Triple Integrals: Applications and Solved Exercises.

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Triple Integrals: Unlocking 3D Space in Calculus

Triple integrals are one of the most powerful tools in multivariable calculus. While single integrals calculate the area under a curve and double integrals calculate the volume under a surface, triple integrals allow us to calculate properties of a solid, three-dimensional region.

Whenever a quantity —like mass, charge, temperature, or probability— varies across a 3D space, triple integrals allow us to sum up those infinite variations into a single, concrete value.

A triple integral extends the concept of integration to three dimensions. Imagine a function f(x,y,z) that represents the temperature at any given point inside a room. If you want to find the average temperature of the entire room, you must sum up the temperature at every microscopic coordinate in that 3D region.

The triple integral of a function $f(x, y, z)$ over a region $R$ in three-dimensional space is expressed as $\int \int \int_{R} f(x, y, z)dV$ where:

1. $f(x, y, z)$ is a function of three variables over a three-dimensional region, representing the density, temperature, or value of some quantity at each point $(x, y, z)$ within the region $R$. For example, if $f(x, y, z)$ represents density, the integral will give you the total mass.
2. $dV$ represents an infinitesimal “small block” or volume element within the region $R$. In Cartesian coordinates, this block is a microscopic box with volume $dV = dxdydz$, but this element can look different in other coordinate systems.
3. The integral is calculated over the region $R$, which can be described by specific bounds on x, y, and z. (Figure 1).

How to Set Up a Triple Integral: The “Shadow” Method

The most challenging part of solving a triple integral is not the integration itself, but finding the bounds of $R$. The most reliable way to do this is the Shadow or Projection Method.

Step 1: Find the Inner Bounds (The “Arrow”). Shoot an imaginary arrow through the region $R$ parallel to one of the axes, usually the z-axis.

• The surface where the arrow enters the solid is your lower bound, e.g., $z = g_1(x,y)$.
• The surface where the arrow exits the solid is your upper bound, e.g., $z = g_2(x,y)$.

Step 2: Find the Outer Bounds (The “Shadow”)

Imagine shining a light down on the solid so it casts a 2D shadow on the coordinate plane below it (the $xy$-plane). This shadow is a 2D region, $D$. You now could set up a standard double integral for $D$ to find the bounds for $x$ and $y$.

If we integrate with respect to $z$ first, the general structure looks like this: $\boxed{\iiint_{R} f(x, y, z) \, dV = \iint_{D} \left( \int_{g_1(x,y)}^{g_2(x,y)} f(x, y, z) \, dz \right) dA}$

There are six possible orders of integration in Cartesian coordinates ($dz \, dy \, dx$, $dz \, dx \, dy$, $dx \, dy \, dz$, etc.). Depending on the shape of the solid, one order is usually vastly easier to compute than the others.

The Power of Coordinate Systems

Symmetry is your best friend in multivariable calculus. Computing the volume of a sphere using $dx \, dy \, dz$ results in a nightmare of square roots. By changing our coordinate system to match the geometry of the region, we simplify the limits of integration.

However, when we warp our grid lines from boxes into cylinders or spheres, the size of our infinitesimal volume element $dV$ changes. We must multiply by a scaling factor known as the Jacobian Determinant.

1. Cartesian Coordinates $(x, y, z)$. Cartesian coordinates are used for regions that align well with the coordinate planes, such as flat planes, rectangular prisms, or boxes. The coordinates $(x, y, z)$ correspond directly to distances along the $x, y$, and $z$ axes, respectively. The differential volume element is: $dV = dxdydz$.

2. Cylindrical Coordinates $(r, \theta, z)$. Cylindrical coordinates are useful for regions with cylindrical symmetry, such as cylinders, cones, and circular regions. In this system, the coordinates are:
   $r$: The radial distance from the $z$-axis in the $xy$-plane, $r = \sqrt{x^2+y^2}$.
   $\theta$: The angle in the xy-plane, measured counterclockwise, from the positive x-axis.
   $z$: The height above the $xy$-plane, the same as in Cartesian coordinates.
   Conversion to Cartesian Coordinates: $z = z, x = r\cos(\theta), y = r\sin(\theta)$, and the differential volume element is: $dV = rdrd\theta dz$.

3. Spherical Coordinates $(\rho, \theta, \phi)$. They are ideal for regions with spherical symmetry such as spheres, hemispheres, or cones. In this system, the coordinates or parameters are:
   $\rho$: The radial distance from the origin to a given point, it ranges from 0 to infinity.
   $\phi$: The polar angle, measured from the positive $z$-axis to the radial line $r$ (the line connecting the origin to the point $P$). It describes how far the point is from the z-axis. It values increases from $0$ (positive $z$-axis), $\frac{\pi}{2}$ (xy-plane), and $\pi$ (negative z-axis), that is, $0 \le \phi \le \pi$.
   $\theta$: The azimuthal angle, measured from the positive x-axis to the projection of the radial line on the xy-plane. It ranges from $0$ to $2\pi$, representing a full rotation around the z-axis. It describes the rotation around the z-axis.

The differential volume element $dV = \rho^2 \sin(\phi) d\rho d\theta d\phi$. Conversion to Cartesian Coordinates: $x = r\cos(\theta) = \rho\sin(\phi)\cos(\theta), y = r\sin(\theta) = \rho\sin(\phi)\sin(\theta)$, and $z = \rho\cos(\phi)$.

Applications of Triple Integrals in Multivariable Calculus

Triple integrals are a powerful tool in multivariable calculus with a wide range of applications in physics, engineering, and other fields. These integrals allow us to calculate properties of three-dimensional objects such as mass, moments of inertia, electric charge, and volumes.

Mass of a Three-Dimensional Object

The mass of a three-dimensional object can be calculated using triple integrals when the object’s density is known. Density, denoted as $\delta$, is a function that describes how mass is distributed within the object. The relationship between mass $m$, volume $V$, and density $\delta$ is given by: $\delta = \frac{m}{V} \implies dm = \delta dV.$

To calculate the mass of a three-dimensional solid object occupying a region R in space, we calculate its total mass by summing up the mass of infinite microscopic volumes and use the triple integral $Mass = \int \int \int_{R} \delta(x, y, z) dV$ where $\delta$ is the density of the object and $dV$ represents an infinitesimal volume.

The integral sums the contributions of mass from each infinitesimal volume element over the entire region $R$.

If the density $\delta$ is constant and equal to 1, the mass is simply the volume of the region, $V = \int \int \int_{R} dV$.

Center of Mass

The center of mass of a three-dimensional body is the point where the entire mass of the body can be considered to be concentrated. It is the average position or balancing point of the object, weighted by its density.

🧭 Core idea

1. The center of mass is the mass‑weighted average position of all the matter in an object. For discrete particles: $\overline{x}=\frac{\sum m_i x_i}{\sum m_i},\quad \overline{y}=\frac{\sum m_i y_i}{\sum m_i},\quad \overline{z}=\frac{\sum m_i z_i}{\sum m_i}.$
2. If the object is continuous, each tiny volume element dV acts like a “particle” with mass $dm = \delta(x,y,z)\, dV.$
3. Replace the sums with integrals: $\overline{x}=\frac{1}{M}\iiint_R x\,\delta(x,y,z)\,dV,\quad \overline{y}=\frac{1}{M}\iiint_R y\,\delta(x,y,z)\,dV,\quad \overline{z}=\frac{1}{M}\iiint_R z\,\delta(x,y,z)\,dV,$ where $ M=\iiint_R \delta(x,y,z)\,dV$ is the total mass.

The coordinates $(\overline {x}, \overline {y}, \overline {z})$ of the center of mass are calculated by the following formulas: $\overline {x} = \frac{1}{Mass(R)}\int \int \int_{R} x \cdot \delta(x, y, z)dV, \overline {y} = \frac{1}{Mass(R)}\int \int \int_{R} y\cdot \delta(x, y, z)dV, \overline {z} = \frac{1}{Mass(R)}\int \int \int_{R} z \cdot \delta(x, y, z)dV$.

These integrals calculate the average position of the mass along each coordinate axis. This allows us to describe the position of the object as a whole, such as the point where the object would balance if placed on a perfectly frictionless support, e.g., the point where you would put your finger to balance your ruler. For a body with uniform density, the center of mass coincides with its geometric center.

Average Value of a Function

The average value of a function of three variables $f(x, y, z)$ in a region $R$, say occupied by a three-dimensional solid, can be calculated using a triple integral. This average value represents the “mean” value of the function across the entire region. It is given by: $\overline {f} = \frac{1}{Vol(R)}\int \int \int_{R} f(x, y, z) dV$ where $\operatorname{Vol(R)}$ is the volume of the region R.

This is literally the continuous version of the discrete average: $\overline{f}=\frac{f_1+f_2+\cdots+f_n}{n}$ where we replace each sample $f_i$ with the value of the function at a point, each “count” 1 with a tiny volume $dv$, and the sum with an integral. If the solid has density $\delta(x,y,z)$, then each tiny volume contributes mass $dm = \delta(x,y,z)\,dV.$ So the weighted average of $f$ is: $\overline{f}=\frac{1}{M}\iiint_R f(x,y,z)\,\delta(x,y,z)\,dV$ where $M=\iiint_R \delta(x,y,z)\,dV.$

This weighted average takes into account how the density varies throughout the region and is also the continuous version of the discrete weighted average: $ \overline{f}=\frac{\sum w_i f_i}{\sum w_i}.$

Moment of Inertia

Moment of Inertia is the tendency of a body to maintain its state of rotation or rest. If the body is at rest, it tries to remain at rest and resist any force trying to bring it into motion. If the body is in rotational motion, it is the tendency which opposes the change in its rotational motion due to external forces, the body tries to continue in motion and resist any force trying to bring it to stop.

The moment of inertia of a body about a given axis is a measure of a body’s resistance to change in its rotational motion about that axis. It depends on both the mass distribution of the object and the distance of each mass element from the axis of rotation. The point-mass formula for the moment of inertia is $I = mr^2$ where $m$ is the mass of the object and $r$ is the distance from the axis of rotation. The general formula for the moment of inertia I about an axis is: I = $\int \int \int_{R} (\text{distance to axis})^2 \cdot \delta \cdot dV$.

Higher moment of inertia implies that it is harder to change the body’s rotational motion about the axis. In other words, more torque is required to change the rotational speed of the object.

The moment of inertia of a body about the z-axis is as follows $I_z = \int \int \int_{R} r^2 \delta dV$ =[Cartesian coordinates] $\int \int \int_{R} (x^2+y^2) \delta dV$. By symmetry, the moment of inertia about the $x$ and $y$ axes are $I_x = \iiint_{R} (y^2 + z^2) \delta \, dV$ and $I_y = \iiint_{R} (x^2 + z^2) \delta \, dV$.

These integrals calculate the distribution of mass relative to each axis, providing a comprehensive understanding of the body’s resistance to rotational changes.

Probability Identity

If $f(x,y,z)$ is a joint probability density function (PDF) for three continuous random variables, then: $f(x,y,z) \ge 0$ and $\iiint_{\mathbb{R}^3} f(x,y,z) \,dV = 1$

The probability that the random vector $(X, Y, Z)$ lies in a region R is: $P\big((X,Y,Z)\in R\big) = \iiint_R f(x,y,z) dV.$

This is the probability‑theory analogue of: $mass = \int \operatorname{density} dV, \operatorname{volume} = \int 1 dV$ where the “density” is probability density.

The expected value of a random variable X with density f is: $E[X] = \iiint_{\mathbb{R}^3} x \, f(x,y,z) \, dV$

The expected value (also called expectation, mean, or first moment) is a generalization of the weighted average. Provided that it is finite, the expected value can be interpreted as the long-run average of results from independent repetitions of the same random experiment. For a random variable that takes finitely many values, the expected value is the weighted average of those values, with their probabilities as weights. More generally, it is the Lebesgue integral (the integral of a non-negative function, i.e., the area between the graph of the function and the $x$-axis) of the random variable with respect to the underlying probability distribution.

The triple integral is a universal averaging machine.

• In physics, it averages mass distribution.
• In calculus, it averages function values.
• In probability, it averages likelihood.

Solved examples (step by step)

• Evaluate $\iiint_{E} z \, dV$, where $E$ is bounded by the cylinder $y^2 + z^2 = 9$ and the planes $x = 0$, $y = 3x$, and $z = 0$ in the first octant Refer to Figure f for a visual representation and aid in understanding it.

Understanding the problem. The goal is to set up the integral for z over the region E using Cartesian coordinates. We need to understand the boundaries of E in three-dimensional space:

1. The cylinder $y^2 + z^2 = 9$. This equation represents a circle of radius 3 in the yz-plane (where x = 0). The cross-section of the cylinder in the yz-plane is a circle with radius 3. The cylinder extends infinitely along the x-axis.
2. The planes $x = 0$ (yz-plane), $y = 3x$ (a plane that passes through the origin, $slope = \tan(\theta)=3 \implies \theta \approx 71.57^\circ$ making a 71-degree angle with the x-axis), and z = 0 (xy-plane).
3. The first octant is the region where x ≥ 0, y ≥ 0, and z ≥ 0.

Limits of Integration

1. In the xy-plane, where $z = 0$, the plane $y = 3x$ intersects the cylinder at $y^2 = 9$ (since z = 0). Solving for y, we get: $y = 3 \implies x = \frac{1}{3}y = \frac{3}{3} = 1$. Therefore, x ranges from 0 to 1.
2. For each x in the interval [0, 1], y varies from the plane y = 3x to the edge of the cylinder, which is y = 3. Hence, y ranges from 3x to 3.
3. For a fixed y, z varies from the xy-plane z = 0 to the top of the cylinder $z= \sqrt{9-y^2}$ which comes from rearranging $y^2 + z^2 = 9$ to solve for z.

Set up the limit (Cartesian Coordinates). Given the ranges for x, y, and z, the triple integral in Cartesian coordinates is: $\int\int\int_{E} zdV = \int_{0}^{1} \int_{3x}^{3} \int_{0}^{\sqrt{9-y^2}} zdzdydx$

Evaluate from the inside out.

Inner integral (with respect to $z$): $\int_{0}^{\sqrt{9-y^2}} z \, dz = \left[ \frac{1}{2}z^2 \right]_{0}^{\sqrt{9-y^2}} = \frac{1}{2}(9 - y^2)$

Middle integral (with respect to $y$): $\int_{3x}^{3} \frac{1}{2}(9 - y^2) \, dy = \frac{1}{2} \left[ 9y - \frac{y^3}{3} \right]_{3x}^{3}$
Evaluate at upper bound $3$: $9(3) - \frac{27}{3} = 27 - 9 = 18$.
Evaluate at lower bound $3x$: $9(3x) - \frac{(3x)^3}{3} = 27x - 9x^3$.
Subtracting gives: $\frac{1}{2} (18 - 27x + 9x^3)$
Outer integral (with respect to $x$): $\int_{0}^{1} \frac{1}{2} (18 - 27x + 9x^3) \, dx = \frac{1}{2} \left[ 18x - \frac{27}{2}x^2 + \frac{9}{4}x^4 \right]_{0}^{1}$
Evaluate at $1$: $\frac{1}{2} \left( 18 - 13.5 + 2.25 \right) = \frac{1}{2}(6.75) = 3.375 = \frac{27}{8}$

Final Answer: The integral evaluates to $27/8$.

• Use a triple integral to find the volume of the solid bounded by $4x^2 + z^2 = 4$, $y = 0$, and $y = z + 2$ Refer to Figure g for a visual representation and aid in understanding it.

Understanding the problem. We need to understand the boundaries of our solid in three-dimensional space:

1. $4x^2 + z^2 = 4$. This equation can be rewritten as $\frac{x^2}{1}+\frac{z^2}{4} = 1$. This describes an elliptic cylinder that extends along the y-axis. The general form of an elliptic cylinder opening along the y-axis is: $\frac{x^2}{a^2}+\frac{z^2}{b^2} = 1$ where $a$ and $b$ are the semi-major and semi-minor axes of the elliptical cross-sections. In our case, the semi-major axis of the ellipse is along the z-axis with a length of 2, and the semi-minor axis is along the x-axis with a length of 1.
2. y = 0 is the xz-plane, which serves as the lower boundary for y.
3. y = z + 2. This plane slopes upwards in the yz-plane. It intersects the xz-plane at z = −2 and rises with a slope of 1 as z increases.

Limits of Integration

1. The elliptic cylinder is centered at $x = 0$ and has a semi-minor axis length of 1. Therefore, $x$ ranges from $−1$ to $1$.
2. For each x in the interval [-1, 1], z varies from the bottom to the top of the elliptical cross-section of the cylinder. Solving the equation $4x^2 + z^2 = 4$ for z, z ranges from $-\sqrt{4-4x^2}$ to $\sqrt{4-4x^2}$.
3. For a given point (x, z), y ranges from the xz-plane (y = 0) to the slanted plane (y = z+2). Thus, y varies from 0 to z+2.

Setting Up the Triple Integral. The volume V of the solid can be found using the following triple integral in Cartesian coordinates:

V = $\int_{-1}^{1} \int_{-\sqrt{4-4x^2}}^{\sqrt{4-4x^2}} \int_{0}^{z+2} dydzdx.$

Evaluate the inner integral with respect to y $\int_{0}^{z+2} dy = y \bigg\vert{}_{0}^{z+2} = z + 2$

Evaluate the middle integral with respect to z $\int_{-\sqrt{4-4x^2}}^{\sqrt{4-4x^2}} (z + 2) \, dz = \left( \frac{z^2}{2} + 2z \right) \Bigg\vert{}_{-\sqrt{4-4x^2}}^{\sqrt{4-4x^2}}$

Because $\frac{z^2}{2}$ is an even function, its evaluation at symmetric limits cancels out to $0$. The $2z$ term gives: $2\sqrt{4-4x^2} - \left( -2\sqrt{4-4x^2} \right) = 4\sqrt{4-4x^2} = 8\sqrt{1-x^2}$

Evaluate the outer integral with respect to x $V = \int_{-1}^{1} 8\sqrt{1-x^2} \, dx$ This integral represents 8 times the area of a semicircle of radius 1 (where Area is $\frac{\pi r^2}{2} = \frac{\pi}{2}$).

$V = 8 \left( \frac{\pi}{2} \right) = 4\pi$

• Find the volume between the cone $z = \sqrt{x^2+y^2}$ and the paraboloid $z = 6-x^2-y^2$ (Figure ii)

Understanding the Problem. The volume we seek is bounded

• Below by the cone $z = \sqrt{x^2+y^2}$ (or $z = r$ in cylindrical coordinates) with its vertex at the origin and its axis along the z-axis (z = 1: $1 = \sqrt{x^2 + y^2} \implies x^2 + y^2 = 1$. This is a circle of radius 1 centered at the origin in the plane z = 1. z = 2: $2 = \sqrt{x^2 + y^2} \implies x^2 + y^2 = 4$. This is a circle of radius 2 centered at the origin in the plane z = 2.)
• And above by the paraboloid opening downwards z = $6-x^2-y^2$. The vertex of the paraboloid is at the point (0, 0, 6) (this is because when x = 0 and y = 0, z = 6) and is symmetric about the z-axis.

  Cross-Section Observations: At $z = 4 \implies 4 = 6 - x^2 - y^2 \implies x^2 + y^2 = 2$. This is a circle of radius $\sqrt{2}$. At $z = 2 \implies 2 = 6 - x^2 - y^2 \implies x^2 + y^2 = 4$. This is a circle of radius $2$.

Determining the Limits of Integration

1. z (height) ranges from the cone $\sqrt{x^2+y^2} = \sqrt{r^2} = r$ to the paraboloid $6-x^2-y^2 = 6 -r^2$
2. r (radial distance) ranges from 0 to the intersection between the cone and the paraboloid: $r = 6 -r^2 \iff r^2 +r -6 = 0 \iff (r+3)(r-2) = 0$
   Since $r$ (radial distance) must be non-negative, $r = 2$.
3. The angle $\theta$ (angular coordinates) sweeps around the entire circle, so it ranges from 0 to $2\pi$.

Setting up the triple integral. The volume V can be expressed as a triple integral in cylindrical coordinates: $V = \int_{0}^{2\pi} \int_{0}^{2} \int_{r}^{6-r^2} rdzdrd\theta$.

Evaluating the Triple Integral

1. Evaluating the inner integral with respect to z: $\int_{r}^{6-r^2} dz = z\bigg|_{r}^{6-r^2}=6-r^2-r$

2. Evaluating the middle integral with respect to r: $\int_{0}^{2} (6-r^2-r)rdr = \int_{0}^{2} (6r-r^3-r^2)dr = 3r^2-\frac{r^4}{4}-\frac{r^3}{3}\bigg|_{0}^{2} = 3·4-\frac{16}{4}-\frac{8}{3} = 12 -4 -\frac{8}{3} = 8 -\frac{8}{3} = \frac{24}{3} -\frac{8}{3} = \frac{16}{3}$

3. Evaluating the outer integral with respect to \theta: $\int_{0}^{2\pi} \frac{16}{3}d\theta = \frac{16}{3}\pi \bigg|_{0}^{2\pi} = \frac{16}{3} 2\pi = \frac{32\pi}{3}$.

Thus, the volume of the region between the cone and the paraboloid is: V = $\frac{32\pi}{3}$.

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