converting vector inside integral into polar coordinate

Question

I am evaluating this function $A(z)$:

$A(z) = \iiint v_x \frac{\partial f_o}{\partial v_x}(1-e^{-\frac{z}{\tau v_z}} ) dv_x dv_y dv_z$

$v$ is a vector in v-space.

$\theta$ is the polar angle between vector $v$ and z-axis. $\phi$ is azimuthal angle.

I need to convert that into a polar coordinate $(v,\theta,\phi)$

my steps:

$A(z) = \int v \sin\theta\cos\phi \frac{\partial f_o}{\partial v \sin\theta\cos\theta}(1-e^{-\frac{z}{\tau v \cos\theta}} ) v^2 \sin\theta dv d\theta d\phi$

$A(z) = \int v \frac{\partial f_o}{\partial v }(1-e^{-\frac{z}{\tau v \cos\theta}} ) v^2 \sin\theta dv d\theta d\phi$

final result:

$A(z) = \int v^3 \frac{\partial f_o}{\partial v }(1-e^{-\frac{z}{\tau v \cos\theta}} ) \sin\theta dv d\theta d\phi$

the expected/correct result:

$A(z) = \int v^3 \frac{\partial f_o}{\partial v }(1-e^{-\frac{z}{\tau v \cos\theta}} ) \sin^3\theta \cos^2\phi dv d\theta d\phi$

It is like my answer multiplied by $\sin^2\theta \cos^2\phi$

I have some doubt in converting $dv_x dv_y dv_z$ into the direction in the polar coordinate $dv d\theta d\phi$

Can someone point my mistake here?

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EDIT:

$\frac{\partial f_o}{\partial v_x} = \frac{\partial f_o}{\partial v} \frac{\partial v}{\partial v_x}$

$v_x = v \sin\theta\cos\phi$

$dv_x = dv \sin\theta\cos\phi $

$\frac{\partial f_o}{\partial v_x} = \frac{\partial f_o}{\partial v \sin\theta\cos\theta}$

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EDIT 2:

1. I wrote $\frac{\partial v_x}{\partial f_o}$

$$\frac{\partial v_x}{\partial f_o} = \sin\theta\cos\phi \frac{\partial v}{\partial f_o} + v\cos\theta\cos\phi \frac{\partial \theta}{\partial f_o}- v\sin\theta\sin\phi \frac{\partial \phi}{\partial f_o}$$

2. (trying) find the inverse of this. I am not sure though.

determinant = $-v^2 \sin^2\theta \cos\theta \sin\phi \cos^2\phi$

Answer 1

It is important to realize that derivatives are not fractions and cannot be treated as such. And as discussed in THIS ANSWER, for the transformation $(x_1,x_2,\dots,x_n) \mapsto (u_1,u_2,\dots,u_n)$, it is not true in general that the partial derivatives $\frac{\partial x_i}{\partial u_j}$ and $\frac{\partial u_j}{\partial x_i}$ are reciprocals.

To wit, the analysis in the edit of the OP is flawed. Instead, with $v=\sqrt{v_x^2+v_y^2+v_z^2}$ we see that

$$ \frac{\partial v}{\partial v_x}=\frac{v_x}{v} =\sin(\theta)\cos(\phi)\ne \frac1{\sin(\theta)\cos(\phi)}=\frac1{\frac{\partial v_x}{\partial v}}$$

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Finaly, with $\frac{\partial \theta}{\partial v_x}=-\frac1v \cos(\theta)\cos(\phi)$ and $\frac{\partial \phi}{\partial v_x}= \frac{\sin(\phi)}{v \sin(\theta)}$, we have

$$\begin{align} \frac{\partial f}{\partial v_x}&=\frac{\partial f}{\partial v}\frac{\partial v}{\partial v_x}+\frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial v_x}+\frac{\partial f}{\partial \phi}\frac{\partial \phi}{\partial v_x}\\\\ &=\sin(\theta)\cos(\phi)\frac {\partial f}{\partial v}-\frac1v \cos(\theta)\cos(\phi)\frac {\partial f}{\partial \theta}+\frac{\sin(\phi)}{v\sin(\theta)}\frac{\partial f}{\partial \phi} \end{align}$$

Now suppose $f$ is independent of $\theta$ and $\phi$. In that case $\frac{\partial f}{\partial \theta}=\frac{\partial f}{\partial \phi}=0$. Then, we see that

$$\begin{align} \frac{\partial f}{\partial v_x}&=\sin(\theta)\cos(\phi)\frac {\partial f}{\partial v} \end{align}$$