Volume of a part of a sphere in defining triple integrals with spherical coordinates

calculusintegrationspherical coordinates

So in my Calculus 3 textbook we have three distinct sub-chapters on double and triple integrals that concern polar substitution, cylindrical substitution and spherical substitution. The goal in each chapter is to try to precisely define the integral using the given substitution. What I mean by that is that we're not using the general theorem for substitutions that just says 'multiply it by the Jacobian', but we're rather trying to derive it for each specific case.

When it came to polar or cylindrical it was easy enough, mostly because it was explained in detail how the areas\volumes of the little parts of the sets were calculated. In the part concerning spherical coordinates I guess the author just got tired and stated this:

It turns out that the volume of the set defined as $$S:=\big\{ (r,
\theta, \varphi)| \quad r\in[r_1,r_2], \theta\in[\theta_1, \theta_2],
\varphi\in[\varphi_1, \varphi_2] \big\}$$
is
$$V(S)=\frac{1}{3}(r_2^3-r_1^3)(\theta_2-\theta_1)(\cos\varphi_2 –
\cos\varphi_1)$$

And then from here it does indeed become easy enough to derive the following formula, which is the part I understand.
$$\iiint_{F(T)}f(x, y, z)dxdydz = \iiint_{T}f(r\cos\theta\sin\varphi, r\sin\theta\sin\varphi, r\cos\varphi)r^2\sin\varphi drd\theta d\varphi
\\
\text{where } F:(r, \theta, \varphi) \to (r\cos\theta\sin\varphi, r\sin\theta\sin\varphi, r\cos\varphi)$$

The thing that I can't figure out is how the volume of the set $S$ was calculated. So, any ideas/hints? Thanks.

Best Answer

Since a radius-$r$ sphere has volume $\frac{4\pi}{3}r^3$, a region defined with a constraint $r\in [r_1,\,r_2]$ with $0\le r_1\le r_2$ is proportional to $r_2^3-r_1^3$. Similarly, the $\theta_2-\theta_1$ factor is trivial. Finally, an increasing function $f(\varphi)$ exists for which the original problem's volume is $(r_2^3-r_2^3)(\theta_2-\theta_1)(f(\varphi_2)-f(\varphi_1))$. The case$$r_1=0,\,\theta_1=0,\,\theta_2=2\pi,\,\varphi_1=0,\,\varphi_2=\pi$$gives$$\frac{4\pi}{3}r_2^3=2\pi r_2^3(f(\pi)-f(0))\implies f(\pi)-f(0)=\frac23.$$It remains to show we can choose $f(\varphi)\propto 1-\cos\varphi$. The above values of $r_1,\,\theta_i$, together with $r_2=1$, allows us to verify this by considering the $\varphi_2=\varphi_1+d\varphi$ case, where we obtain an infinitesimal circular shell of thickness $d(1-\cos\varphi)$ (because, as we can verify with a right-angled triangle, a given value of $\varphi$ comprises a locus in a plane a distance $1-\cos\varphi$ from a parallel plane containing a great circle of the unit sphere centred at $O$).