Finding the area of the surface of a planar region Find the area of the surface bounded by the region $x+y+z=3a/2$, within the bounds of  $0≤x,y,z≤a$
What i tried
I first take the unit normal vector $n=1/3<1,1,1>$.I believe i must do an integral where the bounds are $x+y=a/2$ and $x+y=0$. However im unsure of how to proceed from here. Could anyone explain. Thanks
 A: A simple approach that does not require integration: since $0\leq x,y,z \leq a$, the area is by definition included in a single octant. The equation is of the form $x+y+z=k$, which implies a planar surface. The intersections $A$, $B$, $C$ of the whole plane (not considering the bounds) with the three axes are given by points with coordinates $(\frac{3a}{2},0,0)$, $(0,\frac{3a}{2},0)$, and $(0,0,\frac{3a}{2})$. Also, the intersections with the planes $xy$, $xz$, and $yz$ are given by the lines that connect these points, with $45^o$ degree slopes. So, our area of interest is included in the triangle $ABC$, which is equilateral and whose side is $\frac{3\sqrt{2}a}{2}$.
Knowing this, and taking into account the bounds of $x,y,z$, it is not difficult to identify the resulting regular shape in the central portion of triangle $ABC$. To visualize it, simply trace, in the equilateral triangle, the three lines parallel to each side and whose distance from the parallel side is double than that from the corresponding opposite vertex (this corresponds to applying the bounds $0\leq x,y,z \leq a$ in our triangle). The area of the regular polygon obtained in this way can be easily calculated.
A: Hint: project the surface on the plane XY. Let be $D$ this projection. Parametrize the surface:
$$\Phi:D\longrightarrow\Bbb R^3$$
$$\Phi(x,y,z)=(x,y,?)$$
(what will be $z$?)
an apply the formula for the area of a parametrized surface.
