Volume calculation with change of variables

Question

I am trying to calculate the volume of a solid, given the equations of its bounding surfaces. It is a $3$-dimensional object, so the equations are in $x$, $y$ and $z$. In order to simplify the equations and the volume calculation, I made a substitution, namely $\tilde{x}=x+\frac{kz}{\sigma}$, $\tilde{y}=y\pm\frac{kz}{\sigma}$, and $\tilde{z}=\frac{z}{\sigma k}$.

In these new variables, I carried out a volume calculation, and I got a result; let's call it $V$. I want to know the actual volume, though, back at home in $(x,y,z)$.

I get confused at this part. Is the volume equal to $\sigma k V$, or is it $\frac{V}{\sigma k}$? It must be one of those, because $\sigma k$ and $\frac{1}{\sigma k}$ are the determinants of the matrices that convert between the two coordinate bases. I've just about convinced myself that it's $\sigma k V$, but the answer would make more sense - in the context of the problem - if it were $\frac{V}{\sigma k}$.

What is the right way to think about this kind of problem? Does anyone have a good intuition they can share about how volume changes with changes of variables? Thanks in advance.

Answer 1

As you know, you need to include a “scale factor” when transforming the volume element—the determinant of the Jacobian of the coordinate mapping. There’s a notation sometimes used for the Jacobian that I find helpful in remembering which direction to use when computing this scale factor: $$dV = dx\,dy\,dz = {\partial(x,y,z)\over\partial(\tilde x,\tilde y,\tilde z)}\,d\tilde{x}\,d\tilde{y}\,d\tilde{z}.$$ The term on the right that looks like a fraction represents the determinant of the Jacobian of the map $(\tilde x,\tilde y,\tilde z)\mapsto(x,y,z)$, i.e., the matrix of partial derivatives of $x$, $y$ and $z$ with respect to $\tilde x$, $\tilde y$ and $\tilde z$. (You’ll also see this notation used to represent the Jacobian matrix itself.) As a mnemonic, you can think of the “denominator” of this “fraction” canceling the $d\tilde{x}\,d\tilde{y}\,d\tilde{z}$ and leaving $dx\,dy\,dz$ in the “numerator.”

In your case, we have $dx\,dy\,dz=\sigma k\,d\tilde{x}\,d\tilde{y},d\tilde{z}$, so after making the change of variables you have to multiply by $\sigma k$ to maintain the value of the integral. This makes some intuitive sense: the coordinate transformation shrinks $z$ by a factor of $\sigma k$, so you have to multiply by that to compensate.

Answer 2

In doing variables change, you should change from an orthogonal basis to an orthogonal basis, so that the elementary volumes $dV=dx dy dz$ and $dV'=dx' dy'dz'$ keep the same.
If the new system is orthogonal, but relevant vectors are not unitary, yet have a constant "length" (measured in the base system) then the volume calculated in the two systems will differ by a multiplicative constant, which can be easily calculated.
Otherwise you shall recurse and include the Jacobian, i.e. the determinant of the transformation matrix (which would be $1$ in the first case and $abc$ in the second).
The transformation you applied is not orthogonal, although it is realized through a constant matrix, thus a constant Jacobian.
Then you have $dV'=J*dV$, can you carry on from here ?

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(added)

I think I understood the impasse, as explained in your comment, and actually which is the Jacobian is not standardly defined. Well let me try and laid an Ariadne's thread. In one dimension, taking a simple proportional transformation ($J=const$) $$ \int_{x\; = \;0}^{\;1} {f(x)\,dx} \quad \to \left( {\xi \, = \,J\,x} \right) \to \quad \int_{\xi \; = \;0}^{\;J} {f(J^{\, - 1} \xi )\,J^{\, - 1} d\xi } $$

For example $$ \frac{{27}}{4} = \int_{x\; = \;0}^{\;1} {\left( {3x} \right)^{\,3} \,dx} \quad \to \left( {\xi \, = \,3\,x} \right) \to \quad \int_{\xi \; = \;0}^{\;3} {\xi ^{\,3} \frac{1}{3}d\xi } = \frac{{27}}{4} $$

Now, in 3D the pattern is similar. Taking for simplicity a homogeneous linear constant transformation $$ \begin{array}{l} \left( {\begin{array}{*{20}c} \xi \\ \eta \\ \zeta \\ \end{array}} \right) = {\bf R}\;\left( {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right)\quad \to \left( {{\bf R}\, = \,const} \right) \to \quad \left( {\begin{array}{*{20}c} {d\xi } \\ {d\eta } \\ {d\zeta } \\ \end{array}} \right) = {\bf R}\left( {\begin{array}{*{20}c} {dx} \\ {dy} \\ {dz} \\ \end{array}} \right)\;\quad \to \left( {{\bf R}\, = \,const} \right) \to \\ \to \quad dV' = d\xi d\eta d\zeta = \left( {d{\bf \xi } \times \,d{\bf \eta }} \right)\, \cdot d{\bf \zeta } = \left| {\,{\bf R}\,} \right|\;\left( {d{\bf x} \times \,d{\bf y}} \right)\, \cdot d{\bf z} = J\;dV \\ \end{array} $$

and $$ V = \iiint\limits_{(x,y,z)\, \in \,G} {f(x,y,z)\,dxdydz} = \iiint\limits_{(\xi ,\eta ,\zeta )\, \in \,G'} {g(\xi ,\eta ,\zeta )\,\,J^{\, - 1} d\xi d\eta d\zeta } = J^{\, - 1} \iiint\limits_{(\xi ,\eta ,\zeta )\, \in \,G'} {g(\xi ,\eta ,\zeta )\,\,d\xi d\eta d\zeta } = J^{\, - 1} \,V' $$

Final consideration is that, with some attention and a methodical approach the Jacobian can be managed.
The difficult job, most times, is to convert the integration area from $G$ to $G'$.

For instance, converting a cube when axes are rotated.