About a surface integral of type $1$ 
Compute $\displaystyle F(t)=\iint\limits_{s}f(x,y,z)\ ds$, where $s$ is a surface $x+y+z=t$, and $f(x,y,z)=1-x^2-y^2-z^2$ when $1-x^2-y^2-z^2 \ge 0$,
  else $f=0$.

Here is my attempt:
I think I can rotate the space rectangular coordinate system, let the $x+y+z=t$ be the new $xy$-plane, may be easy to compute, but I don’t know how to determine the new coordinates when changing the variables $x$, $y$ and $z$ to some new $u$, $v$ and $w$.
 A: The integral simplifies with the coordinates you propose. It simplifies even more because the form of the integrand does not change in the new coordinates and, further! the integral in spherical coordinates is very simple.
Following your idea, the vector normal and unitary to the plane is $n_w=\dfrac{1}{\sqrt{3}}(1,1,1)$ we can choose two other vectors being orthogonal to $n_w$ and then use them to determine planes to build the new coordinates. $n_u=\dfrac{1}{\sqrt{6}}(2,-1,-1)$ and $n_v=\dfrac{1}{\sqrt{2}}(0,1,-1)$ will do the job. The change of coordinates is then 
$$
    \begin{matrix}
    u=\dfrac{1}{\sqrt{6}}(2x-y-z) \\
    v=\dfrac{1}{\sqrt{2}}(y-z) \\
    w=\dfrac{1}{\sqrt{3}}(x+y+z) \\
    \end{matrix}
$$
As the vectors are unitary and orthogonal, the jacobian is an unitary matrix. The surface of integration is into the plane $x+y+z=\sqrt{3}w=t$ and the surface element is $\mathbb ds=\mathbb du\mathbb dv$. For the integrand, it's easy to check that $1-x^2-y^2-z^2=1-u^2-v^2-w^2$. So,
$$\iint_S(1-u^2-v^2-t^2/3)\mathbb du\mathbb dv$$
Added
Considering the integrand and the surface symmetry, it get simpler changing to cylindrical coordinates, with $u>0$ as polar axis. The surface is a circle into the plane with $w=t/\sqrt{3}$ (it's the intersection of the sphere of radius $1$ with this plane). The radius of this circle is $\rho(t)=\sqrt{1-w(t)^2}=\sqrt{1-t^2/3}$. The surface element is $\rho\,\mathbb d\rho\,\mathbb d\theta$ and $\rho^2=u^2+v^2$:
$$I(t)=\iint_S(1-u^2-v^2-t^2/3)\mathbb du\,\mathbb dv=$$
$$=\int_0^{2\pi}\int_0^{\sqrt{1-t^2/3}}(1-\rho^2-t^2/3)\rho\,\mathbb d\rho\,\mathbb d\theta=$$
$$=2\pi(1-t^2/3)\dfrac{\left(\sqrt{1-t^2/3}\right)^2}{2}-2\pi\dfrac{\left(\sqrt{1-t^2/3}\right)^4}{4}=$$
$$=\left\{\begin{matrix}
   0&\text{if}&t<-\sqrt{3}\\
   2\pi(1-t^2/3)^2/4&\text{if}&-\sqrt{3}\leq t\leq\sqrt{3}\\
   0&\text{if}&\sqrt{3}<t\\
\end{matrix}\right.$$
