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All those who seem stupid, they are, and also so are half of those who do not, Quevedo.
A double integral allows us to integrate a function of two variables, $f(x, y)$, over a two-dimensional region $R$ in the xy-plane. While a single integral calculates the area under a curve, a double integral calculates the volume under a surface $z = f(x, y)$. It is denoted or written as $\iint_R f(x, y) dA$.
Fubini’s Theorem. If $f(x, y)$ is continuous on a rectangular region $R = [a, b] \times [c, d]$, the double integral can be computed as an iterated integral in either order without changing the result: $\int \int_R f(x, y)dA = \int_{a}^{b}\int_{c}^{d} f(x, y)dydx = \int_{c}^{d}\int_{a}^{b} f(x, y)dxdy$
Double integrals are crucial tools in multivariable calculus, allowing us to calculate fundamental geometric and physical quantities.
Double integrals are powerful mathematical tools that allow us to calculate a variety of important quantities, such as areas of regions, volumes under surfaces, average values of functions over regions, and physical quantities like mass, center of mass, and moment of inertia.
By understanding the techniques for setting up and evaluating double integrals, including the ability to convert to alternative coordinate systems such as polar or spherical coordinates when appropriate, we unlock the ability to solve a wide range of complex problems in multivariable calculus.
When dealing with physical systems or mathematical models, we often encounter situations where we want to find the average value of a function over a certain region. However, if the region has a varying density — meaning that some areas carry more “weight” than others, then we calculate a weighted average.
This approach is particularly useful when the density, denoted as $\delta(x, y)$, is not uniform across the region R.
The weighted average of a function f with a given density function $\delta(x, y)$ over a region R can be computed using a double integral, more specifically, $\text{Weighted Average} = \frac{1}{\text{Mass}(R)} \iint_R f(x, y) \cdot \delta(x, y)dA$, where:
This formula takes into account how the density function distributes mass across the region, providing a more accurate representation of the average value of f(x, y) than a simple arithmetic mean.
It’s a crucial concept in physics because it simplifies the study of the motion of objects by allowing us to treat them as if all their mass were located at a single point. It is effectively the “balance point” of the object, where if you were to support it at this point, the object would remain perfectly balanced.
In discrete systems, we compute $x_{cm} = \frac{\sum m_ix_i}{\sum m_i}$. For a continuous object, each tiny piece of area $dA$ has mass $\delta(x, y)dA$. Replace the sum with an integral: $\frac{\sum m_ix_i}{\sum m_i} \rightarrow \frac{\iint_R x \cdot \delta(x, y) dA}{\iint_R \delta(x, y) dA}$. Thus, the center of mass is the ratio of the total moment to the total mass.
For a region $R$ in the xy-plane with density function $\delta(x, y)$, the coordinates of the center of mass $(\bar{x}, \bar{y})$ are found by calculating the moments about the y-axis and x-axis, respectively: $x_{\text{centreMass}} = \frac{1}{\text{Mass}(R)} \iint_R x \cdot \delta(x, y) dA$, $y_{\text{centreMass}}= \frac{1}{\text{Mass}(R)} \iint_R y \cdot \delta(x, y) dA$
Loosely speaking, the moment of inertia about an axis is how hard it is to change the rotation motion about that axis.
When you want to spin something faster or slower, the moment of inertia determines how much effort (torque) is needed to change the object’s rotational speed. The higher the moment of inertia, the harder it is to change the rotational speed of the object.
The moment of inertia is a fundamental property of a rotating object. It quantifies the object’s resistance to changes in its rotational motion. It depends on both the mass distribution of the object and the distance of each mass element from the axis of rotation.
🎯 The moment of inertia about the origin measures how hard it is to rotate a body around the origin. It depends on how far the mass is from the axis of rotation. Mathematically, $I = \iint r^2 dm$. For a solid with density $\delta(x, y)$, $dm = \delta(x, y)dA$, so $\boxed{I = \iint_{R} r^2 \cdot \delta(x, y)dA}$ where r (in Cartesian coordinates, $r^2 = x^2 + y^2$) represents the perpendicular distance of a mass element from the axis of rotation.
🧠 Moment of inertia is a weighted measure of mass distribution. Each tiny mass element contributes: $\text{Contribution} = \text{distance from axis}^2 \times \text{mass of the element} = r^2\cdot \delta(x, y)dA$. Mass near the origin contributes almost nothing. Mass far away contributes a lot.
$\star$. If the rotation is about the x-axis, the moment of inertia is: $I_x = \iint_{R} y^2 \cdot \delta(x, y)dA.$
$\star$. For rotations about the y-axis, the moment of inertia is: $I_y = \iint_{R} x^2 \cdot \delta(x, y)dA.$
$\star$. For rotation about the origin (the polar moment of inertia): $I_0 = I_x + I_y = \iint_{R} (x^2+y^2) \cdot \delta(x, y)dA.$
$\star$. In polar coordinates, where $r^2 =x^2+y^2$ and the area element dA is given by $rdrd\theta$, this integral can be expressed as: $\boxed{I = \iint_{R} r^2\cdot \delta(r, \theta) \cdot rdrd\theta}$
Kinetic energy is the energy associated with the motion of an object. For an object in linear motion, it can easily be determined by an equation using the mass and velocity of that object.
The kinetic energy is one-half the product of the particle’s mass m and the square of its speed, $\text{KE} = \frac{1}{2}mv^2$ where $m$ is the mass of the object (kg) and $v$ is its velocity (m/s) -Figure iv a-. This equation tells us how much energy an object possesses due to its motion. When something is moving, it has kinetic energy —whether it’s a speeding car, a flying bird, a rolling ball, or even Superman.
For a rotating rigid body, the rotational kinetic energy depends on the moment of inertia I and the angular velocity ω, which is the rate at which the object rotates about a fixed axis.
The angular velocity represents how fast an object rotates about a fixed axis, i.e., the rate of change of the angle with respect to time: $\omega = \frac{\Delta \theta}{\Delta t}$ where $\Delta \theta$ is the change in angle (radians), $\Delta t$ is the time taken for this change.
🧭 Linear velocity ($v$) is the speed of an object moving along a straight path, $v = \frac{d}{t}$.
🧭 If a rigid body completes one rotation in time $T$, then: angle swept: $\Delta \theta = 2\pi$ radians; time $\Delta t = T$ (the period of one complete rotation), so angular velocity is $\omega = \frac{\Delta \theta}{\Delta t} = \frac{2\pi}{T}$. This is the rotational analog of “distance over time.”
🧭 A point on the rotating body at radius $r$ travels a circular arc of length: $\text{distance} = 2\pi r$. Thus, its linear speed is $v = \frac{d}{t} = \frac{2\pi r}{T} = r \omega$ (Figure iv b).
This is the key bridge between linear and rotational motion: Linear speed = radius × angular speed.
🧭 The formula for kinetic energy of a small mass element $dm$ at distance $r$: $K_{\text{element}} = \frac{1}{2}dm v^2 = \frac{1}{2}dm (r \omega)^2 = \frac{1}{2}(dm)r^2 \omega^2$.
🧭 A rigid body is made of infinitely many tiny mass elements. Its total kinetic energy is: $K = \iint \frac{1}{2}r^2 \omega^2 dm$.
Since $\omega$ is the same for every point in a rigid body (they all rotate together), pull it out: $K = \frac{1}{2}\omega^2 \iint r^2 dm$.
Next, we define the moment of inertia as $I = \iint r^2 dm$.
🧩 Why $I = \iint r^2 dm$ makes sense?
It is literally summing up: mass elements, weighted by how far they are from the axis, squared.
It measures how the mass is distributed relative to the axis.
Mass close to the axis contributes very little (small $r$). Mass far from the axis contributes a lot (large $r^2$).
The square makes distant mass dramatically more important.
This matches physical intuition: a long rod is harder to spin than a compact disk of the same mass.
🧭 Substitute $I$ into the kinetic energy expression: The kinetic energy of a rotating rigid body is directly proportional to its moment of inertia and the square of its angular velocity (measured in radians per second), $K = \frac{1}{2}I \omega^2$.
This is the rotational analog of the familiar linear formula: $K = \frac{1}{2}mv^2$. But instead of mass $m$, we have the “rotational mass” $I$, which measures how the mass is distributed relative to the axis.
This is why a figure skater spins faster when pulling their arms inward: they reduce $r$, and thus reduce $I$.
⭐ If density is $\delta(x, y) \implies dm = \delta(x, y)dA$, so $I = \iint r^2 dm = \iint r^2 \delta(x, y)dA = \iint (x^2 + y^2) \delta(x, y)dA$. This is the standard formula used in physics and engineering.
Set up the geometry. For simplicity, we’ll place the square on the coordinate plane such that its center is at the origin $(0, 0)$, and its sides are aligned with the coordinate axes. The square has a side length of $2a$, so its vertices are at $(−a, −a), (a, −a), (−a, a),$ and $(a, a)$.
The total area is $4a^2$. The function for distance from the center is $f(x,y) = \sqrt{x^2 + y^2}$.
Set Up the Integral for the Average Distance.
The average value of a function f(x, y) over a region R is found using the double integral, divided by the area of R. $F_{\text{average}} = \frac{1}{\text{Area of } R} \iint_R f(x,y)dA$ where $f(x, y)$ represents the function whose average value we want to compute, and $R$ represents the region over which we are averaging the function.
In our problem, the function of interest is the distance of a point (x, y) from the center, given by f(x, y) = $\sqrt{x^2+y^2}$.
The average distance $d_{average}$ is then $d_{average} = \frac{1}{4a^2} \iint_R f(x,y)dA = \frac{1}{4a^2} \iint_R \sqrt{x^2+y^2}dxdy$. This integral computes the sum of distances of all points within the square from the center.
Exploit Symmetry. The square is symmetric across both axes and the diagonals. We can integrate over just one octant (the triangle where $0 \le y \le x \le a$) and multiply the result by 8 to account for the entire square. $\text{Average Distance} = \frac{8}{4a^2} \iint_{\text{octant}} \sqrt{x^2 + y^2} dA = \frac{2}{a^2} \iint_{\text{octant}} \sqrt{x^2 + y^2} dxdy$
Convert to Polar Coordinates:
To evaluate the integral, it’s convenient to switch to polar coordinates. In polar coordinates: $x = r\cos(\theta), y = r\sin(\theta), x^2 + y^2 = r^2 \implies \sqrt{x^2 + y^2} = r, dA = rdrd\theta$.
The angle $\theta$ goes from $0$ to $\frac{\pi}{4}$ (one octant).
The radius $r$ goes from $0$ to the vertical line $x = a$. Since $x = r\cos(\theta)$, the upper bound for $r$ is $\frac{a}{\cos(\theta)} = a\sec(\theta)$.
Therefore, the integral becomes: $\text{Avg Dist} = \frac{2}{a^2} \int_{0}^{\frac{\pi}{4}} \int_{0}^{a\sec(\theta)} (r)(r) \, dr \, d\theta = \frac{2}{a^2} \int_{0}^{\frac{\pi}{4}} \int_{0}^{a\sec(\theta)} r^2 dr d\theta$
Evaluate the Inner Integral with respect to r: $\int_{0}^{\frac{a}{\cos(\theta)}}r^2dr = \frac{r^3}{3}\bigg|_{0}^{\frac{a}{\cos(\theta)}} = \frac{1}{3}\frac{a^3}{\cos^3(\theta)} = \frac{a^3}{3\cos^3(\theta)}$.
Evaluate the Outer Integral.
Next, substitute the result into the outer integral and evaluate with respect to $\theta$:
$\text{Avg Dist} = \frac{2}{a^2}·\frac{a^3}{3} \int_{0}^{\frac{\pi}{4}} \sec^3(\theta)d\theta = \frac{2a^3}{3a^2} \int_{0}^{\frac{\pi}{4}} \sec^3(\theta)d\theta = \frac{2a}{3}\int_{0}^{\frac{\pi}{4}} \sec^3(\theta)d\theta$.
Let’s use integration by parts: $\int u dv = uv - \int v du$ where $u = \sec(\theta), dv = \sec^2(\theta)d\theta, du = \sec(\theta)\tan(\theta)d\theta, v = \int \sec^2(\theta)d\theta = \tan(\theta)$
$\int \sec^3(\theta)d\theta = \sec(\theta)\tan(\theta) -\int \tan^2(\theta)\sec(\theta)d\theta$
Since $\tan^2(\theta) = \sec^2(\theta)-1$, we can rewrite the integral as: $\int \sec^3(\theta)d\theta = \sec(\theta)\tan(\theta) -\int (\sec^2(\theta)-1)\sec(\theta)d\theta = \sec(\theta)\tan(\theta) -\int \sec^3(\theta)d\theta + \int \sec(\theta)d\theta$.
Therefore, $2\int \sec^3(\theta)d\theta = \sec(\theta)\tan(\theta) + \int \sec(\theta)d\theta$.
We already know the integral of $\sec(\theta)$, $\int \sec(\theta)d\theta = \ln|\sec(\theta)+\tan(\theta)| + C.$
Thus, $\int \sec^3(\theta)d\theta = \frac{1}{2}\sec(\theta)\tan(\theta) + \frac{1}{2}\ln|\sec(\theta)+\tan(\theta)| + C$.
$\int_{0}^{\frac{\pi}{4}} \sec^3(\theta)d\theta = \frac{1}{2}\sec(\theta)\tan(\theta) + \frac{1}{2}\ln|\sec(\theta) + \tan(\theta)|\bigg|_{0}^{\frac{\pi}{4}}$
Evaluate at $\theta = \frac{\pi}{4}, \sec(\theta) = \sqrt{2}, \tan(\theta) = 1.$
Evaluate at $\theta = 0, \sec(\theta) = 1, \tan(\theta) = 0.$
$\frac{1}{2}\sqrt{2}·1+\frac{1}{2}\ln(1+\sqrt{2}) = \frac{\sqrt{2}}{2}+\frac{1}{2}\ln(1+\sqrt{2}).$
Finally, $\text{Avg Dist} = \frac{2a}{3}(\frac{\sqrt{2}}{2}+\frac{1}{2}\ln(1+\sqrt{2})) = \frac{a}{3}(\sqrt{2}+\ln(1+\sqrt{2}))$
For the unit square $a = \frac{1}{2}$. So the final answer is: $\text{Avg Dist} = \frac{1}{6}(\sqrt{2}+\ln(1+\sqrt{2}))$.
Calculate the total mass. The total mass M of the region R is given by the double integral of the density function over the region: $\text{Mass(R)} = \iint_R \delta(x, y)dA$.
Since the region R is the unit square, where $0 \le x \le 1$, $0 \le y \le 1$, the mass can be computed as: $\int_{0}^{1} \int_{0}^{1} xydydx$.
Compute the inner integral: $\int_{0}^{1} xydy = x\frac{y^2}{2}\bigg|_{0}^{1} = \frac{x}{2}$.
Compute the outer integral: $\text{Mass(R)} = \iint_R \delta(x, y)dA = \int_{0}^{1} \int_{0}^{1} xydydx = \int_{0}^{1} \frac{x}{2}dx = \frac{x^2}{4}\bigg|_{0}^{1} = \frac{1}{4}$. Thus, the total mass M of the unit square with this density function is $\frac{1}{4}$.
Center of Mass. The center of mass of an object is the hypothetical point where its entire mass can be thought of as being concentrated for the purpose of analyzing its motion.
For a region R in the XY-plane with a density function $\delta(x, y)$, the coordinates of the center of mass are given by: $ x_{\text{centreMass}} = \frac{1}{\text{Mass(R)}} \iint_R x \delta(x, y)dA, y_{\text{centreMass}} = \frac{1}{\text{Mass(R)}} \iint_R y \delta(x, y) dA$.
Calculating the Center of Mass: $x_{\text{centreMass}} = 4\int_{0}^{1} \int_{0}^{1} x(xy)dydx, y_{\text{centreMass}} = 4 \int_{0}^{1} \int_{0}^{1} y(xy)dydx$.
$x_{\text{centreMass}} = 4\int_{0}^{1} \int_{0}^{1} x(xy)dydx = 4\int_{0}^{1} \int_{0}^{1} x^2ydydx$
Inner integral: $\int_{0}^{1} x^2ydy = x^2\frac{y^2}{2}\bigg|_{0}^{1} = \frac{x^2}{2}$
Outer integral: $x_{\text{centreMass}} = 4\int_{0}^{1} \int_{0}^{1} x(xy)dydx = 4\int_{0}^{1} \frac{x^2}{2}dx = 2\int_{0}^{1}x^2 = 2\frac{x^3}{3}\bigg|_{0}^{1} = \frac{2}{3}$.
Since this density function is symmetric in x and y, $x_{\text{centreMass}} = y_{\text{centreMass}} = \frac{2}{3}$. So, the center of mass is $(\frac{2}{3}, \frac{2}{3})$
Calculating the Moment of Inertia about the origin $I_0$.
The moment of inertia quantifies how the mass is distributed with respect to an axis, influencing the object’s rotational behavior.
For a region $R$ with density function $\delta(x, y)$, the moment of inertia around the origin is calculated by taking the double integral of the object’s mass distribution with respect to the distance from its axis of rotation $I = \iint r^2\cdot dm$=[For a solid with density $\delta$] $ \iint_{R} r^2\cdot \delta(x, y)dA = \int_{0}^{1} \int_{0}^{1} r^2xydxdy = \int_{0}^{1} \int_{0}^{1} (x^2+y^2)xydxdy = \int_{0}^{1} \int_{0}^{1} x^3y+xy^3dxdy$
Inner integral: $\int_{0}^{1} x^3y+xy^3dx = y\frac{x^4}{4}+y^3\frac{x^2}{2}\bigg|_{0}^{1} = \frac{1}{4}y+\frac{1}{2}y^3$.
Outer integral: $I = \int_{0}^{1} \int_{0}^{1} x^3y+xy^3dxdy = \int_{0}^{1} \frac{1}{4}y+\frac{1}{2}y^3dy = \frac{y^2}{8}+\frac{y^4}{8}\bigg|_{0}^{1} = \frac{1}{8}+\frac{1}{8} = \frac{1}{4}$.
So, the moment of inertia about the origin is $I_0 = \frac{1}{4}$.
Moment of Inertia about the x-axis ($I_x$):
If the rotation is about the x-axis, the moment of inertia is: $I_x = \iint_{R} y^2 \cdot \delta(x, y)dA = \int_{0}^{1} \int_{0}^{1} y^2xydydx = \int_{0}^{1} x(\int_{0}^{1} y^3dy)dx$
Inner integral: $\int_{0}^{1} y^3dy = \frac{y^4}{4}\bigg|_{0}^{1} = \frac{1}{4}$.
Calculate the outer integral: $I_x = \iint_{R} y^2 \cdot \delta(x, y)dA = \int_{0}^{1} \int_{0}^{1} y^2xydydx = \frac{1}{4}\int_{0}^{1} x dx = \frac{1}{4} \frac{x^2}{2}\bigg|_{0}^{1} = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$.
Conclusion. The moment of inertia about the x-axis is $I_x = \frac{1}{8}$.
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