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Everyone with telekinetic powers, raise my hand, Emo Philips
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:
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.
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.
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.
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: $dV = dxdydz$.
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: $dV = rdrd\theta dz$.
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 $z = \rho\cos(\phi)$.
Understanding the problem. The region of interest is bounded by
$\star$ The paraboloid $z = 9 -x^2 -y^2$ opening downwards. The vertex of the paraboloid is at the point $(0, 0, 9)$. This is because when $x = 0$ and $y = 0$, then $z = 9$; and is symmetric about the z-axis.
$\star$ The xy-plane, $z = 0$.
The density function varies with z: $f(r, \theta, z) = 12 - z$.
Determining the Limits of Integration. We will use cylindrical coordinates ($r, \theta, z$):
$\star$ Angular Coordinate, $\theta$: Sweeps a full circle in the xy-plane, ranging from $0$ to $2\pi$
$\star$ Radial Distance, $r$: Ranges from $0$ to the boundary of the surface at the xy-plane ($z=0$): $0 = 9 -x^2 -y^2 = 9 - r^2 \iff r^2 = 9 \implies r = 3$
$\star$ z (Height): Ranges from the xy-plane ($z = 0$) to the surface of the paraboloid $z = 9 - r^2$.
Setting Up the Triple Integral. The mass $M$ is given by: $M = \int_{0}^{2\pi} \int_{0}^{3} \int_{0}^{9-r^2} (12 - z) r \, dz \, dr \, d\theta$
Evaluate the integral.
Evaluate the inner integral with respect to z. $\int_{0}^{9-r^2} (12 - z) \, dz = \left[ 12z - \frac{z^2}{2} \right]_{0}^{9-r^2}$
$= 12(9-r^2) - \frac{(9-r^2)^2}{2} = 108 - 12r^2 - \frac{1}{2}(81 - 18r^2 + r^4)$
$= \frac{135}{2} - 3r^2 - \frac{1}{2}r^4$.
Evaluate the middle integral with respect to r. $\int_{0}^{3} \left( \frac{135}{2} - 3r^2 - \frac{1}{2}r^4 \right) r \, dr = \int_{0}^{3} \left( \frac{135}{2}r - 3r^3 - \frac{1}{2}r^5 \right) \, dr$
$= \left[ \frac{135}{4}r^2 - \frac{3}{4}r^4 - \frac{1}{12}r^6 \right]_{0}^{3} = \frac{135}{4}(9) - \frac{3}{4}(81) - \frac{1}{12}(729)$.
$= \frac{1215}{4} - \frac{243}{4} - \frac{243}{4} = \frac{729}{4}$.
Evaluate the outer integral with respect to $\theta$. $M = \int_{0}^{2\pi} \frac{729}{4} \, d\theta = \frac{729}{4}(2\pi) = \frac{729\pi}{2}$
Understanding the problem. We need to understand the boundaries of our solid in three-dimensional space:
$\star$ $4x^2 + z^2 = 4 \iff \frac{x^2}{1} + \frac{z^2}{4} = 1$. 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-axes of the elliptical cross-sections. In our case, the semi-major axis 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.
$\star$ $y = 0$ is the xz-plane, which serves as the lower boundary for $y$.
$\star$ $y = z + 2$. 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 \
$\star$ The elliptic cylinder is centered at $x = 0$ and has a semi-minor axis length of 1. Therefore, x ranges from $-1$ to $1$.
$\star$ For each x in the interval $[-1, 1]$, z varies from the bottom to the top of the elliptical cross-section. Solving $4x^2 + z^2 = 4$ for z, we find z ranges from $-\sqrt{4-4x^2}$ to $\sqrt{4-4x^2}$.
$\star$ 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: $V = \int_{-1}^{1} \int_{-\sqrt{4-4x^2}}^{\sqrt{4-4x^2}} \int_{0}^{z+2} dy \, dz \, dx$
Evaluate the integral.
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 $\frac{\pi r^2}{2} = \frac{\pi}{2}$).
$V = 8 \left( \frac{\pi}{2} \right) = 4\pi$
Understanding the Problem. The volume we seek is bounded…
$\star$ 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.
Observe that at $z = 1$: $1 = \sqrt{x^2 + y^2} \implies x^2 + y^2 = 1$. This is a circle of radius 1 centered at the origin in the plane $z = 1$.
Similarly at $z = 2$: $2 = \sqrt{x^2 + y^2} \implies x^2 + y^2 = 4$. This is a circle of radius 2 centered at the origin in the plane z = 2.
$\star$ 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
$\star$ $z$ (height) ranges from the cone $\sqrt{x^2+y^2} = \sqrt{r^2} = r$ to the paraboloid $6-x^2-y^2 = 6 -r^2$
$\star$ $r$ (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 radial distance must be non-negative ($r \ge 0$), the intersection occurs at $r = 2$.
$\star$ $\theta$ (angular coordinate) 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: $V = \int_{0}^{2\pi} \int_{0}^{2} \int_{r}^{6-r^2} r \, dz \, dr \, d\theta$
Evaluating the Triple Integral
Evaluating the inner integral with respect to z: $\int_{r}^{6-r^2} dz = z\bigg|_{r}^{6-r^2}=6-r^2-r$
Evaluating the middle integral with respect to r: $\int_{0}^{2} (6-r^2-r)rdr = \int_{0}^{2} (6r-r^3-r^2)dr = 3r^2-\frac{r^4}{4}-\frac{r^3}{3}\bigg|_{0}^{2} = 3·4-\frac{16}{4}-\frac{8}{3} = 12 -4 -\frac{8}{3} = 8 -\frac{8}{3} = \frac{24}{3} -\frac{8}{3} = \frac{16}{3}$
Evaluating the outer integral with respect to $\theta$: $\int_{0}^{2\pi} \frac{16}{3}d\theta = \frac{16}{3}\theta \bigg|_{0}^{2\pi} = \frac{16}{3}·2\pi = \frac{32\pi}{3}$.
Conclusion. Thus, the volume of the region between the cone and the paraboloid is: V = $\frac{32\pi}{3}$.
Understanding the Problem.
The surface $z = ar$ is a cone with its vertex at the origin $(0, 0, 0)$. The value of $a$ determines the slope. Because height $z$ increases linearly with radial distance $r$, rotating this line around the z-axis sweeps out the cone shape.
The moment of inertia $I_z$ measures how mass is distributed relative to the z-axis. For a 3D solid, it is given by: $I_z = \iiint_{R} r^2 \cdot \delta \, dV$
Given $\delta = 1$ and cylindrical volume element $dV = r \, dr \, d\theta \, dz$, this simplifies to: $I_z = \iiint_{R} r^3 \, dr \, d\theta \, dz$
Determining the Limits of Integration
The region $R$ is bounded by the cone $z = ar$ and the plane $z = b$. We need to set up our limits for $r, \theta$, and $z$. Fixing $z$ at a particular value, we get a circle with radius $r = z/a$.
$\theta$ (Angular Coordinate) ranges from $0$ to $2\pi$ because we are integrating around the full circle (the cone has rotational symmetry about the z-axis).
$r$ (Radial Distance) ranges from $0$ to $z/a$. For a given height $z$, the radius $r$ ranges or goes from $0$ ($z$-axis) to where the cone intersects the plane $z = ar$ (cone boundary), that is, $\frac{z}{a}$.
$z$ (Height) ranges from $0$ (at the vertex of the cone) to $b$ (at the base of the cone) as these are the lower and upper bounds for the height of the cone.
Setting Up the Triple Integral. With these limits, the triple integral in cylindrical coordinates is: $I_z = \int_{0}^{b} \int_{0}^{2\pi} \int_{0}^{\frac{z}{a}} r^3drd\theta dz$
Evaluating the Integral
Evaluating the Inner Integral. First, we integrate with respect to r: $\int_{0}^{\frac{z}{a}} r^3dr = \frac{r^4}{4}\bigg|_{0}^{\frac{z}{a}} = \frac{(\frac{z}{a})^4}{4} = \frac{z^4}{4a^4}$
Evaluating the middle integral. Next, we integrate with respect to \theta: $\int_{0}^{2\pi} \frac{z^4}{4a^4} d\theta = \frac{z^4}{4a^4}\int_{0}^{2\pi} d\theta = \frac{z^4}{4a^4}(\theta\bigg|_{0}^{2\pi}) = \frac{z^4\pi}{2a^4}$
Evaluating the Outer Integral. Finally, we integrate with respect to z: $I_z = \int_{0}^{b} \frac{z^4\pi}{2a^4}dz = \frac{z^5\pi}{5·2·a^4}\bigg|_{0}^{b} = \frac{\pi b^5}{10a^4}$
Conclusion. Therefore, the moment of inertia $I_z$ of the solid cone about the z-axis is: $I_z = \frac{πb^5}{10a^4}$∎
Understanding the problem. The solid is above the xy-plane ($z = 0$), under the cone $z = \sqrt{x^2+y^2} = r$ which opens upwards from the origin. This cone has 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 $z = 2$.
It is also inside the sphere centered at the origin with a radius $r = 2, x^2 + y^2 + z^2 = 4$.
Intersection: The cone and the sphere intersect where their boundaries meet: $z^2 = r^2 \implies (x^2 + y^2) + z^2 = r^2 + r^2 = 4 \implies 2r^2 = 4 \implies r^2 = 2 \implies r = \sqrt{2}$, this is a circle of radius $\sqrt{2}$ (Refer to Figure B for a visual representation and aid in understanding it). In other words, the transition between the two upper boundaries happens at a radius of $\sqrt{2}$ in the xy-plane.
Setting Up the Triple Integral. In cylindrical coordinates, where $x = r\cos(\theta), y = r\sin(\theta)$, and $z = z$, the volume element $dV$ is expressed as $dV = rdzdr\theta$. The function to be integrated $f(x, y, z)$ in cylindrical coordinates becomes $f(r, \theta, z)$. The region E can be split into two parts:
Region below the cone and inside the sphere (green area, the inner core):
$\star$ $r$-range: From $0$ to $\sqrt{2}$ (where the cone and sphere intersect).
$\star$ $\theta$-range: From $0$ to $2\pi$ (full rotation around the z-axis).
$\star$ $z$-range: From the xy-plane ($z = 0$) up to the cone ($z = r$).
The integral for this part is: $\int_{0}^{2\pi}\int_{0}^{\sqrt{2}}\int_{0}^{r} f(r, \theta, z) r dz dr d\theta$
Region inside and below the sphere (blue area, the outer ring):
$\star$ $r$-range: From $\sqrt{2}$ (intersection of the cone and sphere) to 2 (the radius of the sphere).
$\star$ $\theta$-range: From $0$ to $2\pi$.
$\star$ $z$-range: From $0$ (the xy-plane) up to the sphere boundary ($z^2=4-x^2-y^2 \implies z = \sqrt{4-r^2}$).
The integral for this part is: $\int_{0}^{2\pi}\int_{\sqrt{2}}^{2}\int_{0}^{\sqrt{4-r^2}} f(r, \theta, z) rdzdrd\theta$
Combining both parts, the total volume integral over the region E is: $\int\int\int_{E} f(x, y, z)dV = \int_{0}^{2\pi}\int_{0}^{\sqrt{2}}\int_{0}^{r} f(r, \theta, z) rdzdrd\theta + \int_{0}^{2\pi}\int_{\sqrt{2}}^{2}\int_{0}^{\sqrt{4-r^2}} f(r, \theta, z) rdzdrd\theta$
Understanding the problem. The unit sphere $x^2+ y^2+z^2 = 1$ is a sphere centered at the origin with a radius of $1$. The region of interest lies above the plane $z = 1 −y$ and inside the sphere.
Plane $z = 1 -y$. $z$-intercept: When x = 0 and y = 0, z = 1. This gives us the point $(0, 0, 1)$. \
$y$-intercept: When x = 0 and z = 0, we have 0 = 1 - y, so y = 1. This gives us the point $(0, 1, 0)$.
The equation shows that z decreases by 1 unit for every 1 unit increase in y. This means the plane has a slope of -1 in the y-direction.
Because the sphere is centered at the origin with radius 1, this plane essentially slices off a cap from the top-front of the sphere.
Determining the Limits of Integration:
$\star$ For a fixed $y$, z ranges from the plane $z = 1 - y$ up to the surface of the sphere $z=\sqrt{1-x^2-y^2}.$
Intersection Boundary (xy-plane projection): Set the sphere and plane equal to find the footprint of the slice: $x^2 + y^2 + (1 - y)^2 = 1 \implies x^2 + y^2 + 1 - 2y + y^2 = 1 \implies x^2 + 2y^2 - 2y = 0$
Solving for x, the boundary in the xy-plane is $x^2 = 2y - 2y^2$.
$\star$ For each y, x ranges inside $1 - y \lt \sqrt{1-x^2-y^2} \implies (1-y)^2 \lt 1-x^2-y^2 \implies 1 +y^2 -2y \lt 1-x^2-y^2 \implies x^2 \lt 2y -2y^2$, so x ranges between $-\sqrt{2y -2y^2}$ and $\sqrt{2y -2y^2}$
$\star$ For $x$ to remain a real number, the term under the square root must be non-negative: $2y - 2y^2 \ge 0$, which restricts $y$ to $[0, 1]$.
The integral in Cartesian coordinates becomes: $V = \int_{0}^{1} \int_{-\sqrt{2y -2y^2}}^{\sqrt{2y -2y^2}} \int_{1-y}^{\sqrt{1-x^2-y^2}} dzdxdy$, this is not an easy integral, so it will be wise to use symmetry and it is left to the reader to do so.
Given the complexity of evaluating these integrals directly, using symmetry and potentially switching to cylindrical coordinates ($x = r\cos(\theta), y = r\sin(\theta), z = z$) simplifies the problem.
Applying the standard cylindrical conversion $x = r\cos(\theta)$ and $y = r\sin(\theta)$ to the footprint $x^2 + 2y^2 - 2y \le 0$ ( the region in the $xy$-plane over which the vertical slice between plane and sphere exists) yields: $r^2\cos^2(\theta) + 2r^2\sin^2(\theta) - 2r\sin(\theta) \le 0$
$r^2(\cos^2(\theta) + \sin^2(\theta) + \sin^2(\theta)) \le 2r\sin(\theta)$
$r^2(1 + \sin^2(\theta)) \le 2r\sin(\theta)$
Dividing by $r$ establishes the radial boundary: $r \le \frac{2\sin(\theta)}{1 + \sin^2(\theta)}$
Because $r$ represents a non-negative distance, $\sin(\theta)$ must also be non-negative. This restricts the angular sweep strictly to the upper half-plane, meaning $\theta$ ranges from $0$ to $\pi$.
Set Up the New Integral: $V = \int_{0}^{\pi} \int_{0}^{\frac{2\sin(\theta)}{1+\sin^2(\theta)}} \int_{1-r\sin(\theta)}^{\sqrt{1-r^2}} r \, dz \, dr \, d\theta$
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