# Volume of region bounded by planes and parabolic cylinder

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.

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

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

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 $$z= 3-x^2$$.