Surface Integral over a unit sphere

Question

How can I prove that $$ \int_{|z|=1} F(x+ t z) \cdot z d S_z = \frac{1}{t^2}\int_{|x-y|=t} F(y) \cdot n dS_y $$ Here, $z,x \in \mathbb R^3$ , $F : \mathbb R^3 \to \mathbb R^3 $, $t >0 $. Here why $1/t^2$ comes out? I think the substitution $$ y = x + t z, \;\; dS_y = t^2 d S_z $$ but here, why $dS_y = t^2 S_z$ ?

Answer 1

You may assume $x=0$. Then the left side of your equation is $$\int_{S_1} F(t\,z)\cdot z\ {\rm d}\omega(z)\ ,\qquad(1)$$ where $S_1$ is the $2$-dimensional unit sphere in ${\mathbb R}^3$ and ${\rm d}\omega$ denotes the euclidean surface element.

Let $S_t$ denote the sphere of radius $t>0$ centered at $0$. Then the unit normal at a point $y\in S_t$ is given by $n={y\over t}$, and the right side of your equation is $${1\over t^2}\int_{S_t} F(y)\cdot {y\over t}\ {\rm d}\omega(y)\ .$$ When we parametrize $S_t$ by means of $y:= t\, z$ $\ (z\in S_1)$ then it is geometrically obvious that ${\rm d}\omega(y)=t^2\,{\rm d}\omega(z)$. Therefore your right side becomes $$={1\over t^2}\int_{S_1} F(t z)\cdot z\ \ t^2\,{\rm d}\omega(z)\ ,$$ and after cancelling $t^2$ this is equal to $(1)$.

Answer 2

Yes, in the integral on the left hand let us introduce a new variable $y$ defined by $z=\tfrac{1}{t}(y-x).$
During this substitution we have that:

1. the domain of integration becomes $|y-x|=t;$
2. the integration form changes from $dS_z$ to $\tfrac{1}{t^2}dS_y;$ here there is the factor $\tfrac{1}{t^2}$ because it is the jacobian determinant for our change of variable $y\in (x+\tfrac{1}{t}S^2)\mapsto z=\tfrac{1}{t}(y-x)\in S^2;$
3. the integrating function $F(x+tz)\cdot z$ is replaced by $F(y)\cdot\tfrac{1}{t}(y-x)\equiv F(y)\cdot n_y,$ where $n_y$ is the normal versor to the surface of integration at $y$.