Using a triple integral to find the volume of a solid bounded by $y=0, \;\; z=0, \;\; y=x, \;\; and \;\;z=4-x^2-y^2$ in the first octant.

Question

I'm having a hard time setting that integral up. Here's what I've done so far.

With these questions, it is essential to graph them: Here is z$=4-x^2-y^2$

Since it is in the first octant then only the top half and the section in quadrant I are needed, but it will be intersected by the $y=x$ plane. After making a bunch of graphs I come up with the following bounds: $$0\le x\le\sqrt2\;\;\;\;x\le y \le \sqrt{4-x^2}\;\;\;\; 0 \le z \le4-x^2-y^2$$ I think those are correct. Now, to find the volume I first integrate $1$ for $dz$ as shown: $$\int \int \int dzdydx=V$$, correct? If those bounds are correct, then I should be able to do the rest, that is, unless I should have used cylindrical or spherical coordinate systems.

Sorry for the lackluster shown work, but these problems are very hard for me.

I really do appreciate any help/confirmation on my values, thank you.

Answer 1

Though, this way is not generic, neither too formal.

But using principle of symmetry, by finding full volume in cylinderical coordinates in quadrant 1,2,3,4 then dividind volume by 8 should do it.

Total volume seems to be

$$ \int_0^{\pi/4}\int_0^4 \sqrt{4-z}d\theta dz $$

Answer being $ 4\pi/3 $

Answer 2

Why don't you try cylindrical coordinates?

$$V=\int _0^{\pi/4} \int _0^2 \int_0^{4-r^2} dz r dr d\theta $$