<s>[INST] How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$? How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$ ? I can't even figure out what the integrand will be ( should it be $\sqrt{z^2-x^2}$ ?
 ) and not even the limits .  Please help . thanks in advance 
 [/INST] The slope of the plane $x+2z=3$ is smaller than the slope of the cone, so their intersection curve is an ellipse, drawn with brown in the figure. The plane $z=0$ passes through the apex. Therefore, the region we are interested in is a ''hat''. 
The projection of the surface region onto to the $x-y$ co-ordinate plane is another ellipse (the red one).

Between the two planes we have $0\le z\le \frac{3-x}2$, so the red ellipse is detemined by
$$
x^2+y^2\le\left(\frac{3-x}2\right)^2 \\
\frac{(x+1)^2}4 + \frac{y^2}3 \le 1.
$$
So, the area of the red ellipse is $2\sqrt3\pi$.
The slope of the cone is $1$, so the are of the two areas is $\sqrt2$. Therefore, the area of the conic region is $\sqrt2\cdot 2\sqrt3\pi=2\sqrt6\pi$.
(The question did not involve the area of the brown ellipse; just in order to verify the previous answer, from the slope of the plane  $z=\frac{3-x}2$ it is $\sqrt{1^2+(\frac12)^2}\cdot 2\sqrt3\pi=\sqrt{15}\pi$.)
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