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?
 A: It's always a good idea to try and figure out how your shape looks, this helps with setting up the right integral bounds.

Note that calculating the volume using an integral can be written as:
$$ \int_{V}dV = \iiint 1\,dx\,dy\,dz$$
or
$$ \int_{A}f(A)\,dA = \iint f(x,y)\,dx\,dy$$
So using $\iiint z\,dz\,dy\,dx$ is not how you calculated the volume.
The volume bounded by these surfaces could be calculated as:
$$\int_{0}^{1}\int_{y/2}^{3}\int_{0}^{1-y^2} 1\,dz\,dx\,dy = \int_{0}^{1}\int_{y/2}^{3}{1-y^2 \,dx\,dy}   $$
Note that if you want to integrate $x$ at the end, you should split up the integrals as such:
$$\int_{0}^{0.5}{\int_{0}^{2x} 1-y^2}\,dy\,dx + \int_{0.5}^{3}{\int_{0}^{1} 1-y^2}\,dy\,dx$$
A: 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.
