.

Slicing turns the double integral into two ordinary integrals. For x-slices, the blue knife is the plane x = s. Its slice is the blue panel, and the panel’s signed area $A(s)$ is an ordinary integral in y. The graph under the view plots A against the knife’s place. With Columns off, the signed area shaded yellow up to the knife is the volume swept so far, and the signed area over the whole range is the double integral. This is the iterated integral. When every slice of R is a single interval, it reads

The inner limits are where the knife enters and leaves R, the blue points on the floor. On the triangle $0 \le y \le x \le 2$ the slice at x runs from y = 0 to y = x, so the inner integral is $\int_0^x$: the upper limit depends on x. On the ring $1 \le x^2 + y^2 \le 4$, a knife with $|x| < 1$ crosses the hole, the slice falls into two pieces, and the inner integral becomes a sum of two integrals, one for each piece.

Changing the order means reading the limits from the region again, not swapping them. On the same triangle a slice at fixed y runs from x = y to x = 2, so

Both orders shows the two slice areas side by side. They are different functions, $A(x) = x + x^2/2$ rising steadily and $B(y) = (1 + y)(2 - y)$ rising and falling, yet the areas under them agree. A finite sum of columns gives the same total added row by row or column by column, and the two iterated integrals are what those two ways of adding become as the grid gets finer. If f is continuous on a closed, bounded region whose boundary is made of finitely many smooth curves, both orders give the double integral. This is Fubini’s theorem, and every preset on this page satisfies it.

Sometimes only one order can be done with elementary functions. Over the triangle $0 \le x \le y \le 1$, fixing x leaves $\int_x^1 e^{y^2}\,dy$, and $e^{y^2}$ has no elementary antiderivative. Fixing y instead, each slice is a rectangle y wide and $e^{y^2}$ tall, and the volume is $\int_0^1 y\,e^{y^2}\,dy = (e - 1)/2 \approx 0.859$.