Math 344: Calculus III

15.4 Applications of Double Integrals

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15.4.1 Applications

Integrals are studied so extensively because of its interpretations and applications in mathematics, physics, and statistics. The basic geometric interpretation of a double integral as the volume under a surface was covered in previous sections, but now let us turn our attention to some physical and statistical idea with which we associate double integrals:

1. Density and Mass
2. Center of Mass
3. Moment of Inertia
4. Probability
5. Expected Value

15.4.2 Density and Mass

Double integrals allow for the computation of the mass of lamina (thin plates) that do not have uniform density. Suppose we have a flat lamina of density $\rho(x,y)$ where the mass per unit area is a function of $x$ and $y$. If we wanted to find the mass of the entire lamina (or a section of it) defined by a region $D$, we assume that the plate lies in the $xy$-plane and simply need to integrate its density function over $D$.

$$ m = \iint_D \rho(x,y)\,dA $$

Example 1

Let $D$ by the triangular region bounded by the lines $y = 0$, $y = 2x$, and $x + 2y = 1$. Find the mass of the lamina with density given by $\rho(x,y) = x$.

Solution

The region is depicted below.

The shading here indicates that the plate's density increases as we move to the right. We see that this is a type II region and will set up the integral

$$ m = \iint_D \rho(x,y)\,dA $$
accordingly and solve.

$$ \begin{align*} m &= \int_0^{0.4}\int_{x\, =\, \frac{y}{2}}^{x\, =\, 1-2y} x\,dxdy \\ \\ &= \int_0^{0.4}\left.\left[\frac{1}{2}x^2\right]\right|_{\,\frac{y}{2}}^{\,1-2y}\,dy \\ \\ &= \int_0^{0.4}\left[\frac{1}{2}\left((1-2y)^2 - \left(\frac{y}{2}\right)^2\right) \right]\,dy \\ \\ &= \frac{1}{2}\int_0^{0.4} 1 - 4y + 4y^2 - \frac{y^2}{4}\,dy \\ \\ &= \frac{1}{2}\int_0^{0.4} \frac{15}{4}y^2 - 4y + 1\,dy \\ \\ &= \frac{1}{2}\left.\left[\frac{5}{4}y^3 - 2y^2 + y\right]\right|_{\,0}^{\,0.4} \\ \\ &= \frac{1}{2}\left[ \frac{2}{25} - \frac{8}{25} + \frac{2}{5}\right] \\ \\ &= \frac{2}{25} \end{align*} $$
Therefore the mass the lamina is $\frac{2}{25}$ units.

Charge Density

In addition to mass, we can do a similar computation involving electric charge. The idea is that the charge density function $\sigma(x,y)$ takes the place of the mass density function, and if you integrate this function over a region $D$ the value of the integral

$$ q = \iint_D \sigma(x,y)\,dA $$
yields the total charge $q$ on the lamina defined by the region $D$.

15.4.3 Center of Mass

The center of mass of a lamina $D$ is determined by computing the first moment of the lamina about the $x$ and $y$ axes. With our knowledge of double integrals, we can handle the center of mass for lamina of non-uniform density $\rho(x,y)$. The formulas for these moments are

$$ M_y = \iint_D x\rho(x,y)\,dA \qquad\qquad\qquad M_x = \iint_D y\rho(x,y)\,dA $$

Notice that the letters here are "mismatched" between the subscript and what shows up in the integrand.

The location of the lamina's center of mass is given by $(\overline{x},\overline{y})$ where the coordinates are determined by taking the appropriate moment divided by the total mass of the lamina $m$.

$$ \overline{x} = \dfrac{M_y}{m} = \dfrac{1}{m}\iint_D x\rho(x,y)\,dA \qquad\qquad\qquad \overline{y} = \dfrac{M_x}{m} = \dfrac{1}{m}\iint_D y\rho(x,y)\,dA $$
The mass $m$ is computed using the method above.

Example 2

Find the center of mass of the region from Example 1 .

Solution

From our work above, we know that $m$ is $\frac{2}{25}$, so all we need to compute are the first moments $M_y$ and $M_x$.

$$ \begin{align*} M_y &= \iint x\rho(x,y)\,dA \\ \\ &= \int_0^{0.4}\int_{x\, =\, \frac{y}{2}}^{x\, =\, 1-2y} x^2\,dxdy \\ \\ &= \int_0^{0.4}\left.\left[\dfrac{x^3}{3}\right]\right|_{\,x\, =\, \frac{y}{2}}^{\,x\, =\, 1-2y}\, dy \\ \\ &= \frac{1}{3}\int_0^{0.4}\left[(1-2y)^3 - \left(\frac{y}{2}\right)^3\right]\, dy \\ \\ &= \frac{1}{3}\left[-\frac{1}{2}\frac{(1-2y)^4}{4} - 2\left(\frac{y}{2}\right)^4\frac{1}{4} \right]_0^{0.4} \\ \\ &= \frac{1}{3}\left[\left(-\frac{(0.2)^4}{8} - \frac{(0.4)^4}{32}\right) - \left(-\frac{1^4}{8} - \frac{0^4}{32}\right)\right] \\ \\ &= \frac{31}{750} \end{align*} $$
So the $x$-coordinate of the center of mass is


$$ \overline{x} = \frac{M_y}{m} = \frac{31}{750}\cdot\frac{25}{2} = \frac{31}{60} $$

For the other moment,

$$ \begin{align*} M_x &= \iint y\rho(x,y)\,dA \\ \\ &= \int_0^{0.4}\int_{x\, =\, \frac{y}{2}}^{x\, =\, 1-2y} xy\,dxdy \\ \\ &= \frac{1}{2}\int_0^{0.4} \left.\left[x^2 y\right]\right|_{\,x\, =\, \frac{y}{2}}^{\,x\, =\, 1-2y}\, dy \\ \\ &= \frac{1}{2}\int_0^{0.4} y(1-2y)^2 - y\left(\frac{y}{2}\right)^2\,dy \\ \\ &= \frac{1}{2}\int_0^{0.4} y(1-4y+4y^2) - \frac{y^3}{4}\,dy \\ \\ &= \frac{1}{2}\left.\left[\frac{y^2}{2} - \frac{4y^3}{3} + y^4 - \frac{y^4}{16}\right]\right|_{\,0}^{\,0.4} \\ \\ &= \frac{7}{750} \end{align*} $$
Which gives that the $y$-coordinate of the center of mass is

$$ \overline{y} = \frac{M_x}{m} = \frac{7}{750}\cdot\frac{25}{2} = \frac{7}{60} $$
Here is there region depicted with the center of mass.

15.4.4 Moments of Inertia

A moment of inertia is the rough equivalent of mass when discussing angular momentum. Mass for linear momentum is the quantity that indicates how much a body resists being accelerated. The more massive an object is, the larger amount of energy it takes to speed it up or slow it down. A moment of inertia works the same way for rotating about an axis. We start by considering a lamina of density $\rho(x,y)$ given by region $D$ and either the $x$- or $y$-axis. Next, we compute a quantity known as the second moment , whose formulas are closely related to the first moment formulas above.

$$ I_x = \iint_D y^2\rho(x,y)\,dA \qquad\qquad\qquad I_y = \iint_D x^2\rho(x,y)\,dA $$ $I_x$ is the moment about the $x$-axis and $I_y$ is the moment about the $y$-axis. Physically, this represents trying to rotate the lamina about the axis in question. For example, if we take the region from the previous examples and think about rotating it about $x$-axis we visualize "spinning" the lamina about the axis as indicated.

Additionally, we are also interested in the moment of inertia about the origin $I_0$, which physically represents the objects resistance to rotating about the origin in the $xy$-plane.

$$ I_0 = \iint_D \left(x^2 + y^2\right)\rho(x,y)\,dA = I_x + I_y $$

Exercise 1

Compute the moment about the $x$-axis $I_x$, the moment about the $y$-axis $I_y$, and the moment about the origin $I_0$ for a uniform density disk $\rho(x,y) = \rho$ of radius $a$ centered at the origin.

Check Your Work

$$ \begin{align*} I_0 &= \frac{\pi\rho a^4}{2} \\ \\ I_x = I_y &= \frac{\pi\rho a^4}{4} \end{align*} $$

Follow Along

Since this is a disk centered at the origin, polar coordinates are an obvious choice.

$$ \begin{align*} I_0 &= \iint_D \left(x^2 + y^2\right)\rho\,dA = \rho\int_0^{2\pi}\int_0^a r^2 r\,dr d\theta \\ \\ &= \rho\int_0^{2\pi}\,d\theta\int_0^a r^3\,dr = 2\pi\rho\left.\left[\frac{r^4}{4}\right]\right|_{\,0}^{\,4} = \frac{\pi\rho a^4}{2} \end{align*} $$
Lastly, by symmetry we see that $I_0 = I_x + I_y$ implies $I_x = I_y = \dfrac{I_0}{2} = \dfrac{\pi\rho a^4}{4}$.

15.4.5 Probability

A probability density function is used in the computations of probabilities. A probability density function $f(x)$ has the property that

$$ \int_{-\infty}^\infty f(x)\,dx = 1 $$
This represents the conservation of probability , as an event $X$ must occur if we examine all possibilities between $[\infty,\infty]$. If we want to find the probability of $X$ occurring within an interval $[a,b]$, we change the limits of integration

$$ P\left(a\leq X\leq b\right) = \int_a^b f(x)\,dx $$

This is the one-dimensional case for a single random variable. If we have two random variables, double integrals get involved. If the random variables are $X$ and $Y$, then the probability that $(X,Y)$ is in a region $D$ is given by

$$ P\left(\left(X,Y\right)\in D\right) = \iint_D f(x,y)\,dA $$
where is called a joint density function . Due to the conservation of probability, $f(x,y)$ must have the property that

$$ \iint_{\mathbb{R}^2} f(x,y)\,dA = \int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y)\,dxdy = 1 $$

15.4.6 Expected Value

An expected value is essentially an average value for one of the joint density function's variables. If $X$ and $Y$ are the random variables of the function $f(x,y)$, then the expected values of $X$ and $Y$ are given by

$$ \mu_1 = \iint_{\mathbb{R}^2} xf(x,y)\,dA \qquad\qquad\qquad \mu_2 = \iint_{\mathbb{R}^2} yf(x,y)\,dA $$
These formulas are the same as the first moment formulas from above and serve the same role as the center of mass values for a lamina with $\overline{x} = \mu_1$ and $\overline{y} = \mu_2$ (since the joint density function property $\iint_{\mathbb{R}^2} f(x,y)\,dA = 1$ is equivalent to $m=1$). The way to think about these values is that if you look over all space $\left(\mathbb{R}^2\right)$ $\mu_1$ is the most likely value for $X$ to be found. Likewise with $\mu_2$ and $Y$.

Expected values are important in physics, such as with quantum mechanics ) . Since we cannot pinpoint the exact location of a particle (as it moves like a wave), we rely on expected values to estimate the most likely locations for the particle to be based on its wave function (the name used in quantum mechanics for a joint density function).

Exercise 2

1. Find $C$ so that $f(x,y)$ will be a joint density function
   
   $$ f(x,y) = \left\{\begin{array}{ll} Cxy & \text{when } 0\leq x\leq 1,\ \ 0\leq y\leq 1 \\ 0 & \text{otherwise} \end{array}\right. $$
   
2. If $X$ and $Y$ are random variables whose joint density function if $f$, find
   1. $P\left(X\geq\frac{1}{2}\right)$
   2. $P\left(X\geq\frac{1}{2},Y\leq\frac{1}{2}\right)$
3. Find the expected values of $X$ and $Y$.

Check Your Work

1. $C = 4$
2. 1. $ P\left(X\geq\frac{1}{2}\right) = \frac{3}{4} $
   2. $ P\left(X\geq\frac{1}{2},Y\leq\frac{1}{2}\right) = \frac{3}{16} $
3. $\mu_1 = \mu_2 = \frac{2}{3}$

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