Question

How are triple integrals defined in cylindrical and spherical coordinates?
Why might one prefer working in one of these coordinate systems to working in rectangular coordinates?

Answer 1

Step 1
The equation relating to rectangular (x, y, z) and cylindrical \(\displaystyle{\left({r},\theta,{z}\right)}\) coordinates are, \(\displaystyle{x}={r} \cos{\theta},{y}= \sin{\theta},{z}={z}\)
\(\displaystyle{r}^{2}={x}^{2}+{y}^{2}\)
Here, \(\displaystyle \tan{\theta}=\frac{y}{{x}}\)
Hence , the solution is \(\displaystyle \tan{\theta}=\frac{y}{{x}}.\)
The equation relating spherical coordinates to Cartesian and cylindrical coordinates is, \(\displaystyle{r}=\delta \sin{\theta},{x}={r} \cos{\theta}=\delta \sin{\cancel{\circ}} \cos{\theta}\)
\(\displaystyle{z}=\delta \cos{\cancel{\circ}},{y}={r} \sin{\theta}=\delta \sin{\cancel{\circ}} \sin{\theta}\)
Step 2
Here, \(\displaystyle\delta=\sqrt{{{x}^{2}+{y}^{2}+{z}^{2}}}\)
From equation 1, we know \(\displaystyle{x}^{2}+{y}^{2}={r}^{2}.\)
Therefore,
\(\displaystyle\delta=\sqrt{{{r}^{2}+{z}^{2}}}\)
Hence, the solution is \(\displaystyle\delta=\sqrt{{{r}^{2}+{x}^{2}}}.\)
The diagram regarding the cylindrical coordinates \(\displaystyle{\left({r},\theta,{z}\right)}\) is given below: