The Change of Variables in Surface Integral Actually, I can't understand the question. In the hint they say first to pick a coordinate with z1 perpendicular to x+y+z=t . Why do we have to change the coordinate with (x1,y1,z1)? I mean can we directly calculate without any change? Can someone give me a brief proof of this? 

 A: First, we write the plane $P:x+y+z = t$ with unit normal vector. This gives $P:\frac{x}{\sqrt3} + \frac{y}{\sqrt3}+\frac{z}{\sqrt3} = \frac{t}{\sqrt3}.$ Let's denote the unit normal vector of this plane by $u$, so $u = (\frac{1}{\sqrt3},\frac{1}{\sqrt3},\frac{1}{\sqrt3}).$
Now we pick a new orthogonal coordinate system with coordinates $(x_1,y_1,z_1)$ where $u$ is the direction of the positive $z_1$-axis. Such a coordinate system exists. There are lots of choices. Alternatively, you can think about it as follows. Find a rotation $R$ that rotates $u$ into the positive $z$-axis, i.e. $Ru = (0,0,1).$ Again, such a rotation exists, and there are lots of choices. Since $R$ is orthogonal, it preserves all the Euclidean structure. From this, it follows that $R$ also preserves surface area, volume, and measure. Moreover, $R$ maps $S^1$ onto itself, where $S^1$ denotes the unit sphere. Also, $R$ preserves the function $\Phi,$ i.e. $\Phi(v) = \Phi(Rv)$ for any vector $v.$ That's because $x^2+y^2+z^2$ is just $\|v\|^2,$ which is preserved by $R.$ Finally, $R$ maps the plane $P$ to the plane $Q:z = \frac{t}{\sqrt3}.$ All this shows that, by applying $R$, or equivalently changing coordinates in an orthogonal manner, we have transformed the original problem into something much more amenable - without doing any cumbersome calculations :-). Namely, if we denote by $K$ the part of $Q$ inside of $S^1,$ Then we have $R(S) = K$ and moreover
$$
\int\int_S\Phi(x,y,z)dS = \int\int_K\Phi(x,y,z)dK.
$$
From the equation for $Q$ it's easy to see that
$$
K = \left\{(x,y,z)\ |\ z = \frac{t}{\sqrt3},\ x^2+y^2 \leq 1 - \frac{t^2}{3} \right\} \quad for\ |t| \leq \sqrt3
$$
and $K = \emptyset$ for $|t| > \sqrt3.$ That latter case corresponds exactly to the result $0$ in your question and is uninteresting, so let's assume $|t| \leq \sqrt3$ from now on. Then we can express $x$ and $y$ in polar coordinates
$$
x = r \cos\psi\quad and \quad y = r\sin\psi
$$
and get
$$
K = \left\{(r,\psi,z)\ |\ z = \frac{1}{\sqrt3},\ 0 \leq r \leq \sqrt{1 - \frac{t^2}{3}},\ 0 \leq \psi < 2\pi \right\}.
$$
Moreover we have
$$
\Phi(x,y,z) = 1-x^2-y^2-z^2 = 1-\frac{t^2}{3}-r^2,
$$
and finally recall the "well known" differential transformation
$$
dK = dxdy = rdrd\psi.
$$
With all this, we calculate
$$
\begin{align}
\int\int_K\Phi(x,y,z)dK & = \int_0^{2\pi}\int_0^{\sqrt{1-\frac{t^2}{3}}}(1-\frac{t^2}{3}-r^2)rdrd\psi \\
& = 2\pi\left(\frac{1}{2}\left(1-\frac{t^2}{3}\right)r^2 - \frac{1}{4}r^4\right)_0^{\sqrt{1-\frac{t^2}{3}}} \\
& = 2\pi\frac{1}{4}\left(1-\frac{t^2}{3}\right)^2 \\
& = \frac{\pi}{18}\left(3-t^2\right)^2,
\end{align}
$$
as desired.
