Compute the volume of the region bounded by the surfaces $z=0, \ y^2=x^3 \ and \ \sqrt{x} +z=2$.
I have previously computed similar questions, however there has been an integral already given, so I am confused as to how I would find the boundaries. I think a sketch would help, but difficult with it being 3d. Edit:
$\text{Limits of z: } 0 \leq z \leq 2-\sqrt{x} \\ z = 0 \ gives \ \sqrt{x} = 2, x = 4 \text{ on xy plane} \\ \text{hence }, 0 \leq y^2 \leq x^3 \leq 64 \\ 0 \leq y \leq 8 \\ \text{Then this gives the integral:} \\ V = \int_{y=0}^8\int_{x=y^{2/3}}^4\int_{z=0}^{2-\sqrt{x}}dzdxdy \\ \text{which solves to: }64/15$
Would this be correct? I feel like I am at least moving in the right direction with this answer?
Yes your edited work is correct except note that the volume is double of what you wrote.
$y^2 = x^3 = 64 \implies y = \pm 8$, so $-8 \leq y \leq 8$. Using symmetry above and below xz-plane, we can write it as
$ \displaystyle V = 2 \int_{ 0}^8\int_{y^{2/3}}^4\int_{0}^{2-\sqrt{x}}dz ~ dx ~ dy = \frac{128}{15}$
Alternatively, we can change the order of integration of $dy$ and $dx$.
$ \displaystyle V = 2 \int_{ 0}^4\int_0^{x^{3/2}}\int_{0}^{2-\sqrt{x}}dz ~ dy ~ dx = \frac{128}{15}$