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Lessons for Multivariable Calculus

Mark Fitch

Section 4.3 Triple Integrals in Spherical

The fundamental shapes for integrating in each coordinate system along with the formula are shown in Figure 4.3.1. The derivation of the volume formula for the spherical shape is found in Section 4.4.
Cartesian
Cylindrical
Spherical
\(dx \: dy \: dz \)
\(r \; dr \: d\theta \: dz \)
\(\rho^2 \sin\phi \; d\rho \: d\phi \: d\theta \)
Figure 4.3.1. Fundamental Shapes for Triple Integrals

Example 4.3.2.

Find the volume inside
\begin{equation*} \rho = (1-\cos\phi)(1+\sin(3\theta)) \end{equation*}
There is nothing irregular and this is a single surface so we can integrate in any order we desire. \(\rho\) goes from inside (0) to the surface, so the initial integral is
\begin{equation*} \int_0^{(1-\cos\phi)(1+\sin(3\theta))} \rho^2 \sin\phi \; d\rho \end{equation*}
\(\phi\) runs from straight up to straight down without impact from \(\theta\text{.}\) Thus the next integral is
\begin{equation*} \int_0^{\pi} \int_0^{(1-\cos\phi)(1+\sin(3\theta))} \rho^2 \sin\phi \; d\rho \; d\phi \; d\theta \end{equation*}
For \(\theta\) we note that the solid consists of three copies of the same shape. We can therefore calculate just one of these and multiply that result by 3. The final integral is
\begin{equation*} 3\int_0^{2\pi/3} \int_0^{\pi} \int_0^{(1-\cos\phi)(1+\sin(3\theta))} \rho^2 \sin\phi \; d\rho \; d\phi \; d\theta \end{equation*}

Example 4.3.3.

Find the volume inside both \(\rho = 1-\cos\phi\) and \(\rho = 1\text{.}\)
The cardiod has a point at the origin (\(\phi=0\)) and then grows past a radius of 1 (reaching radius 2 at \(\phi=\pi\text{.}\) Thus it will intersect the sphere \(\rho=1\) which will form the bottom of this solid. We will need to calculate the volume as two integrals which we wil add.
First we must find the value of \(\phi\) at which the two surfaces intersect.
\begin{align*} 1-\cos\phi & = 1. \\ -\cos\phi & = 0. \\ \cos\phi & = 0. \\ \phi & = \pi/2, 3\pi/2, \ldots \end{align*}
We need only the first value for this calculation.
Neither surface has any interaction between \(\phi\) and \(\theta\text{,}\) so we can integrate in the order we want. The integrals are
\begin{equation*} \int_0^{2\pi} \int_0^{\pi} \int_0^{1-\cos\phi} \rho^2 \sin\phi \; d\rho \; d\phi \; d\theta + \int_0^{2\pi} \int_0^{\pi} \int_0^1 \rho^2 \sin\phi \; d\rho \; d\phi \; d\theta \end{equation*}
Note the bottom half is just the bottom half of a sphere we can calculate that from the known volume of a sphere which is \(\frac{4}{3}\pi r^3\text{.}\) This would be
\begin{equation*} \int_0^{2\pi} \int_0^{\pi} \int_0^{1-\cos\phi} \rho^2 \sin\phi \; d\rho \; d\phi \; d\theta + \frac{1}{2} \cdot \frac{4}{3}\pi (1)^3 \end{equation*}

Exercises Exercises

Setup the following triple integrals. Use technology as needed for showing all steps of the calculations.

1.

Find the volume inside \(\rho=\sin\theta\text{.}\)

2.

Find the volume inside \(\rho=\phi\text{.}\)

3.

Find the volume inside \(\rho=\cos\phi\text{.}\)
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