Applications of Double Integrals, multivariable and vector calculus lesson

Balance a cardboard cutout on a fingertip and there is exactly one spot where it stays level. If the cardboard is thicker on one side, that spot moves. Finding it means adding up tiny pieces of mass, each weighted by where it sits, and a double integral does exactly that. The same sums, read with probability instead of mass, give averages and correlations of two random quantities.

Key terms

Area

Definition Let $D$ be a bounded region in the plane whose boundary has zero area, for example a boundary made of finitely many smooth curves. Its area is

A(D) = \iint_D 1 \, dA.

In plain words Integrating the constant height $1$ gives a solid whose volume has the same number as the area of its base.

Example Let $D$ be the region between $y = x^2$ and $y = x + 2$ . The curves meet where $x^2 = x + 2$ , at $x = -1$ and $x = 2$ , and the line is on top. So

\begin{aligned} A(D) &= \int_{-1}^{2} \int_{x^2}^{x + 2} 1 \, dy \, dx \\[0.4em] &= \int_{-1}^{2} (x + 2 - x^2) \, dx = \frac{9}{2}. \end{aligned}

Density and mass

Definition A lamina is a thin plate occupying a region $D$ as above, with area density $\sigma(x, y) \ge 0$ , measured in mass per unit area. Assume $\sigma$ is continuous on the closed bounded region $D$ , or more generally bounded and continuous except on a set of zero area. A small patch of area $dA$ around $(x, y)$ then has mass $dM = \sigma(x, y) \, dA$ , and the total mass is

m = \iint_D \sigma(x, y) \, dA.

In plain words Cut the plate into tiny patches, multiply each patch’s area by how heavy the material is there, and add.

Example A disk of radius $2$ meters has density $\sigma = \sqrt{x^2 + y^2}$ kilograms per square meter, heavier toward the rim. In polar coordinates

\begin{aligned} m &= \int_0^{2\pi} \int_0^2 r \cdot r \, dr \, d\theta \\[0.4em] &= \frac{16\pi}{3} \text{ kilograms,} \end{aligned}

about $16.8$ kilograms.

Moments

Definition The moments of the lamina about the coordinate axes are

\begin{aligned} M_x &= \iint_D y \, \sigma \, dA, \\[0.4em] M_y &= \iint_D x \, \sigma \, dA. \end{aligned}

The moment about the $x$ axis weights each patch by its signed distance $y$ from that axis, and the moment about the $y$ axis uses $x$ .

In plain words A moment measures how strongly the mass would tip the plate around an axis. Mass far from the axis counts more, and mass on opposite sides cancels.

Example Take the triangle $x \ge 0$ , $y \ge 0$ , $x + y \le 2$ with density $\sigma = x$ . For fixed $x$ the variable $y$ runs from $0$ to $2 - x$ , so

\begin{aligned} M_y &= \int_0^2 \int_0^{2 - x} x^2 \, dy \, dx = \frac{4}{3}, \\[0.4em] M_x &= \int_0^2 \int_0^{2 - x} xy \, dy \, dx = \frac{2}{3}. \end{aligned}

Center of mass

Definition If the lamina has mass $m > 0$ , its center of mass is the point $(\bar{x}, \bar{y})$ with

\bar{x} = \frac{M_y}{m}, \qquad \bar{y} = \frac{M_x}{m}.

When $\sigma$ is constant this point depends only on the shape and is called the centroid of $D$ .

In plain words The center of mass is the density weighted average position. A plate supported at that single point balances.

Example For the triangle with $\sigma = x$ the mass is

\begin{aligned} m &= \int_0^2 \int_0^{2 - x} x \, dy \, dx \\[0.4em] &= \int_0^2 x(2 - x) \, dx = \frac{4}{3}. \end{aligned}

So $\bar{x} = (4/3)/(4/3) = 1$ and $\bar{y} = (2/3)/(4/3) = 1/2$ . The uniform triangle has centroid $(2/3, 2/3)$ , and the extra density toward large $x$ pulls the balance point right.

Moment of inertia

Definition The moments of inertia, or second moments, of the lamina about the $x$ axis, the $y$ axis and the origin are

\begin{aligned} I_x &= \iint_D y^2 \sigma \, dA, \\[0.4em] I_y &= \iint_D x^2 \sigma \, dA, \\[0.4em] I_0 &= \iint_D (x^2 + y^2) \, \sigma \, dA \\[0.4em] &= I_x + I_y. \end{aligned}

The last one is the moment about the axis through the origin perpendicular to the plate.

In plain words Moment of inertia measures how hard the plate is to start spinning about an axis. Distance enters squared, so mass near the rim matters much more than mass near the axis.

Example For the triangle with $\sigma = x$ ,

\begin{aligned} I_y &= \int_0^2 \int_0^{2 - x} x^3 \, dy \, dx = \frac{8}{5}, \\[0.4em] I_x &= \int_0^2 \int_0^{2 - x} x y^2 \, dy \, dx = \frac{8}{15}, \end{aligned}

and $I_0 = 8/5 + 8/15 = 32/15$ .

Radius of gyration

Definition For a lamina of mass $m > 0$ with moment of inertia $I$ about a given axis, the radius of gyration about that axis is the number $R \ge 0$ with $mR^2 = I$ , that is

R = \sqrt{\frac{I}{m}}.

In plain words If you squeezed all the mass into a single point at distance $R$ from the axis, it would be exactly as hard to spin as the real plate.

Example For the triangle, the radius of gyration about the $y$ axis is

\sqrt{\frac{8/5}{4/3}} = \sqrt{\frac{6}{5}} \approx 1.095.

This is larger than $\bar{x} = 1$ . That is no accident, since the mass weighted average of $x^2$ is never smaller than the square of the mass weighted average of $x$ .

Joint probability density

Definition A joint probability density function of a pair of random variables $(X, Y)$ is a function $f$ on $\mathbb{R}^2$ with $f \ge 0$ , integrable over the whole plane as an improper integral, and

\iint_{\mathbb{R}^2} f(x, y) \, dA = 1,

such that for every region $D$ whose boundary has zero area

P\big((X, Y) \in D\big) = \iint_D f(x, y) \, dA.

In plain words A joint density is a probability mass spread over the plane. The chance of landing in a region is the amount of that mass sitting over the region.

Example Let $f(x, y) = x + y$ on the unit square $0 \le x \le 1$ , $0 \le y \le 1$ , and $f = 0$ elsewhere. It is nonnegative and its total integral is $1/2 + 1/2 = 1$ . The chance that $X + Y \le 1$ is

\begin{aligned} &\int_0^1 \int_0^{1 - x} (x + y) \, dy \, dx \\[0.4em] &\quad = \int_0^1 \frac{1 - x^2}{2} \, dx = \frac{1}{3}. \end{aligned}

Marginal densities

Definition If $(X, Y)$ has joint density $f$ , then $X$ alone has the marginal density

f_X(x) = \int_{-\infty}^{\infty} f(x, y) \, dy,

and similarly $f_Y(y)$ is obtained by integrating out $x$ . The variables $X$ and $Y$ are independent when

\begin{aligned} &P(X \in A, \ Y \in B) \\[0.4em] &\quad = P(X \in A) \, P(Y \in B) \end{aligned}

for all intervals $A$ and $B$ . For a joint density this holds exactly when $f(x, y) = f_X(x) f_Y(y)$ for all $(x, y)$ outside a set too thin to change any double integral, such as finitely many curves.

In plain words To forget about $Y$ , pile all the probability above each $x$ onto that $x$ .

Example For $f(x, y) = x + y$ on the unit square, and for $0 \le x \le 1$ ,

f_X(x) = \int_0^1 (x + y) \, dy = x + \frac{1}{2},

with $f_X(x) = 0$ outside $[0, 1]$ . By symmetry $f_Y(y) = y + 1/2$ on $[0, 1]$ . Since $(x + 1/2)(y + 1/2)$ differs from $x + y$ everywhere except on the lines $x = 1/2$ and $y = 1/2$ , the variables are not independent.

Expectation and covariance

Definition For a continuous function $g$ , the expected value of $g(X, Y)$ is

E[g(X, Y)] = \iint_{\mathbb{R}^2} g \, f \, dA,

with $g$ and $f$ evaluated at $(x, y)$ , provided the integral of $\lvert g \rvert f$ is finite, otherwise the expectation does not exist. In particular $\mu_X = E[X]$ and $\mu_Y = E[Y]$ . When $E[X^2]$ and $E[Y^2]$ are finite, the covariance is

\begin{aligned} &\operatorname{Cov}(X, Y) \\[0.4em] &\quad = E\big[(X - \mu_X)(Y - \mu_Y)\big] \\[0.4em] &\quad = E[XY] - \mu_X \mu_Y. \end{aligned}

These conditions make $E[X]$ , $E[Y]$ and $E[XY]$ finite, by the Cauchy-Schwarz inequality. Independent variables with finite $E[X^2]$ and $E[Y^2]$ have covariance $0$ , but covariance $0$ does not imply independence.

In plain words An expectation is a weighted average with the density as the weight. Covariance is positive when large values of $X$ tend to come with large values of $Y$ , and negative when they tend to come with small ones.

Example For $f(x, y) = x + y$ on the unit square,

\begin{aligned} E[X] &= \int_0^1 x\left(x + \frac{1}{2}\right) dx \\[0.4em] &= \frac{7}{12}, \\[0.4em] E[XY] &= \int_0^1 \int_0^1 xy(x + y) \, dy \, dx \\[0.4em] &= \frac{1}{3}, \end{aligned}

and $E[Y] = 7/12$ by symmetry. So

\operatorname{Cov}(X, Y) = \frac{1}{3} - \frac{49}{144} = -\frac{1}{144}.

The covariance is slightly negative, so the two variables lean a little in opposite directions.

Check yourself

Find the center of mass of a half disk $x^2 + y^2 \le a^2$ , $y \ge 0$ , with constant density $\sigma$ .

Show the answer

By symmetry $\bar{x} = 0$ . The mass is $m = \sigma \pi a^2 / 2$ , and in polar coordinates

\begin{aligned} M_x &= \sigma \int_0^\pi \int_0^a r\sin\theta \cdot r \, dr \, d\theta \\[0.4em] &= \frac{2\sigma a^3}{3}. \end{aligned}

So $\bar{y} = M_x / m = 4a/(3\pi)$ , about $0.42a$ .

A uniform rectangular plate $0 \le x \le 3$ , $0 \le y \le 2$ has density $1$ . Find its moment of inertia about the $x$ axis and its radius of gyration about that axis.

Show the answer

The mass is $m = 6$ and

I_x = \int_0^3 \int_0^2 y^2 \, dy \, dx = 3 \cdot \frac{8}{3} = 8.

The radius of gyration is $\sqrt{8/6} = 2/\sqrt{3} \approx 1.155$ .

Let $f(x, y) = 2$ on the triangle $0 \le y \le x \le 1$ and $0$ elsewhere. Check that it is a joint density, find the marginal density of $X$ , and find $E[Y]$ .

Show the answer

The function is nonnegative and the triangle has area $1/2$ , so the total integral is $2 \cdot 1/2 = 1$ . For $0 \le x \le 1$ the marginal is $f_X(x) = \int_0^x 2 \, dy = 2x$ , and $f_X = 0$ elsewhere. Then

\begin{aligned} E[Y] &= \int_0^1 \int_0^x 2y \, dy \, dx \\[0.4em] &= \int_0^1 x^2 \, dx = \frac{1}{3}. \end{aligned}

Applications of Double Integrals

Key terms

Area

Density and mass

Moments

Center of mass

Moment of inertia

Radius of gyration

Joint probability density

Marginal densities

Expectation and covariance

Check yourself

How this lesson connects