I was given int a test the following exercise:
$\int \:\int \:\int _D\:dA$
Solid limited by:
$z=x^2+y^2$
and the plane:
$x+y=1$
And I am asked to calculate it in the first octant.
Is it possible to calculate the area with triple integrals, didn't he mean dV for volume? I am somehow confused.
I understand $D$ is a solid 'above' $XY$ plane and below the paraboloid (whose apex is the origin of coord system and the axis is $OZ$), truncated with three 'vertical' planes: $x=0$, $y=0$ and $x+y=1$.
In $XYZ$ we can describe it e.g. as $$D=\left\{(x,y,z):\ 0\le x\le1,\ 0\le y\le 1-x,\ 0\le z\le x^2+y^2\right\}$$ hence the integral can be written as $$\int_0^1 \int_0^{1-x} \int_0^{x^2+y^2} \mathrm dz\,\mathrm dy\,\mathrm dx$$
EDIT
To answer the question, the use of $\mathrm dA$ does not necessarily mean area, although ususally we use A for area and V for volume.
If the question displays a triple integral and talks about a solid and does not clearly require you to calculate the area, I would assume it's a typo in the expression and calculate the volume. It should be possible to add some short comment to the answer, e.g., 'a volume defined by the given triple integral is...' or, even bettter, 'assuming $\mathrm dA$ denotes a differential of volume, the volume of the solid is...'
The latter points out the ambiguity you find in the problem formulation and your interpretation. Of course the interpretation may be wrong, but the explanation shows it's not your misreading but an actual problem in the test itself.