Multiple integrals & coordinates

Cut the reachable set into cells and add them up

Put the arm's base at the origin. With link lengths $L_1 = 1.3$ m and $L_2 = 0.7$ m, the set of positions the hand can reach is the annulus $0.6 \le \sqrt{x^2+y^2} \le 2.0$ — an inner circle the fully folded arm cannot get inside, an outer circle it cannot pass. Suppose the arm carries a payload whose areal density rises with distance from the base, $\rho(x,y) = 1 + 0.4\,(x^2+y^2)$ kg/m². The total mass is a double integral:

The definition is the right-hand side: chop $D$ into small cells, evaluate the density at a point of each cell, multiply by the cell's area, and add. Pick a region shape, raise the grid resolution, and watch the sum converge to the exact value underneath. A cell counts as inside when its centre is inside; at low resolution that is a crude boundary, and the crude boundary is the whole point.

Why an area element is not simply $dr\,d\theta$

For a region with circular symmetry — an annulus is the obvious one — the natural grid is polar: a radial coordinate $r$ and an angle $\theta$. The tempting area element is $\Delta r\,\Delta\theta$, but that is wrong, and the picture shows why. A cell at radius $r$ spanning $dr$ and $d\theta$ is, to first order, a rectangle with sides $dr$ and $r\,d\theta$. Its area is

The factor $r$ is the local stretch of a change of variables, and it is exactly the Jacobian determinant from Part 5: the map $(r,\theta)\mapsto(x,y)$ has determinant $r$. Slide the sweep open and watch the ring cells fill the disc; the highlighted element carries its own $r\,dr\,d\theta$ label. The readout compares the sum against the exact integral, which for a full sweep is $\pi R^2 + 0.4\,\pi R^4/2$.

Ball, cylinder, box — and the element each one wants

The same construction one dimension up: a triple integral is a limit of sums over little boxes, $\iiint_V f\,dV = \lim \sum f\,\Delta V$. Pick the solid and the coordinates that fit it. A cylinder is cut naturally into shells $r$, angles $\theta$ and heights $z$, and the box cut by $dr,d\theta,dz$ has volume $r\,dr\,d\theta\,dz$, so

A ball wants spherical coordinates $(\rho,\varphi,\theta)$, where $\varphi$ is the angle down from the $z$-axis. The small coordinate box has sides $d\rho$, $\rho\,d\varphi$ and $\rho\sin\varphi\,d\theta$, giving the classic element

Switch solid and coordinate system below. The canvas draws an $(r,z)$ cross-section with a representative cell highlighted; the readout reports the numeric triple integral using the chosen element, against the exact volume. Any coordinate system can compute any solid if you carry the right Jacobian — but the natural pairing converges fastest.

Sum without $r$ or $\rho^2\sin\varphi$ and get the wrong solid

It is worth seeing the factors fail. Compute the volume of a solid with the coordinate cells but drop the Jacobian: sum $\Delta r\,\Delta\theta\,\Delta z$ for a cylinder, or $\Delta\rho\,\Delta\varphi\,\Delta\theta$ for a ball. The sums still converge — to the wrong numbers, and to numbers you can predict. For a cylinder of radius $R$ and height $H$, the correct sum tends to $\pi R^2 H$; the naive one tends to $R\cdot 2\pi\cdot H = 2\pi R H$, which even has the wrong units. For a ball the naive sum tends to $2\pi^2 R$. The factor is not bookkeeping, it is the geometry: near the axis a coordinate cell is thin, near the rim it is stretched, and $r$ (or $\rho^2\sin\varphi$) measures that stretch.

Add the cells column by column, or row by row

A double integral over a rectangle is an iterated integral, and you may integrate in either order. This is Fubini's theorem: for a function continuous on the rectangle,

Both sides sum exactly the same cells, merely in a different sequence, so the totals agree. The demo makes that visible: switch the order and the shaded cells light up column-first instead of row-first, while the two totals in the readout stay equal. The progress slider sums cells in the chosen order and shows the partial total climbing toward the same value along the two different routes.

Volume elements are everywhere a density is

The change-of-variables machinery underneath these factors is in Part 5, and it pays off directly in Part 11: a probability density is normalised by integrating it to one, and changing variables for a random vector means multiplying by exactly the Jacobian determinant you watched stretch grids here. The same spherical element is how you integrate a radiance field over directions, which is the continuous version of the sampling in Multi-view geometry; and every expectation, marginal and normalisation constant in a model is a multiple integral over the space the variables live in.

• MIT 18.02, Multivariable Calculus — the lectures that derive these volume elements from double sums.
• Paul's Online Math Notes, Multiple Integrals — worked examples in every coordinate system.
• Khan Academy, Integrating multivariable functions — extra practice on iterated integrals and polar areas.