In $\mathbb{R}^3$ I have to find the volume of the domain which lies inside both the cylinder of equation $x^2+y^2=1$ and the ellipsoid with equation $4x^2+4x^2+z^2=64$, where $(x,y,z)$ are Cartesian Coordinates in $\mathbb{R}^3$.
The cylinder and the ellipsoid are shown at the graph. The domain in question is a tube domain with the side surface being the surface of the cylinder, and the top and the bottom parts of its boundary being parts of the surface of the ellipsoid.
Approach:
First, I substitute with cylindrical coordinates as follows:
$r^2=x^2+y^2, \tan(θ)=y/x$ and $z=z$
From the equation of the cylinder I get that $r^2=1$ and from the equation of the ellipsoid that $z^2=64-4r^2 \Rightarrow z=\pm \sqrt{64-4r^2}$. Replacing with $r^2=1$ in the equation of the ellipsoid i get the solutions $z=\pm 2\sqrt{15}$.
Question:
In order to find the volume in question I am considering the triple integral(s) for $-1\leq r \leq 1$, $0 \leq θ \leq 2π$ and for $-2\sqrt{15} \leq z \leq 2\sqrt{15}$.
Are these boundaries correct?
Since $z=\pm \sqrt{64-4r^2}$, it is broken into two functions, namely $z= \sqrt{64-4r^2}$ for which the boundaries of $r$ and $θ$ will remain the same but the boundaries of $z$ would be $ 0 \leq z \leq 2\sqrt{15}$ and $z= -\sqrt{64-4r^2}$ for which the boundaries of $r$ and $θ$ will remain the same but the boundaries of $z$ would be $ -2\sqrt{15}\leq z \leq 0$. And thus I will calculate the triple integrals. Is this approach correct? Am I missing something?
Best Answer
The integral you're looking for is
$$\int_0^{2\pi} \int_0^1 \int_{-\sqrt{64-4r^2}}^{\sqrt{64-4r^2}} r \, dz \, dr \, d\theta$$
For any point $(x,y,z)$ in the region of interest, the $z$-coordinate is bounded above and below by the halves of an ellipsoid, $z=\pm\sqrt{64-4r^2}$. The bounds of $z$ cannot be constant, since the $z$-coordinate depends on both $x$ and $y$.