Integral with more-dimensional substitution variables Good day,
In the lecture of partial differential equations we had the following transformation:
$$\int_{||\nu||=1} h(x+\nu c t, \tau) d\nu = \frac{1}{c^2 t^2}  \int_{||y-x||=ct} h(y,\tau) dy$$
for $y=(y_1,y_2,y_3)^T, x=(x_1,x_2,x_3)^T, \nu=(\nu_1,\nu_2,\nu_3)^T \in \mathbb{R}^3, h: \mathbb{R}^3 \times \mathbb{R} \to \mathbb{R}.$
The subsition should work like $x+\nu ct=y$ and the new boundaries are $1=||\nu||=||\frac{y-x}{ct}||$. From the one-dimensional case I would only get the additional $\frac{1}{ct}$ from $d\nu = \frac{1}{ct} dy$.
So I looked up the formula for more-dimensional substitution and got this formula:
$$ \int_{g(M)} h(\nu) d \nu = \int_M h(g(y)) | \det g'(y) | dy $$
In my case, I suppose, I have $g(y)=\frac{y-x}{ct}=\nu.$ But then 
$$g'(y)=\begin{pmatrix} \frac{1}{ct} & 0 & 0 \\ 0 & \frac{1}{ct} & 0 \\ 0 & 0 & \frac{1}{ct} \end{pmatrix}$$
and the determinant would give $\frac{1}{c^3 t^3}$.
Can someone please help me clear my confusion? Do I get an additional $ct$ from the boundardy maybe?
Thanks a lot,
Marvin
 A: First recall how a surface integral of first kind is computed when you have a parameterization of the surface $\Gamma$
$$\int\limits_{\Gamma}{f(\vec{x})ds}=\int\limits_{(u,v)\in D}{f(\vec{x}(u,v))|\frac{\partial \vec{x}}{\partial u}\times\frac{\partial \vec{x}}{\partial v}|dudv}$$ where $D$ is the domain where the parameters $u,v$ change.
Now for your problem: Let us denote with $\Gamma$ the unit sphere $\|\nu\|=1$ in the $\nu-$ space, and with $\hat\Gamma$ the sphere $\|y-x\|=ct$ in the $y-$ space after the transformation
$$F: y\in \mathbb R^3\to \nu\in\mathbb R^3$$ defined by $F(y)=\frac{y-x}{ct}=\nu$. Here $F^{-1}(\nu)=x+ct\nu$.  So we have $\hat\Gamma \xrightarrow{F}\Gamma$ and $\Gamma \xrightarrow{F^{-1}}\hat\Gamma$. Now let us make an arbitrary parameterization of $\Gamma$:
$$\nu=\nu(u,v),\quad (u,v)\in D\subset \mathbb R^2$$
\begin{align}
\begin{array}{|l}
&\nu_1=\nu_1(u,v)\\
&\nu_2=\nu_2(u,v)\\
&\nu_3=\nu_3(u,v)
\end{array}
\end{align}
Then your surface integral becomes:
$$\int\limits_{\Gamma}{h(x+ct\nu,\tau)d\nu}=\int\limits_{(u,v)\in D}{h(x+ct\nu(u,v),\tau)|\frac{\partial \nu}{\partial u}\times\frac{\partial \nu}{\partial v}|dudv}\quad (1)$$. Having the parameterization of $\Gamma$ and the transformation $F$, you automatically get a parameterization of $\hat\Gamma$ by:
$$y=F^{-1}(\nu(u,v))\Leftrightarrow y=x+ct\nu(u,v)$$
 So the integral on the right becomes:
$$\int\limits_{\hat\Gamma}{h(y,\tau)dy}=\int\limits_{(u,v)\in D}{h(F^{-1}(\nu(u,v)))|\frac{\partial y}{\partial u}\times\frac{\partial y}{\partial v}|dudv}\\
=c^2t^2\int\limits_{(u,v)\in D}{h(x+ct\nu(u,v),\tau)|\frac{\partial\nu}{\partial u}\times\frac{\partial \nu}{\partial v}|dudv}\\
\overset{(1)}=c^2t^2\int\limits_{\Gamma}{h(x+ct\nu,\tau)d\nu}$$
Here, we used
$$\frac{\partial y}{\partial u}=ct\frac{\partial\nu}{\partial u}\\
\frac{\partial y}{\partial v}=ct\frac{\partial\nu}{\partial v}$$
$$\Rightarrow |\frac{\partial y}{\partial u}\times\frac{\partial y}{\partial v}|=c^2t^2|\frac{\partial \nu}{\partial u}\times\frac{\partial \nu}{\partial v}|$$
Remark: In the case of a general transformation (not translation and scaling) the harder thing is to express $$|\frac{\partial y}{\partial u}\times\frac{\partial y}{\partial v}|\text{  by  } |\frac{\partial \nu}{\partial u}\times\frac{\partial \nu}{\partial v}|$$
