The three regions in $R^3$ to consider are:
1) $R_1 = \{(x,y,z): x^2 + y^2 + z^2 \le 1\}$
2) $R_2 = \{(x,y,z): x^2 + y^2 \le a^2\}, 0<a<1$
3) $R_3 = \{(x,y,z): z\ge b(x^2 + y^2) \}, b>0$
I want to compute the volume of the common region.
The problem statement also says,
distinguish two cases:
$ba^2\le \sqrt{1-a^2}$
$ba^2 > \sqrt{1-a^2}$
My thoughts:
So, I think that the common region is the intersection of
1) $x^2 + y^2 \le 1 - z^2$
2) $x^2 + y^2 \le a^2, 0<a<1$
3)$x^2 + y^2 \le \frac{z}{b}, b>0$
So that $x^2 + y^2 \le min \{1-z^2, a^2, \frac{z}{b}\}$
But how can I actually compute the $min$?
After getting the $min$, then I think I can just go to cylindrical coordinates and integrate the 1-function. But what about the bounds for the height $z$?
Solving (1) for $z$ gives $z\le +/- \sqrt{1-(x^2+y^2)}$
And (3) gives that $z\ge b(x^2 + y^2)$
So...I'm not really sure about the limits of integration for $z$.
Any ideas are welcome.
Thanks,
Here's the first thing you should observe: the regions of interest are all rotationally symmetric about the $z$-axis, because with the cylindrical coordinate transformation $$(x,y,z) = (r \cos \theta, r \sin \theta, z),$$ we find $x^2 + y^2 = r^2$, therefore, all three equations are independent of the angle $\theta$. Hence it suffices to look at the $rz$-plane inequalities $$r^2 + z^2 \le 1, \quad r \le a, \quad z \ge br^2$$ in the first quadrant (i.e., $r, z \ge 0$). The first is bounded by a circle (red curve). The second region is bounded by a vertical line at a distance $a$ from the origin (green curve). The third is quite clearly bounded by a parabola through the origin (blue curve).
Now, the common region of intersection to these three inequalities comprises the region inside the circle, above the parabola, and to the left of the line. This requires a consideration of different cases, because if the parabola intersects the circle at a distance less than $a$, then the line is irrelevant. We solve for this condition by noting that if $r^2 + z^2 = 1$ and $z = br^2$, then the intersection occurs at a radius $\rho$ satisfying $$\rho^2 + b^2 \rho^4 = 1.$$ The volume integral is perhaps most easily evaluated by the method of cylindrical shells; i.e., $$V = \int_{r = 0}^R 2\pi r \left(\sqrt{1-r^2} - br^2\right) \, dr,$$ where $R = \min\{\rho,a\}$.