What does a triple integral represent in this problem? I need to determine whether the following statement is true or false:
For $a>0$, the region of integration of the triple iterated integral $$\int_{0}^{a}\int_{0}^{a}\int_{0}^{x}f(x,y) dzdydx$$ lies above a square in the xy-plane and below a plane
The answer is true because
$0\leq z \leq x$
$0 \leq y \leq a$
$0 \leq x \leq a$
So here the region of integration is below the plane $z=x$ and above the square $[0,a] \times [0,a]$
My question is: Is this true because this is a function $f(x,y)$ and not $f(x,y,z)$? So this integral 'collapses' into the double integral since z is a constant. Because I thought that a triple integral $$\iiint_{D}dV$$ represents the volume inside the domain and $$\iint_{R}f(x,y) dA$$ represents the volume under the surface but above the region $R$
So if the question asked about the integral
$$\int_{0}^{a}\int_{0}^{a}\int_{0}^{x}f(x,y,z) dzdydx$$
Then the answer would not be true because it would be something of the form $$\iint_{D} f(x,y,z) dV$$ which represents the mass of a solid with density function $f$?
Is this correct?
 A: The function is irrelvant.  The integral is $\int_0^a\int_0^a\int_0^x f dzdydx$.
The outermost itegral is with respect to x and goes from x= 0 to x= a.  While the outermost integral must always have constant limits, the next integral, with respect to y, may have functions of x as limits but here the limits are again constants, y= 0 to y= a.  The outer two integrals are over the square, in the xy plane, with vertices at (0,0), (a,0), (0,a) and (a,a).
The innermost integral, with respect to z, has limits z= 0 and z= x. In three dimensions those are planes. The region over which we are integrating lies above the plane z= 0, within the square in that plane, up to the plane z= x.
Again the function being integrated is irrelevant.  If it were a constant, not depending on any of x, y, or z, it would be that constant times the volume of that figure:
$\int_0^a\int_0^a\int_0^x C dzdydz= \int_0^a\int_0^a Cx dydx= \int_0^a aC xdx= \frac{aC}{2}a^2$
Or a function of, say, x, $x^3$, $\int_0^a\int_0^a\int_0^x x^3 dxdydz= \int_0^a\int_0^a x^4 dydx= \int_0^a ax^4dx= \frac{a}{5}a^5$.
Or a function of, say, x and y, xy^2, $\int_0^a\int_0^a\int_0^x xy^3 dxdydz= \int_0^a\int_0^a x^2y^3 dydx= \int_0^a \frac{1}{4} a^4 x^2 dx= \frac{1}{20}a^7$.
Or a functoin of all three variables, say x^2ye^z: $\int_0^a\int_0^a\int_0^z x^2ye^z dxdydz= \int_0^a\int_0^a  x^2ye^x dxdy= \frac{a^2}{2}\int_0^a x^2e^x dx= a^2e^a- ae^a- e^a+ 1$.
