volume between the planes $z = ax + by$, $z=0$ and the cylinder of radius 1 Find the volume between the planes $z = ax + by$, $z=0$ and the cylinder of radius 1, whose axis of symmetry is the z axis in the first octant $0≤x, 0≤y, 0≤z$.

I´m studying integral in 3 variables, but i stuck with this problem , i tried to solve this with doble integral i can't  found the correct integral to solve this. 

The solid lies above the region $D$ in the $xy$-plane bounded by  $x^{2} + y^{2} = r^{2}$, so the volume is given by the integral $$\int\int\limits_{D} f(x,y) \ dA = \int\limits_{-r}^{r}\int\limits_{-\sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}} f(x,y) \ dx dy$$ 

Its that correct? Can you help me please.
is there any restriction with $a$ and $b$? 

You need to find the volume in that region,

$\int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{0}^{ax+by}1dzdydx$.

When you have a triple integral with $f(x,y,z)=1$, 

$\int\int\int f(x,y,z)dzdydx$ it is the volume of the region where you are  trying to solve the triple integral.
is there any restriction with $a$ and $b$? 

You need to find the volume in that region,

$\int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{0}^{ax+by}1dzdydx$.

When you have a triple integral with $f(x,y,z)=1$, 

$\int\int\int f(x,y,z)dzdydx$ it is the volume of the region where you are  trying to solve the triple integral.