Volume of region bounded by planes and parabolic cylinder

Question

Find the volume of the region in the first octant enclosed by the planes $x=0$, $z=0$, $y=0$, $y=2$ and the parabolic cylinder $z=3-x^2$

I found the region to be bounded by $0≤y≤2$,$0≤x≤\sqrt{3-z}$,$0≤z≤3-x^2$

However, when i put the y bounds (constants) on the outside, the other two bound are not dependent on y. I don't understand what to do.

Answer 1

All depends upon the way you "slice" the domain:

• if we consider $0≤y≤2$ in $x-z$ plane $0≤x≤\sqrt{3}$ and we have

$$0≤y≤2\,,\quad 0≤x≤\sqrt{3}\,,\quad 0≤z≤3-x^2$$

• if we consider $0≤z≤3$ in $x-y$ plane $0≤y≤2$ and we have

$$0≤z≤3\,,\quad 0≤y≤2\,,\quad 0≤x≤\sqrt{3-z} $$

• if we consider $0≤x≤\sqrt{3}$ in $y-z$ plane $0≤y≤2$ and we have

$$0≤x≤\sqrt{3}\,,\quad 0≤y≤2\,,\quad 0≤z≤3-x^2 $$

Answer 2

You are correct in observing that the bounds of $x$ and $z$ do not depend on $y$, because along the $y$ direction, the cross section of the parabolic cylinder, given by $z= 3-x^2$, remains the same regardless $y$.

So, the volume integral is effectively a double integral over $x$ and $z$,

$$V= 2\int_0^3\int_0^{\sqrt{3-z}}dxdz $$

Note that the lower and upper bounds of $z$ is 0 and 3, respectively. The upper bound 3 is obtained by setting $x=0$ in $z= 3-x^2$.