Finding volume for triple integrals using spherical coordinates

Volume formula in spherical coordinates

We can use triple integrals and spherical coordinates to solve for the volume of a solid sphere. The volume formula in rectangular coordinates is

$V=\int\int\int_Bf(x,y,z)\ dV$

where $B$ represents the solid sphere and $dV$ can be defined in spherical coordinates as

$dV=\rho^2\sin\ d\rho\ d\theta\ d\phi$

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To convert in general from rectangular to spherical coordinates, we can use the formulas

$x=\rho\sin$

$y=\rho\sin$

$z=\rho\cos$

$\rho^2=x^2+y^2+z^2$

Remember, rectangular coordinates are given as $(x,y,z)$, and spherical coordinates are given as $(\rho,\theta,\phi)$.

In order to find limits of integration for the triple integral, we’ll say that $\phi$ is defined on the interval $[0,\pi]$ and that $\theta$ is defined on the interval $[0,2\pi]$. Then we only have to find an interval for $\rho$.

Using triple integrals in spherical coordinates to find volume

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Finding volume given by a triple integral over the sphere, using spherical coordinates

Example

Use spherical coordinates to find the volume of the triple integral, where $B$ is a sphere with center $(0,0,0)$ and radius $4$.

$\int\int\int_Bx^2+y^2+z^2\ dV$

Using the conversion formula $\rho^2=x^2+y^2+z^2$, we can change the given function into spherical notation.

$\int\int\int_Bx^2+y^2+z^2\ dV=\int\int\int_B\rho^2\ dV$

Then we’ll use $dV=\rho^2\sin\ d\rho\ d\theta\ d\phi$ to make a substitution for $dV$.

$\int\int\int_B\rho^2\left(\rho^2\sin\ d\rho\ d\theta\ d\phi\right)$

$\int\int\int_B\rho^4\sin\ d\rho\ d\theta\ d\phi$

Now we’ll find limits of integration. We already know the limits of integration for $\phi$ and $\theta$, since they are always the same if we’re dealing with a full sphere, so we get

$\int_0^\pi\int_0^\ d\rho\ d\theta\ d\phi$

Since $\rho$ defines the radius of the sphere, and we’re told that this sphere has its center at $(0,0,0)$ and radius $4$, $\rho$ is defined on $[0,4]$, so

$\int_0^\pi\int_0^\ d\rho\ d\theta\ d\phi$

We can use triple integrals and spherical coordinates to solve for the volume of a solid sphere.

We always integrate inside out, so we’ll integrate with respect to $\rho$ first, treating all other variables as constants.

$V=\int^{\pi}_0\int^{2\pi}_0\frac15\rho^5\sin{\phi}\Big|^{\rho=4}_\ d\theta\ d\phi$

$V=\int^\ d\theta\ d\phi$

$V=\int^\ d\theta\ d\phi$

Now we’ll integrate with respect to $\theta$, treating all other variables as constants.

$V=\frac{1,024}{5}\int^{\pi}_0\theta\sin{\phi}\Big|^{\theta=2\pi}_\ d\phi$

$V=\frac\ d\phi$

$V=\frac\ d\phi$

Finally, we’ll integrate with respect to $\phi$.

$V=\frac{2,048\pi}{5}\left(-\cos{\phi}\right)\Big|^{\pi}_0$

$V=-\frac{2,048\pi\cos{\phi}}{5}\Big|^{\pi}_0$

???V=-\frac\right]???

???V=-\frac(1)???

???V=\frac???

???V=\frac???

This is the volume of the region bounded beneath the surface ???x^2+y^2+z^2??? and above the sphere defined by ???B???.

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