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16.5 Triple Integrals in Spherical Coordinates

In spherical, we locate a point by \((\rho, \phi, \theta)\).

Greek "rho" (\(\rho\))

distance from origin to point

Greek "phi" (\(\phi\))

angle measured from z-axis down

\(\theta\)

the same \(\theta\) as in polar or cylindrical

3D coordinate system showing a point (rho, phi, theta) with its projection on the xy-plane.
\[ \rho \ge 0 \]\[ 0 \le \theta \le 2\pi \]\[ 0 \le \phi \le \pi \]

doesn't need to go to \(2\pi\)

"shadow" on xy-plane

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converting from/to Cartesian

3D diagram decomposing spherical coordinates into vertical and horizontal triangles.

note: \(x^2 + y^2 + z^2 = \rho^2\)

top triangle

Right triangle with hypotenuse rho, vertical side z, and angle phi.

\(\cos \phi = \frac{z}{\rho}\)

\[ z = \rho \cos \phi \]

\(\sin \phi = \frac{l}{\rho} \quad \text{so} \quad l = \rho \sin \phi \)

lower triangle

Right triangle in the xy-plane with hypotenuse l, sides x and y, and angle theta.

\(\cos \theta = \frac{x}{l}\)

\(x = l \cos \theta\)

\(\sin \theta = \frac{y}{l}\)

\(y = l \sin \theta\)

\[ x = \rho \sin \phi \cos \theta \]
\[ y = \rho \sin \phi \sin \theta \]
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Example (Cartesian → Spherical)

Given Cartesian coordinates \( (x, y, z) = (1, -1, \sqrt{2}) \), find the spherical coordinates \( (\rho, \phi, \theta) = ? \)

From the position of the point, we can observe:

  • \( \frac{3\pi}{2} \le \theta \le 2\pi \)
  • \( 0 \le \phi \le \frac{\pi}{2} \)
3D Cartesian axes with a point plotted. Angle phi is from z-axis, theta is in the xy-plane.

Calculating \( \rho \):

\[ x^2 + y^2 + z^2 = \rho^2 \]\[ 1 + 1 + 2 = \rho^2 \]
\( \rho = 2 \)

Calculating \( \phi \):

\[ z = \rho \cos \phi \]\[ \sqrt{2} = 2 \cos \phi \]\[ \cos \phi = \frac{\sqrt{2}}{2} \]
\( \phi = \frac{\pi}{4} \)

Calculating \( \theta \):

\[ \begin{cases} x = \rho \sin \phi \cos \theta \\ y = \rho \sin \phi \sin \theta \end{cases} \]\[ \frac{y}{x} = \tan \theta = \frac{-1}{1} = -1 \]

In Quadrant IV, find \( \theta \) such that \( \tan \theta = -1 \):

\( \theta = \frac{7\pi}{4} \)
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Volume Element in Triple Integrals

In triple integrals, the volume element \( dV \) varies by coordinate system:

  • Cartesian: \( dV = dz \, dy \, dx \)
  • Cylindrical: \( dV = r \, dz \, dr \, d\theta \)
  • Spherical: \( dV = ? \)

Deriving the Spherical Volume Element

Take a small spherical shell from small changes in \( \rho, \theta, \phi \).

When \( d\rho, d\phi, d\theta \) are small, the shell is approximately a rectangular box.

3D axes showing a spherical wedge element with arc lengths rho sin(phi) d-theta and rho d-phi.
Detailed view of the spherical shell element with dimensions d-rho, rho d-phi, and rho sin(phi) d-theta.
Rectangular box representing the volume element dV with sides rho d-phi, d-rho, and rho sin(phi) d-theta.

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

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Example: Triple Integral in Cartesian Coordinates

\[ \int_{0}^{6} \int_{0}^{\sqrt{36-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{72-x^2-y^2}} dz \, dy \, dx \]

terrible in Cartesian

the upper bound of \( z \) is part of a sphere:

\[ z = \sqrt{72-x^2-y^2} \]\[ z^2 = 72-x^2-y^2 \]\[ x^2+y^2+z^2 = 72 \]

when dealing with spheres or spheres-like, spherical is good

so, go to spherical: bounds for \( \rho, \phi, \theta \)?

\( 0 \le x \le 6 \)

\( 0 \le y \le \sqrt{36-x^2} \)

A quarter circle in the first quadrant of the xy-plane with radius 6, showing angle theta.

this is the shadow onto xy-plane

\[ 0 \le \theta \le \frac{\pi}{2} \]
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\[ \underbrace{\sqrt{x^2+y^2}}_{\text{cone}} \le z \le \underbrace{\sqrt{72-x^2-y^2}}_{\text{sphere radius } \sqrt{72}} \]
3D diagram showing a cone inside a sphere in the first octant.

we want everything inside

we accumulate from origin \( (\rho = 0) \)

out to boundary \( (\rho = \sqrt{72}, \text{ sphere}) \)

\[ 0 \le \rho \le \sqrt{72} \]

onto the yz-plane

2D projection onto the yz-plane showing a sector bounded by a vertical axis and a line at 45 degrees.

\( \phi \) from 0 to where edge of cone is

cone: \( z = \sqrt{x^2+y^2} = \sqrt{y^2} = y \)

slope 1 \( \rightarrow \) angle is \( \frac{\pi}{4} \)

so,

\[ 0 \le \phi \le \frac{\pi}{4} \]
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Volume in Spherical Coordinates

The volume is calculated using spherical coordinates as follows:

\[ \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sqrt{72}} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta = \dots = \boxed{12\sqrt{72} \pi \left( 1 - \frac{1}{\sqrt{2}} \right)} \]

Note: The differential volume element is \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \). The integration limits are specified for \( \theta \), \( \phi \), and \( \rho \) respectively.

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Example: Volume of a Solid

Problem Statement

Find the volume of the solid outside \( \rho = 1 \) and inside \( \rho = 2 \cos \phi \).

\( \rho = 1 \): Sphere radius 1, center \( (0,0,0) \)

\( \rho = 2 \cos \phi \): What is this shape?

Derivation

Starting with \( \rho = 2 \cos \phi \):

\[ \sqrt{x^2 + y^2 + z^2} = 2 \cos \phi \cdot \frac{\rho}{\rho} = \frac{2 \rho \cos \phi}{\rho} = \frac{2z}{\sqrt{x^2 + y^2 + z^2}} \]

Multiplying through:

\[ x^2 + y^2 + z^2 = 2z \implies x^2 + y^2 + (z-1)^2 = 1 \]

This is a sphere with center \( (0,0,1) \) and radius 1.

Visualizing "Toad's Hat"

The intersection of these two spheres creates a shape referred to as "Toad's hat".

\( 0 \le \theta \le 2\pi \)

This indicates the volume goes all the way around the z-axis.

3D coordinate graph showing two intersecting spheres on the z-axis, labeled as Toad's hat.
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xz-plane projection

The projection in the xz-plane shows the region bounded by two spheres in spherical coordinates.

\[ 1 \le \rho \le 2 \cos \phi \]
Graph in the xz-plane showing two overlapping circular regions labeled rho=2 cos phi and rho=1.

Determining the angle \(\phi\)

\(\phi\): from z-axis (\(\phi = 0\)) to intersection.

At the intersection:

\[ \rho = 1, \quad \rho = 2 \cos \phi \]
\[ 1 = 2 \cos \phi \implies \cos \phi = \frac{1}{2} \rightarrow \phi = \frac{\pi}{3} \]
\[ 0 \le \phi \le \frac{\pi}{3} \]
Diagram in the xz-plane showing the angle phi measured from the positive z-axis to the intersection of two curves.

Volume Calculation

The volume is calculated using a triple integral in spherical coordinates:

\[ \text{volume} = \int_{0}^{2\pi} \int_{0}^{\pi/3} \int_{1}^{2 \cos \phi} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta = \dots = \boxed{\frac{11\pi}{12}} \]
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the volume of my hat is \(\frac{11\pi}{12}\) !

A 3D render of the character Toad from the Mario franchise, standing and looking forward.