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# Triple Integral – Definition, General Forms, and Examples

Through**triple integrals**, we can now extend our understanding of iterated integrals to expressions and functions with three variables. We can use triple integrals to calculate the volume and mass of different objects â€“ which are important components we need in physics, structural engineering, and other fields. This is why understanding triple integrals are important if we want to work with setups where there are now three variables components.

**$\boldsymbol{f(x, y, z)}$.**

*Triple integrals represent the iterated of functions with three variables,***At this point, it is important that you are familiar with how we set up and evaluate iterated and double integrals. Open the embedded links in a separate tab or window in case you need an immediate refresher. Through this article, weâ€™ll show you how we can visualize and understand what triple integrals represent. Weâ€™ll also provide a thorough explanation of how to evaluate different triple integrals. By the end of the discussion, you should also learn how to solve problems that involve triple integrals. Are you ready? Letâ€™s begin by visualizing triple integrals in three-dimensional coordinate systems then use these representations to define them.**

*Instead of integrating our function over a given area, weâ€™re now integrating it over a three-dimensional figure that has volume.*## What Is a Triple Integral?

A triple integral is an iterated integral with three variables and over a three-dimensional region. We can treat triple integralsâ€™ definition as an extension of that of the double integrals, but this time, weâ€™re integrating over a volume instead of an area.Order of Integration |
Triple Integral |
Order of Integration |
Triple Integral |

$x \rightarrow y \rightarrow z$ | $\int_{z_1}^{z_2} \int_{y_1(z)}^{y_2(z)}\int_{x_1(y, z)}^{x_2(y, z)} f(x, y, z)\phantom{x}dx dy dz$ | $y \rightarrow z \rightarrow x$ | $\int_{x_1}^{x_2} \int_{z_1(x)}^{z_2(x)}\int_{y_1(x, y)}^{y_2(x, y)} f(x, y, z)\phantom{x}dx dy dz$ |

$x \rightarrow z \rightarrow y$ | $\int_{y_1}^{y_2} \int_{z_1(y)}^{z_2(y)}\int_{x_1(y, z)}^{x_2(y, z)} f(x, y, z)\phantom{x}dx dz dy$ | $z \rightarrow x \rightarrow y$ | $\int_{z_1}^{z_2} \int_{x_1(z)}^{x_2(z)}\int_{y_1(x, z)}^{y_2(x, z)} f(x, y, z)\phantom{x}dx dz dy$ |

$y \rightarrow x \rightarrow z$ | $\int_{z_1}^{z_2} \int_{x_1(z)}^{x_2(z)}\int_{y_1(x, z)}^{y_2(x, z)} f(x, y, z)\phantom{x}dy dx dz$ | $z \rightarrow y \rightarrow x$ | $\int_{x_1}^{x_2} \int_{y_1(x)}^{y_2(x)}\int_{z_1(x, z)}^{z_2(x, z)} f(x, y, z)\phantom{x}dy dx dz$ |

## How To Do Triple Integrals?

Use the iterated form of the triple integral to evaluate the function with respect to $x$, $y$, and $z$. Here are some general pointers to remember when working with triple integrals:- Set up the three pairs of limits of integration correctly.
- Integrate the function with respect to the innermost variable â€“ innermost to the outermost.
- When integrating the functions, only work with one variable while treating the remaining variables as constants.
- Evaluate the resulting expression at the given bounds then repeat the process until we have integrated it three times in a row.

### Finding the Triple Integral Over a Rectangular Box

When working with a triple integral defined by a rectangular box, we simply use the limits based on the boxâ€™s bounds with respect to $x$, $y$, and $z$, respectively. Since weâ€™re working with a rectangular box, $S$, we can define its limits as $[a, b] \times [c, d] \times [m, n]$. \begin{aligned}\int \int \int_{S} f(x, y, z) \phantom{x} dV &= \int_{m}^{n} \int_{c}^{d} \int_{a}^{b} f(x, y, z) \phantom{x} dxdydz \end{aligned} Letâ€™s say we want to evaluate $\int \int \int_{S} 2xyz \phantom{x} dV$ over the region, $S = [1, 2] \times [0, 4] \times [1, 3]$. Set up the triple integral by using $[1, 2]$, $[0,4]$, and $[1, 3]$ as the limits of integrations for $x$, $y$, and $z$, respectively. \begin{aligned}\int \int \int_{S} 2xyz \phantom{x} dV &= \int_{1}^{3} \int_{0}^{4} \int_{1}^{2} 2xyz \phantom{x} dxdydz \end{aligned} In fact, we can switch the orders for $dx$, $dy$, $dz$, and the value of the triple integral should remain the same. Just make sure that we use the correct limits of integration for each variable! Now, letâ€™s evaluate our triple integral, $S = \int_{1}^{3} \int_{0}^{4} \int_{1}^{2} 2xyz \phantom{x} dxdydz $, by integrating the function with respect to $x$ first since itâ€™s the innermost function. Keep in mind that weâ€™ll be treating $y$ and $z$ as constants. \begin{aligned}\int_{1}^{3} \int_{0}^{4} \int_{1}^{2} 2xyz \phantom{x} dxdydz &= \int_{1}^{3} \int_{0}^{4} \left[\int_{1}^{2} 2xyz \phantom{x} dx \right ]dydz\\&= \int_{1}^{3} \int_{0}^{4} 2yz\left[\int_{1}^{2} x \phantom{x} dx \right ]dydz\\&= \int_{1}^{3} \int_{0}^{4} 2yz\left[\dfrac{x^2}{2}\right]_{1}^{2}dydzÂ \\&= \int_{1}^{3} \int_{0}^{4} 2yz\left(\dfrac{2^2}{2} – \dfrac{1^2}{2}\right)dydz\\&= \int_{1}^{3} \int_{0}^{4} 3yz \phantom{x}dydzÂ \end{aligned} As we have expected, the resulting function is defined by $y$ and $z$ only. We continue integrating our double integral, now with respect to $y$ this time. \begin{aligned}\int_{1}^{3} \int_{0}^{4} 3yz \phantom{x}dydz &= \int_{1}^{3} \left[\int_{0}^{4} 3yz \phantom{x}dy \right ]dz\\&= \int_{1}^{3} 3z\left[\int_{0}^{4} y \phantom{x}dy \right ]dz\\&= \int_{1}^{3} 3z\left[\dfrac{y^2}{2}\right ]_{0}^{4}dz\\&= \int_{1}^{3} 3z\left[\dfrac{4^2}{2} – \dfrac{0^2}{2} \right ] dz \\&= \int_{1}^{3} 24z \phantom{x}dz\end{aligned} Weâ€™re now left with a single integral, so letâ€™s continue to evaluate and simplify our expression. \begin{aligned}\int_{1}^{3} 24z \phantom{x}dz &= 24\left[\dfrac{z^2}{2} \right ]_{1}^{3}\\&= 24 \left[\dfrac{3^2}{2} – \dfrac{1^2}{2}\right ]\\&= 96 \end{aligned} Hence, weâ€™ve shown you how to evaluate $\int_{1}^{3} \int_{0}^{4} \int_{1}^{2} 2xyz \phantom{x} dxdydz$ and in fact, it is equal to $96$. Letâ€™s now explore a more complex example â€“ evaluating triple integrals over a more complex region.### Finding the Triple Integral Over a General Region

When working with triple integrals that involve more complex regions, it is important that we account for the projection the region forms with respect to the $xy$, $xz$, or $yz$ planes- depending on what works for our problem. Let us work on $\int \int \int_{S} z \phantom{x}dV$, where $S$ is a solid figure bounded by the following planes: $xÂ = 0$, $y = 0$, $z = 0$, and $x + y + z = 4$. It is most helpful to sketch two graphs: 1) the actual solid region of $S$ and 2) the projection of $S$ on the plane (for our case, weâ€™ll use the $xy$-plane).**Calculate the triple integral, $\int_{0}^{1} \int_{0}^{2} \int_{0}^{4} (xz + y) \phantom{x}dxdydz$.**

*Example 1*__Solution__The triple integral has constants as limits of integrations, so we know that weâ€™re integrating it over a rectangular box. There is no need for us to sketch the region, so weâ€™ll go ahead and integrate the expression three times in a row with respect to $x \rightarrow y \rightarrow z$- in that particular direction. \begin{aligned}\int_{0}^{1} \int_{0}^{2} \int_{0}^{4} (xz + y) \phantom{x}dxdydz &= \int_{0}^{1} \int_{0}^{2} \left[\int_{0}^{4} (xz + y) \phantom{x}dx \right ]dydz\\&=\int_{0}^{1} \int_{0}^{2} \left[\dfrac{zx^2}{2} + yx\right ]_{0}^{4}dydz\\&=\int_{0}^{1} \int_{0}^{2} \left[\dfrac{z(4^2)}{2} + y(4)\right ] dydz \\&= \int_{0}^{1} \int_{0}^{2} (8z + 4y) \phantom{x}dydz\\&= \int_{0}^{1} \left[8zy + \dfrac{4y^2}{2}\right]_{0}^{2} dz\\&=\int_{0}^{1} (16z + 8) \phantom{x}dz\\&= \left[\dfrac{16z^2}{2} + 8z \right ]_{0}^{1}\\&= 8 + 8 – 0 \\&= 16 \end{aligned} From this, we can see that $\int_{0}^{1} \int_{0}^{2} \int_{0}^{4} (xz + y) \phantom{x}dxdydz$ is equal to $16$. Apply a similar process when evaluating triple integrals over a rectangular box of any dimensions.

**Calculate the triple integral, $\int_{0}^{1} \int_{0}^{z^2} \int_{0}^{2} y\sin(z^5) \phantom{x}dxdydz$.**

*Example 2*__Solution__Weâ€™ll apply a similar process when integrating the expression, $\int_{0}^{1} \int_{0}^{z^2} \int_{0}^{2} y\sin(z^4) \phantom{x}dxdydz$. First, letâ€™s integrate the expression with respect to $x$ first and treating the rest of the variables constant. \begin{aligned}\int_{0}^{1} \int_{0}^{z^2} \int_{0}^{2} y\sin(z^5) \phantom{x}dxdydz &= \int_{0}^{1} \int_{0}^{z^2} \left[\int_{0}^{2} y\sin(z^5) \phantom{x}dx \right ]dydz\\&= \int_{0}^{1} \int_{0}^{z^2} \left[ y\sin(z^5)\int_{0}^{2} \phantom{x}dx \right ]dydz\\&= \int_{0}^{1} \int_{0}^{z^2} \left[ y\sin(z^5)(2 – 0) \right ]dydz\\&= \int_{0}^{1} \int_{0}^{z^2} \left[ 2y\sin(z^5)\right ]dydz\end{aligned} Now that we have a double integral, we apply a similar process but this time, we integrate with respect to $y$ then with respect to $z$. Apply the $u$-substitution method to integrate the single integral expression. When you let $u = z^5$, we have $du = 5z^4 \phantom{x}dz$. \begin{aligned}u &= z^5 \\ du &= 5z^4 \phantom{x}dz\\\\ \int z^4 \sin(z^5) \phantom{x}dz &= \dfrac{1}{5}\int \sin u \phantom{x}du\\&= -\dfrac{1}{5} \cos u\\&= -\dfrac{1}{5}\cos z^5\end{aligned} Use this expression to simplify our triple integral further and find its actual value. \begin{aligned}\int_{0}^{1} \int_{0}^{z^2} \int_{0}^{2} y\sin(z^5) \phantom{x}dxdydz &= \int_{0}^{1}z^4 \sin(z^5) \phantom{x}dz\\ &= – \dfrac{1}{5} \left[\cos z^5Â \right ]_{0}^{1}\\&= -\dfrac{1}{5}[\cos 1 – 1]\\&\approx 0.0919\end{aligned} This means that the triple integralâ€™s value is approximately equal to $0.0919$.

**Evaluate the triple integral, $\int \int \int_{S} \sqrt{x^2 + z^2} \phantom{x}dV$, where $S$ is a solid figure bounded by the following: \begin{aligned}\textbf{paraboloid}&: y = x^2 + z^2\\\textbf{plane}&: y = 9\end{aligned}**

*Example 3*__Solution__Weâ€™ve discussed projecting solids on the $xy$-plane and if we apply this technique to this problem, weâ€™ll end up with a complex triple integral shown below. \begin{aligned}\int \int \int_{S} \sqrt{x^2 + z^2} \phantom{x}dV &= \int_{-3}^{3} \int_{9}^{x^2} \int_{\sqrt{y â€“ x^2}}^{-\sqrt{y â€“ x^2}} sqrt{x^2 + z^2} \phantom{x}dzdydx \end{aligned} This example will highlight why itâ€™s helpful to also consider the other projections. Since weâ€™re given a paraboloid, $y = x^2 + z^2$, itâ€™s much easier to project on the $xz$-plane.

**One of the important applications of triple integrals is finding a solid objectâ€™s mass, moments, and center of mass. We can represent these components using the triple integrals shown below.**

*Example 4*Mass |
\begin{aligned}m &= \int \int \int_{S} \rho(x, y, z) \phantom{x}dV\end{aligned} |

Moments |
\begin{aligned}M_{yz} &= \int \int \int_{S} x\rho(x, y, z) \phantom{x}dV\\M_{xz} &= \int \int \int_{S} y\rho(x, y, z) \phantom{x}dV\\M_{xy} &= \int \int \int_{S} z\rho(x, y, z) \phantom{x}dV\end{aligned} |

Center of Mass |
\begin{aligned}\overline{x} &= \dfrac{M_{yz}}{m} \\\overline{y} &= \dfrac{M_{xz}}{m}\\ \overline{z} &= \dfrac{M_{xy}}{m}Â \end{aligned} |

__Solution__First, letâ€™s visualize how the solid (itâ€™s in fact a parabolic cylinder) looks like and its project on the $xy$-plane.

\begin{aligned}\boldsymbol{M_{yz} = \int \int \int_{S} x\rho(x, y, z)dV}\end{aligned} | \begin{aligned}\int_{-1}^{1}\int_{4y^2}^{4}\int_{0}^{x/2} x\rho \phantom{x}dzdxdy &= \int_{-1}^{1}\int_{4y^2}^{4}\left[\int_{0}^{x/2} x\rho \phantom{x}dz \right ]dxdyÂ \\&= \rho \int_{-1}^{1}\int_{4y^2}^{4} \dfrac{x^2}{2} \phantom{x}dxdy \\&= \dfrac{\rho}{2} \int_{-1}^{1}\left[\dfrac{x^3}{3}\right]_{4y^2}^{4}Â \phantom{x}dy \\&=\dfrac{\rho}{6} \int_{-1}^{1}\left[4^3 – (4y^2)^3\right] \phantom{x}dy \\&= 64\cdot\dfrac{\rho}{6} \int_{-1}^{1} (1 – y^6) \phantom{x}dy\\&= \dfrac{32\rho}{3}\left[y – \dfrac{y^7}{7} \right ]_{-1}^{1}\\&= \dfrac{128\rho}{7}Â Â \end{aligned} |

\begin{aligned}\boldsymbol{M_{xz} = \int \int \int_{S} y\rho(x, y, z)dV}\end{aligned} | \begin{aligned}\int_{-1}^{1}\int_{4y^2}^{4}\int_{0}^{x/2} y\rho \phantom{x}dzdxdy &= \int_{-1}^{1}\int_{4y^2}^{4}\left[\int_{0}^{x/2} y\rho \phantom{x}dz \right ]dxdyÂ \\&= \rho \int_{-1}^{1}\int_{4y^2}^{4} \dfrac{yx}{2} \phantom{x}dxdy \\&= \dfrac{\rho}{2} \int_{-1}^{1}\left[\dfrac{yx^2}{2}\right]_{4y^2}^{4}Â \phantom{x}dy \\&=\dfrac{\rho}{4} \int_{-1}^{1}\left[4^2y – (4y^2)^2y\right] \phantom{x}dy \\&= 4\rho\int_{-1}^{1} (y – y^3) \phantom{x}dy\\&= 4\rho\left[\dfrac{y^2}{2} – \dfrac{y^4}{4} \right ]_{-1}^{1}\\&= 0\end{aligned} |

\begin{aligned}\boldsymbol{M_{xy} = \int \int \int_{S} z\rho(x, y, z)dV}\end{aligned} | \begin{aligned}\int_{-1}^{1}\int_{4y^2}^{4}\int_{0}^{x/2} z\rho \phantom{x}dzdxdy &= \int_{-1}^{1}\int_{4y^2}^{4}\left[\int_{0}^{x/2} z\rho \phantom{x}dz \right ]dxdyÂ \\&= \rho \int_{-1}^{1}\int_{4y^2}^{4} \dfrac{x^2/4}{2} \phantom{x}dxdy \\&= \dfrac{\rho}{8} \int_{-1}^{1}\left[\dfrac{x^3}{3}\right]_{4y^2}^{4}Â \phantom{x}dy \\&=\dfrac{\rho}{24} \int_{-1}^{1}\left[4^3 – (4y^2)^3\right] \phantom{x}dy \\&= 64\cdot\dfrac{\rho}{24} \int_{-1}^{1} (1 – y^6) \phantom{x}dy\\&= \dfrac{8\rho}{3}\left[y – \dfrac{y^7}{7} \right ]_{-1}^{1}\\&= \dfrac{32\rho}{7}Â Â \end{aligned} |

\begin{aligned} \boldsymbol{\overline{x}}\end{aligned} | \begin{aligned} \boldsymbol{\overline{y}}\end{aligned} | \begin{aligned} \boldsymbol{\overline{z}}\end{aligned} |

\begin{aligned}\overline{x}&= \dfrac{M_{yz}}{m}\\&= \dfrac{\dfrac{128\rho}{7}}{\dfrac{32\rho}{5}}\\&= \dfrac{20}{7}\end{aligned} | \begin{aligned}\overline{x}&= \dfrac{M_{xz}}{m}\\&= \dfrac{0}{\dfrac{32\rho}{5}}\\&= 0\end{aligned} | \begin{aligned}\overline{z}&= \dfrac{M_{xy}}{m}\\&= \dfrac{\dfrac{32\rho}{7}}{\dfrac{32\rho}{5}}\\&= \dfrac{5}{7}\end{aligned} |

\begin{aligned}\boldsymbol{(\overline{x}, \overline{y}, \overline{z})} &= \left(\dfrac{20}{7}, 0, \dfrac{5}{7}\right) \end{aligned} |

### Practice Questions

1. Calculate the triple integral, $\int_{0}^{2} \int_{0}^{4} \int_{0}^{1} (xy -z) \phantom{x}dxdydz$. 2. Calculate the triple integral, $\int_{0}^{2} \int_{0}^{z^2} \int_{0}^{3} y\cos(z^5) \phantom{x}dxdydz$. 3. Evaluate the triple integral, $\int \int \int_{S} z \phantom{x}dV$, where $S$ is a solid figure bounded by the following planes: $x = 0$, $y = 0$, $z = 0$, and $x + y + z = 9$. 4. Evaluate the triple integral, $\int \int \int_{S} \sqrt{x^2 + z^2} \phantom{x}dV$, where $S$ is a solid figure bounded by the following: \begin{aligned}\textbf{paraboloid}&: y = x^2 + z^2\\\textbf{plane}&: y = 16\end{aligned} 5. Set up a triple integral to determine the volume of the tetrahedron, $S$, bounded by the following planes: $x = 0$, $x = 4y$, $z = 0$, and $x + 4y + z = 4$. 6. Determine the center of mass of a solid with a constant density, $\rho(x, y, z)$, bounded by the solid, $x = y^2$, and the following planes: $x = z$, $z = 0$, and $x = 1$.### Answer Key

1. $\int_{0}^{9} \int_{0}^{9 – x} \int_{0}^{9 – x – y} z \phantom{x}dzdydx = \dfrac{2187}{8}$ 2. $\int_{0}^{2} \int_{0}^{z^2} \int_{0}^{3} y\cos(z^5) \phantom{x}dxdydz = \dfrac{3\sin 32}{10} \approx 0.165$ 3. $\int_{0}^{2} \int_{0}^{4} \int_{0}^{1} (xy -z) \phantom{x}dxdydz = 0$ 4.$\int \int_{S_3} \left[\int_{x^2 + z^2}^{16} \sqrt{x^2 + z^2}\phantom{x} dy\right] dA = \dfrac{4096\pi}{15} \approx 857.86$ 5. $\int_{0}^{2} \int_{x/4}^{1- x/4} \int_{0}^{4 â€“ x – 4y} \phantom{x}dzdydx = \dfrac{4}{3}$ 6. $(\overline{x}, \overline{y}, \overline{z}) = \left(\dfrac{5}{7}, 0, \dfrac{5}{14}\right)$*Images/mathematical drawings are created with GeoGebra.*

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