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Concept Version 5
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Triple Integrals in Spherical Coordinates

When the function to be integrated has a spherical symmetry, change the variables into spherical coordinates and then perform integration.

Learning Objective

  • Evaluate triple integrals in spherical coordinates


Key Points

    • Switching from Cartesian to spherical coordinates, the function is transformed by this relation: $f(x,y,z) \longrightarrow f(\rho \cos \theta \sin \varphi, \rho \sin \theta \sin \varphi, \rho \cos \varphi)$.
    • For the transformation, the $dx\, dy\, dz$ differentials in the integral are transformed to $\rho^2 \sin \varphi \, d\rho \,d\varphi \,dz$.
    • Therefore, an integral evaluated in Cartesian coordinates can be switched to an integral in spherical coordinates as $\iiint_D f(x,y,z) \, dx\, dy\, dz = \iiint_T f(\rho \sin \varphi \cos \theta, \rho \sin \varphi \sin \theta, \rho \cos \varphi) \rho^2 \sin \varphi \, d\rho\, d\theta\, d\varphi.$

Terms

  • spherical coordinate

    a coordinate system for three-dimensional space where the position of a point is specified by three numbers: the radial distance of that point from a fixed origin, its polar angle measured from a fixed zenith direction, and the azimuth angle of its orthogonal projection on a reference plane that passes through the origin and is orthogonal to the zenith

  • Jacobian determinant

    the determinant of the Jacobian matrix


Full Text

When the function to be integrated has a spherical symmetry, it is sensible to change the variables into spherical coordinates and then perform integration.

Spherical Coordinates

Spherical coordinates are useful when domains in $R^3$ have spherical symmetry.

In $R^3$ some domains have a spherical symmetry, so it's possible to specify the coordinates of every point of the integration region by two angles and one distance. It's possible to use therefore the passage in spherical coordinates; the function is transformed by this relation: 

$f(x,y,z) \longrightarrow f(\rho \cos \theta \sin \varphi, \rho \sin \theta \sin \varphi, \rho \cos \varphi)$

Points on $z$-axis do not have a precise characterization in spherical coordinates, so $\theta$ can vary from $0$ to $2 \pi$. The better integration domain for this passage is obviously the sphere.

Integrals in Spherical Coordinates

The Jacobian determinant of this transformation is the following: 

$\displaystyle{\frac{\partial (x,y,z)}{\partial (\rho, \theta, \varphi)}} = \begin{vmatrix} \cos \theta \sin \varphi & - \rho \sin \theta \sin \varphi & \rho \cos \theta \cos \varphi \\ \sin \theta \sin \varphi & \rho \cos \theta \sin \varphi & \rho \sin \theta \cos \varphi \\ \cos \varphi & 0 & - \rho \sin \varphi \end{vmatrix} = \rho^2 \sin \varphi$

The $dx\, dy\, dz$ differentials therefore are transformed to $\rho^2 \sin \varphi \, d\rho \,d\varphi \,dz$.

Finally, you obtain the final integration formula: It's better to use this method in case of spherical domains and in case of functions that can be easily simplified, by the first fundamental relation of trigonometry, extended in $R^3$; in other cases it can be better to use cylindrical coordinates.

Example

Integrate $f(x,y,z) = x^2 + y^2 + z^2$ over the domain $D = x^2 + y^2 + z^2 \le 16$. 

In spherical coordinates: 

$f(\rho \sin \varphi \cos \theta, \rho \sin \varphi \sin \theta, \rho \cos \varphi) = \rho^2$

while the intervals of the transformed region $T$ from $D$:

$0 \leq \rho \leq 4, 0 \leq \varphi \leq \pi, 0 \leq \theta \leq 2\pi$

Therefore:

$\begin{aligned} \iiint_D (x^2 + y^2 +z^2) \, dx\, dy\, dz &= \iiint_T \rho^2 \ \rho^2 \sin \theta \, d\rho\, d\theta\, d\phi, \\ &= \int_0^{\pi} \sin \phi \,d\phi \int_0^4 \rho^4 d \rho \int_0^{2 \pi} d\theta \\ &= 2 \pi \int_0^{\pi} \sin \phi \left[ \frac{\rho^5}{5} \right]_0^4 \, d \phi \\ &= 2 \pi \left[ \frac{\rho^5}{5} \right]_0^4 \left[- \cos \phi \right]_0^{\pi}= \frac{4096 \pi}{5} \end{aligned}$

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