Find the volume of the solid in the first octant bounded by the three surfaces $z = 1-y^2$, $y=2x$, and $x=3$ I want to find the volume of the solid in the first octant bounded by the three surfaces $z = 1-y^2$, $y=2x$, and $x=3$. It seems that would simply be to calculate the following triple integral:
$\int_0^3 \int_0^{2x} \int_0^{1-y^2} z\,dz\,dy\,dx$
This is pretty straight-forward to do without any variable substitutions etc. which makes me think it's almost too simple (for a home assignment).
Am I missing something or is the above correct?

I ended up doing the following (omitting some algebra steps):
$1 - y^2 \geq 0$ when $y \leq 1$ which gives an integration interval $0 \leq y \leq 1$. By dividing the integral into two parts with the intervals $0 \leq x \leq \frac{1}{2}$ and $\frac{1}{2} \leq x \leq 3$ we get the two simpler integrals:
$V =
V_1 + V_2 =
\int_0^{1/2} \int_0^{2x} \int_0^{1-y^2} dz\,dy\,dx +
\int_{1/2}^3 \int_0^1 \int_0^{1-y^2} dz\,dy\,dx$
$V_1 = \int_0^{1/2} \int_0^{2x} \int_0^{1-y^2} dz\,dy\,dx =
\int_0^{1/2} \int_0^{2x} (1-y^2)\,dy\,dx =
\int_0^{1/2} \left(2x - \frac{8}{3}x^3\right) dx = 
\frac{5}{24}$
$V_2 = \int_{1/2}^3 \int_0^1 \int_0^{1-y^2} dz\,dy\,dx =
\int_{1/2}^3 \int_0^1 (1-y^2) dy\,dx =
\int_{1/2}^3 \frac{2}{3} dx =
\frac{5}{3}$
$V = V_1 + V_2 = \frac{5}{24} + \frac{5}{3} = \frac{15}{8}$
Makes sense? Thank you for the comments.
