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Triple Integrals in Spherical Coordinates

There are certain double integrals that are easier to evaluate in polar coordinates, if the region over which we are integrating is more easily described in polar coordinates than rectangular coordinates. Similarly, there are certain triple integrals that are easier to evaluate in spherical coordinates, if the region over which we are integrating is most easily described in the three-dimensional system.

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Triple Integrals in Spherical Coordinates Definition

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The spherical coordinates $(\rho, \theta, \phi)$ of a point in a three dimensional plane is related by the relations of $x = \rho \sin \phi \cos \theta, y = \phi sin \rho \sin \theta$ and $z = \rho \cos \phi$, where $\rho ^2$ = x2 + y2 + z2.

To integrate over some region in 3-dimensional space that is described by the spherical coordinates, and especially, if we see the expression x2 + y2 + z2  in the integrand, we can definitely switch to spherical coordinates.

The formula for triple integration in spherical coordinates is given as follows:

$\int \int \int f(x,y,z)dV=\int_{\delta }^{\gamma }\int_{\alpha }^{\beta }\int_{a}^{b}\rho ^{^{2}}\sin\phi f(\rho \sin\phi \cos\theta ,\rho \sin\phi \sin\theta ,\rho \cos\phi )d\rho d\theta d\phi $

The volume of a spherical coordinate can be interpreted from the following diagram:

Triple Integrals in Spherical Coordinates

Calculating Triple Integrals in spherical coordinates:
We can use the substitution method of variables to calculate triple integrals, that is, taking an integral in terms of x's and changing it in terms of v's. Triple integral implies, integrating the function three times, and in fact, we already used to when we converted triple integrals to spherical coordinates.
 
Generally, the reason for changing variables is to find an integral that we can use it with the new variables. Another reason for changing variables is to convert the dimensions into a nice diagram to work with it. When we were converting the spherical coordinates, we need not worry about this change as it is easy enough to find the new limits based on that given dimension.

Solved Examples

Question 1: Apply the method for spherical coordinates to find the formula for the volume of a sphere which is centered at the origin and has radius a.
Solution:
For a sphere, we have, 0 $\leq $ $\rho$ $\leq $ a, 0 $\leq $ $\theta$ $\leq $ $2\pi$, 0 $\leq $ $\phi$ $\leq $ $\pi$. Therefore, volume will be calculated by putting the values in the below formula,

V = $\int_{0}^{\Pi }\int_{0}^{2\Pi }\int_{0}^{a}\rho ^{2} \sin\varphi d\rho d\theta d\varphi $

By solving, we will get V = $\frac{4\Pi a^{3}}{3}$

Question 2: Find the volume that lies inside the sphere x2 + y2 + z2and outside the cone z2 = x2 + y2.
Solution:
After converting to spherical coordinates, we will get the equation of the form, $2 \times \rho ^2 \cos ^2 \phi = \phi ^2$.
Thus, $\phi$ = $\frac{180}{4}$ or $\frac{3\Pi }{4}$, and hence, the volume would be dV = $\rho ^2 \sin \phi d \rho d\theta d\phi$ or V = $\int_{0}^{\Pi }\int_{\frac{\Pi }{4}}^{\frac{3\Pi }{4}}\int_{0}^{\sqrt{2}}\rho ^{2}\sin\varphi d\rho d\theta d\varphi $ for the respective value of $(\rho, \theta, \phi)$ and on solving the integration one by one of the three times we will get the answer as $\frac{8\Pi }{3}$.


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