Lesson 23: Triple Integrals in Spherical Coordinates (16.5)

Accessible transcription generated on 3/11/2026

Original Notes

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Lesson 23: Triple Integrals in Spherical Coordinates (16.5)

Review: Cylindrical Coordinates

The page contains two diagrams illustrating coordinate systems:

  • First diagram: A 3D Cartesian coordinate system with \(x\), \(y\), and \(z\) axes. A point is plotted in the first octant and labeled \((x, y, z)\). A vertical dashed line, labeled \(z\), descends from this point to its projection on the \(xy\)-plane at coordinate \((x, y, 0)\). In the \(xy\)-plane, a radial line of length \(r\) connects the origin to the point \((x, y, 0)\). The angle measured from the positive \(x\)-axis to this radial line is labeled \(\theta\). Shaded diagonal lines indicate the region in the \(xy\)-plane between the \(x\)-axis and the radial vector.
  • Second diagram: A 2D right triangle representing the geometry in the \(xy\)-plane. The horizontal base is labeled \(x\), the vertical side is labeled \(y\), and the hypotenuse is labeled \(r\). The interior angle between the base and the hypotenuse is labeled \(\theta\). The interior of the triangle is shaded with diagonal lines.

\[ \begin{aligned} x &= r \cos \theta \\ y &= r \sin \theta \\ z &= z \end{aligned} \quad \longleftrightarrow \quad \begin{aligned} r &= \sqrt{x^2 + y^2} \\ \theta &= \tan^{-1}(y/x) \\ z &= z \end{aligned} \]

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Spherical Coordinates

A three-dimensional Cartesian coordinate system shows the \(x\), \(y\), and \(z\) axes. A point in space is labeled \((x, y, z)\). A red line segment, labeled with the Greek letter \(\rho\), connects the origin \((0, 0, 0)\) to the point \((x, y, z)\). The angle between the positive \(z\)-axis and this red line segment is labeled \(\phi\). A vertical dashed line descends from point \((x, y, z)\) to its projection on the \(xy\)-plane at \((x, y, 0)\). A dashed line segment in the \(xy\)-plane connects the origin to \((x, y, 0)\); the angle from the positive \(x\)-axis to this line segment is labeled \(\theta\).

\(\rho =\) distance of point from origin \(> 0\)

\(\phi =\) smallest Angle from the +ve \(z\)-axis

\[0 \le \phi \le \pi\]

\(\theta =\) Angle from +ve \(x\)-axis to the shadow line from \((0,0,0)\) to \((x,y,0)\)

\[0 \le \theta \le 2\pi\]

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Spherical Coordinate Conversions

A three-dimensional Cartesian coordinate system is shown with \(x, y,\) and \(z\) axes. A point \((x, y, z)\) is plotted in space. A vector of length \(\rho\) (rho, colored red) connects the origin to this point. The angle between the positive $z$-axis and the vector \(\rho\) is \(\phi\) (phi, colored green). A vertical line drops from the point \((x, y, z)\) to\((x, y, 0)\) in the \(xy\)-plane. A line segment of length \(r\) connects the origin to \((x, y, 0)\). The angle between the positive \(x\)-axis and the segment \(r\) is \(\theta\) (theta, colored blue). Two right-angled triangles are highlighted with diagonal hatch marks: a vertical one with sides \(z\) and \(r\) and hypotenuse \(\rho\), and a horizontal one in the \(xy\)-plane with legs \(x\) and \(y\) and hypotenuse \(r\).
A right-angled triangle extracted from the vertical plane. The vertical leg is labeled \(z\), the horizontal leg is labeled \(r\), and the hypotenuse is \(\rho\) (colored red). The angle between the \(z\)-leg and the hypotenuse is \(\phi\) (colored green). The interior of the triangle is shaded with pink diagonal lines.
\[ \frac{z}{\rho} = \cos \phi \rightsquigarrow z = \rho \cos \phi \] \[ \frac{r}{\rho} = \sin \phi \rightsquigarrow r = \rho \sin \phi \]
A right-angled triangle representing the projection in the $xy$-plane. The vertical leg is labeled $x$ and the horizontal leg is labeled $y$. The hypotenuse is $r$. The angle between the $x$-leg and the hypotenuse is $\theta$ (colored blue). The interior of the triangle is shaded with purple diagonal lines.
\[ x = r \cos \theta = \rho \sin \phi \cos \theta \] \[ y = r \sin \theta = \rho \sin \phi \sin \theta \]

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Example: Converting Spherical to Cartesian Coordinates

Convert \( (\rho, \phi, \theta) = \left(4, \frac{3\pi}{4}, \frac{\pi}{3}\right) \) to Cartesian Coordinates.

A three-dimensional coordinate system shows a point plotted in spherical coordinates. The radial distance \(\rho\) is labeled as 4. The polar angle \(\phi\) is shown as \(\frac{3\pi}{4}\), measured from the positive z-axis down towards the negative z-axis. The azimuthal angle \(\theta\) is shown as \(\frac{\pi}{3}\), measured from the positive x-axis in the xy-plane. Dashed lines project the point to the Cartesian coordinate \( (\sqrt{2}, \sqrt{6}, -2\sqrt{2}) \).

\( \theta = \frac{\pi}{3} \implies x > 0, y > 0 \)
\( \phi = \frac{3\pi}{4} \implies z < 0 \)

Solution:

\[ x = \rho \sin \phi \cos \theta = 4 \sin\left(\frac{3\pi}{4}\right) \cos\left(\frac{\pi}{3}\right) \] \[ = 4 \cdot \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \] \[ = \sqrt{2} \]

\[ y = \rho \sin \phi \sin \theta = 4 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \sqrt{6} \]

\[ z = \rho \cos \phi = 4 \cdot \cos\left(\frac{3\pi}{4}\right) = -2\sqrt{2} \]

The resulting Cartesian point is \( (\sqrt{2}, \sqrt{6}, -2\sqrt{2}) \).

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Summary:

A diagram of a 3D Cartesian coordinate system illustrating the relationship between rectangular and spherical coordinates. A point in space is labeled \((x, y, z)\). A red line segment representing the distance \(\rho\) (rho) connects the origin to this point. The polar angle between the positive z-axis and the line segment \(\rho\) is labeled \(\phi\) (phi). A vertical dashed line connects the point \((x, y, z)\) to its projection on the xy-plane at \((x, y, 0)\). A purple line segment connects the origin to this projected point. The azimuthal angle between the positive x-axis and the purple projection line in the xy-plane is labeled \(\theta\) (theta).

Conversion from spherical coordinates \((\rho, \phi, \theta)\) to rectangular coordinates \((x, y, z)\):

\[x = \rho \sin \phi \cos \theta\] \[y = \rho \sin \phi \sin \theta\] \[z = \rho \cos \phi\]
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Conversion from rectangular coordinates \((x, y, z)\) to spherical coordinates \((\rho, \phi, \theta)\):

\[\rho = \sqrt{x^2 + y^2 + z^2}\] \[\phi = \cos^{-1}\left(\frac{z}{\rho}\right) = \cos^{-1}\left(\frac{z}{\sqrt{x^2 + y^2 + z^2}}\right)\] \[\theta = \tan^{-1}\left(\frac{y}{x}\right)\]

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Graphs in Spherical Coordinates

Example: \(\rho = 5\)

\[ \sqrt{x^2 + y^2 + z^2} = 5 \]

\[ x^2 + y^2 + z^2 = 25 \]

a sphere of radius 5

Visual Description: A 3D coordinate system with \(x\), \(y\), and \(z\) axes. A sphere is centered at the origin. A red radius vector extends from the origin to the surface of the sphere, labeled with a length of 5. The angle \(\phi\) is shown between the positive \(z\)-axis and the radius vector. Blue curves represent lines of latitude and longitude on the sphere's surface to illustrate its three-dimensional volume.

in general \(\rho = R\) is a sphere of radius \(R\)

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Example 2

Example 2: \(\phi = \frac{\pi}{4}\)
A three-dimensional coordinate system showing a cone opening upward from the origin. The vertex of the cone is at the origin \((0,0,0)\). The angle \(\phi\) between the positive z-axis and the outer surface of the cone is labeled as \(\frac{\pi}{4}\). The cone is outlined in red, and horizontal circular cross-sections are represented by dashed blue lines.
\[ \cos \phi = \frac{z}{\rho} \] \[ \cos\left(\frac{\pi}{4}\right) = \frac{z}{\sqrt{x^2+y^2+z^2}} > 0 \rightsquigarrow z > 0 \] \[ \frac{1}{\sqrt{2}} = \frac{z}{\sqrt{x^2+y^2+z^2}} \] \[ x^2+y^2+z^2 = 2z^2 \] \[ x^2+y^2 = z^2, \quad z > 0 \quad \} \text{ Single Cone} \]

\(\phi = \frac{\pi}{2} \rightsquigarrow z = 0 \rightsquigarrow\) xy plane

\(\phi = \phi_0 \neq \frac{\pi}{2} \rightsquigarrow\) Single Cone

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Example 3: \(\theta = \pi/4\)

A coordinate diagram drawn on grid paper showing a ray originating from the origin and extending into the first quadrant. The angle between the horizontal axis and the ray is labeled as \(\pi/4\). There are several parallel blue lines drawn diagonally across the first quadrant. A red line segment is drawn perpendicular to these blue lines, originating from the ray.
\[ \tan \theta = \frac{y}{x} \] \[ \tan \frac{\pi}{4} = \frac{y}{x}, \quad y > 0, x > 0 \] \[ y = x \]

half plane

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Triple integrals:

\[ \iiint_{D} f(x, y, z) \, dV \]

Cartesian:

A diagram of a small 3D cube representing an infinitesimal volume element in the Cartesian coordinate system.

\[ \begin{aligned} dV &= dx \, dy \, dz \\ &= dy \, dz \, dx \end{aligned} \]

Cylindrical:

A diagram of a differential volume element in cylindrical coordinates, shaped like a curved wedge. The base of the wedge is shaded in red and labeled with the area element \(r \, dr \, d\theta\).

\[ dV = r \, dz \, dr \, d\theta \]

Spherical: Spherical sectors

A diagram of a differential volume element in spherical coordinates, showing a small spherical wedge with curved faces.

\[ dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]

\[ \iiint_{D} f(x, y, z) \, dV = \int_{\theta} \int_{\phi} \int_{\rho} f(\rho \sin \phi \cos \theta, \rho \sin \phi \sin \theta, \rho \cos \phi) \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta . \]

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Example 1

Find Volume of a sphere of Radius \(R\) \(\quad (\frac{4}{3}\pi R^3)\)
A diagram of a sphere centered at the origin of a 3D coordinate system with axes labeled \(x\), \(y\), and \(z\). A radial vector extends from the origin to the surface, labeled \(\rho = R\). The polar angle \(\phi\) is shown opening from the positive \(z\)-axis. The azimuthal angle \(\theta\) is indicated in the \(xy\)-plane. A blue circular cross-section is drawn in the upper hemisphere.
\[V = \iiint_D 1 \, dV\] \[0 \le \rho \le R\] \[0 \le \phi \le \pi\] \[0 \le \theta \le 2\pi\] \[V = \int_0^{2\pi} \int_0^{\pi} \int_0^{R} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\] \[= \int_0^{2\pi} \int_0^{\pi} \frac{R^3}{3} \sin \phi \, d\phi \, d\theta = \int_0^{2\pi} \frac{2R^3}{3} \, d\theta = \frac{4\pi R^3}{3}\]

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Example 2

Find volume of Region inside \(\rho = 4 \cos \phi\) & Outside \(\rho = 2\).

A 3D coordinate system shows two intersecting spheres. The first sphere, centered at the origin with radius 2, is defined by \(\rho = 2\). The second sphere is shifted upwards along the positive z-axis, defined by \(\rho = 4 \cos \phi\). Converting to rectangular coordinates shows this is a sphere centered at \((0, 0, 2)\) with radius 2. The region of interest, shaded with diagonal lines, is located inside the upper sphere and outside the lower sphere. A red line segment representing \(\rho\) extends from the origin to the outer boundary of the upper sphere. An angle \(\phi\) is marked between the positive z-axis and this radius vector. A horizontal blue dashed circle indicates the plane of intersection between the two spheres.

To define the region, we first convert the spherical equations to rectangular form:

\[ \rho = 4 \cos \phi \] \[ \rho^2 = 4\rho \cos \phi \] \[ x^2 + y^2 + z^2 = 4z \] \[ x^2 + y^2 + z^2 - 4z + 4 = 4 \] \[ x^2 + y^2 + (z - 2)^2 = 4 \]

For the second boundary:

\[ \rho = 2 \] \[ x^2 + y^2 + z^2 = 4 \]

Setting the limits for the triple integral in spherical coordinates:

\[ 2 \le \rho \le 4 \cos \phi \] \[ 0 \le \theta \le 2\pi \]

\(0 \le \phi \le\) intersection point:

\[ 4 \cos \phi = 2 \] \[ \cos \phi = \frac{1}{2} \] \[ \Rightarrow \phi = \frac{\pi}{3} \]

The setup for the volume is:

\[ \text{Volume} = \int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{2}^{4 \cos \phi} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]

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Example 3

Find Volume of the Region between \(\rho=6\) (Sphere) and \(\phi=\pi/6\) (Cone).
A three-dimensional Cartesian coordinate system is illustrated with \(x\), \(y\), and \(z\) axes. A sphere is centered at the origin, and an inverted cone with its tip at the origin opens upward around the positive \(z\)-axis. The region of interest is the intersection of the sphere and the cone, creating a wedge-like "ice cream cone" shape. This volume is shaded with diagonal hatching. The angle from the \(z\)-axis to the surface of the cone is labeled \(\phi = \pi/6\), and the radius of the sphere is indicated as \(\rho = 6\).
The bounds for the region are: \[0 \le \rho \le 6\] \[0 \le \phi \le \pi/6\] \[0 \le \theta \le 2\pi\] \[\text{Volume} = \int_{0}^{2\pi} \int_{0}^{\pi/6} \int_{0}^{6} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\] Integrating with respect to \(\rho\): \[= \int_{0}^{2\pi} \int_{0}^{\pi/6} \frac{6^3}{3} \sin \phi \, d\phi \, d\theta\] Integrating with respect to \(\phi\): \[= \int_{0}^{2\pi} \frac{6^3}{3} \left[-\cos\left(\frac{\pi}{6}\right) + 1\right] \, d\theta\] Integrating with respect to \(\theta\): \[= 2\pi * \frac{6^3}{3} * \left[1 - \frac{\sqrt{3}}{2}\right].\]