Why is the Jacobian ${\rho}$

Question

Just a little confused. When I find the volume of a cone (or a sphere) for that matter I multiply the partial derivatives by the Jacobian. ${\rho}$ for a cone. and ${\rho^2 \sin \phi}$ for a sphere. But I integrate ${\delta \rho}$ from 0 to R... So then ${\delta \rho}$ is the rate of change in the radius... or is it the rate of change in the jacobian? I am a little confused.

Answer 1

The volume of a region $R$ is given by the integral $\int_R1{\rm d}x_1\cdots{\rm d}x_n$. However in order to actually calculate this integral the region needs to be parametrized in some way; say $f:S\to R$ where $S$ is another region under consideration (and $f$ is a homeomorphism). Then we have

$$\int_R1\,{\rm d}x_1\cdots{\rm d}x_n=\int_S~\left|\det\left[\frac{\partial x_i}{\partial y_j}\right]\right|~{\rm d}y_1\cdots{\rm d}y_n$$

by the multivariable substitution formula, where $(x_1,\cdots,x_n)=f(y_1,\cdots,y_n)$. This formula is believable if you expand each ${\rm d}x_i=\sum\frac{\partial x_i}{\partial y_j}{\rm d}y_j$ and then use the fact that differential forms are alternating bilinear forms to rewrite ${\rm d}x_1\cdots{\rm d}x_n$ (via combining like terms) as the Leibniz formula for the determinant times ${\rm d}y_1\cdots{\rm d}y_n$. This is somewhat high-altitude stuff though. A simpler way to understand this formula is as a generalization of $\int_{a}^{b}f(u(t))u'(t)dt=\int_{u(a)}^{u(b)}f(u)du$, where the appropriate generalization of $u'(t)dt=du$ to multiple variables is with the Jacobian.

Let's consider cylindrical coordinates. We have

$$\begin{cases} x = \rho\cos\varphi \\ y=\rho\sin\varphi \\ z=z\end{cases} $$

The Jacobian is

$$\begin{array}{ll} \left|\begin{matrix}\frac{\partial x}{\partial\rho} & \frac{\partial x}{\partial \varphi} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial \rho} & \frac{\partial y}{\partial \varphi} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial \rho} & \frac{\partial z}{\partial\varphi} & \frac{\partial z}{\partial z}\end{matrix}\right| & =\left|\begin{matrix}\cos\varphi & -\rho\sin\varphi & 0 \\ \sin\varphi & \rho\cos\varphi & 0 \\ 0 & 0 & 1 \end{matrix}\right| \\ & =(1)((\cos\varphi)(\rho\cos\varphi)-(\sin\varphi)(-\rho\sin\varphi)) \\ & =\rho. \end{array}$$

Therefore ${\rm d}x{\rm d}y{\rm d}z=\rho\,{\rm d}\rho{\rm d}\varphi{\rm d}z$.

Now consider spherical coordinates $(\rho,\theta,\phi)$, where

$$\begin{cases}x=\rho\cos\theta\cos\phi \\ y =\rho\cos\theta\sin\phi \\ z=\rho\sin\theta\end{cases} $$

Can you calculate the Jacobian yourself? You should end up with $\rho^2\sin\phi$.

Back to cylindrical coordinates, let's try to do a cone of height $H$ and base radius $R$. We may parametrize the cone as $(\rho,\varphi,z)$ with $0\le \rho\le R$ and $0\le z\le (1-\rho/R)H$ (this latter bound can be derived using geometry). Therefore the volume integral is

$$\int_0^R\int_0^{(1-\rho/R)H}\int_0^{2\pi}\rho\,{\rm d}\varphi{\rm d}z{\rm d}\rho=2\pi\int_0^R \rho(1-\rho/R)H\,{\rm d}\rho=\frac{\pi}{3}R^2H.$$