Proof of polar coordinates theorem in Evans' PDE Book Evans used this theorem '4' to derive the fundamental solutions of Poisson's equation.
in the appendix it says that this is a special case of a more general theorem called 'Coarea Formula'. I found proofs of that one but they involve some advanced level Geometry that I don't have good command of.
Is there a way to prove this theorem '4'just by using multivariable calculus?
And I think I know the intuition behind this, it's like a generalized disk method, right?
In dimensions over than 3 it's quite hard to imagine.

 A: Part (i) is essentially integration in spherical coordinates.
If $n=2$, spherical coordinates are called polar coordinates.
In general, a point $x\in\mathbb R^n\setminus0$ can be represented uniquely as $x=r\omega$, where $r>0$ and $\omega\in S^{n-1}$.
These two "coordinates" $(r,\omega)$ are found simply as $r=|x|$ and $\omega=x/|x|$.
These are the spherical coordinates.
You can think of the theorem as a generalization of integration in polar coordinates in the plane, and the proof is morally the same.
This identification gives a bijection $\phi\colon\mathbb R^n\setminus0\to(0,\infty)\times S^{n-1}$.
Instead of integrating over $\mathbb R^n\setminus0$, you can use your multidimensional change of variables formula to get
$$
\int_{\mathbb R^n\setminus0}f(x)dx
=
\int_0^\infty\left(
\int_{S^{n-1}}f(r\omega)dS(\omega)
\right)r^{n-1}dr,
$$
where $S$ is the usual measure on the sphere $S^{n-1}=\partial B(0,1)$.
The usual change of variables formulas hold for open sets of Euclidean spaces, and the sphere is not one.
To make the proof technically accurate does require some effort, and a single answer would be too little to cover it.
However, it should be possible to reason heuristically what the Jacobian determinant (the stretch factor) should be.
I assume Evans actually defines the integral of a continuous function over a sphere in his book.
If you look up the definition, it should be clear that
$$
r^{n-1}\int_{S^{n-1}}f(r\omega)dS(\omega)
=
\int_{\partial B(0,r)}f(x)dS.
$$
I don't want to give a definition of the integral here, as it might well be different from yours and therefore confusing.
Work with the conventions of your book.
In addition, you can shift the origin freely by a linear change of coordinates, and you can exclude a single point without changing the value of the integral.
For part (ii), the argument of (i) can be used to show that for any $R>0$
$$
\int_{B(0,R)\setminus0}f(x)dx
=
\int_0^R
\left(
\int_{\partial B(0,r)}f(r\omega)dS(\omega)
\right)
dr.
$$
Now you can take the derivative with respect to $R$.
On the left-hand side you don't need to do anything, and on the right-hand side you need to apply the fundamental theorem of calculus to the outer integral.
