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Math Analysis 2

Double integrals: normal domains, change of variables and polar coordinates

Definition, reduction formulas and swapping the order, the Jacobian and change of variables, polar coordinates, the Gaussian integral, centres of mass and moments of inertia.

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The integral of Analysis 1 computes the area under a graph. The double integral does the same one floor up: it computes the volume under a surface, but also the mass of a plate, the charge on an electrode, the centroid of a cross-section or the moment of inertia of a beam. In this lesson we see how it is defined, how it is computed by reducing it to two single integrals and how variables are changed, with polar coordinates.

The definition

Split the region D of the plane into many small rectangles of area \Delta A_k, choose a point in each and form the sum \sum f(x_k, y_k)\,\Delta A_k. If, as the rectangles get smaller, the sums tend to a limit, that is the double integral:

    \[ \iint_D f(x, y)\,dx\,dy \]

With f \ge 0 it is the volume of the solid between the region D and the graph of f. With f = 1 it is the area of D. Continuous functions on bounded regions with a regular boundary are always integrable, and linearity, monotonicity and additivity over regions hold.

Computing it: normal domains

The idea is to slice the solid. A region is normal with respect to the x axis if it can be described as

    \[ D = \{(x, y) : a \le x \le b,\ g_1(x) \le y \le g_2(x)\} \]

that is, if every vertical line cuts it in a single segment.

A domain normal with respect to the x axis between two curves with a vertical strip
A normal domain: for each x between a and b, y runs from g_1(x) to g_2(x). The double integral is computed by first summing along the vertical strip (in y) and then adding up all the strips (in x).

Reduction formula (Fubini’s theorem):

    \[ \iint_D f\,dx\,dy = \int_a^b\left(\int_{g_1(x)}^{g_2(x)}f(x, y)\,dy\right)dx \]

The inner integral, with x fixed, gives the area of the vertical slice; integrating the slice areas gives the volume. It is Cavalieri’s principle. In the same way you integrate first in x and then in y if the domain is normal with respect to the y axis.

Example. \iint_D xy\,dx\,dy over the triangle with vertices (0, 0), (1, 0), (1, 1), that is 0 \le x \le 1, 0 \le y \le x:

    \[ \int_0^1\left(\int_0^x xy\,dy\right)dx = \int_0^1 x\cdot\frac{x^2}{2}\,dx = \frac18 \]

Swapping the order of integration

Sometimes one order leads to an impossible integral and the other to an easy one. \int_0^1\int_y^1 e^{x^2}dx\,dy cannot be computed as it stands, because e^{x^2} has no elementary antiderivative. The domain is the same triangle as before, seen as 0 \le y \le x, 0 \le x \le 1. Swapping:

    \[ \int_0^1\left(\int_0^x e^{x^2}dy\right)dx = \int_0^1 x\,e^{x^2}dx = \frac{e - 1}{2} \]

Change of variables

Switching to new variables (u, v) with x = x(u, v), y = y(u, v), the small rectangles du\,dv become curved parallelograms of area |\det J|\,du\,dv, where J is the Jacobian matrix of the transformation. It is the determinant as the area scaling factor, seen in Geometry and Algebra:

    \[ \iint_D f(x, y)\,dx\,dy = \iint_{D'} f(x(u, v), y(u, v))\,|\det J(u, v)|\,du\,dv \]

Polar coordinates

With x = r\cos\theta, y = r\sin\theta:

    \[ J = \begin{pmatrix}\cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta\end{pmatrix},\qquad \det J = r\cos^2\theta + r\sin^2\theta = r \]

    \[ dx\,dy = r\,dr\,d\theta \]

Area element in polar coordinates, a sector of an annulus
The small “rectangle” of polar coordinates is a piece of an annulus with sides dr and r\,d\theta: its area is r\,dr\,d\theta. The factor r explains why regions far from the origin weigh more.

Polar coordinates simplify everything with circular symmetry: discs, annuli, sectors, and functions depending on x^2 + y^2.

Example: the area of a circle. \int_0^{2\pi}\int_0^R r\,dr\,d\theta = 2\pi\cdot\frac{R^2}{2} = \pi R^2.

The Gaussian integral

I = \int_{-\infty}^{\infty}e^{-x^2}dx cannot be computed with antiderivatives. The trick is to square it and move to the plane:

    \[ I^2 = \iint_{\mathbb R^2}e^{-(x^2 + y^2)}dx\,dy = \int_0^{2\pi}\int_0^\infty e^{-r^2}r\,dr\,d\theta = 2\pi\cdot\frac12 = \pi \]

so I = \sqrt\pi. It is the result that normalises the Gaussian distribution, used everywhere from statistics to thermal noise in circuits.

Physical applications

For a plate with surface density \rho(x, y):

  • mass M = \iint_D\rho\,dA;
  • centre of mass x_G = \frac1M\iint_D x\rho\,dA, y_G = \frac1M\iint_D y\rho\,dA;
  • moment of inertia about the origin I_0 = \iint_D(x^2 + y^2)\rho\,dA.

Example. The moment of inertia of a uniform disc of mass M and radius R (density \rho = M/\pi R^2): I_0 = \rho\int_0^{2\pi}\int_0^R r^2\cdot r\,dr\,d\theta = \rho\cdot 2\pi\frac{R^4}{4} = \frac12 MR^2. It is the value in the table of the Physics 1 lesson on the rigid body, now proved.

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