Find the volume of the solid bounded by the cone: frac
Find the volume of the solid bounded by the cone: $$ \frac{x^2+y^2}2=(z-\sqrt2)^2 $$ and the ellipsoid: $$ \frac{x^2+y^2}6+\frac{z^2}2=1. $$
Explanation of Method
Writing both surfaces in terms of $r^2=x^2+y^2$ as functions of $z$, the cone has vertex $(0,0,\sqrt2)$ and $r^2=2(z-\sqrt2)^2$, while the ellipsoid gives $r^2=6-3z^2$. Solving for their intersection heights and slicing the solid perpendicular to the $z$-axis (cylindrical shells) reduces the volume to a single integral in $z$.
- Find the intersection heights by equating the two expressions for $r^2$.
- Determine which surface is the outer boundary between those heights.
- Set up and evaluate the volume as a single integral in $z$.
Step 1 · Find the intersection heights
Setting $2(z-\sqrt2)^2=6-3z^2$: $2z^2-4\sqrt2z+4=6-3z^2\Rightarrow5z^2-4\sqrt2z-2=0$. By the quadratic formula:
$$ z=\frac{4\sqrt2\pm\sqrt{32+40}}{10}=\frac{4\sqrt2\pm6\sqrt2}{10}, $$
giving $z=\sqrt2$ (the cone's vertex, where $r^2=0$) or $z=-\dfrac{\sqrt2}5$ (where $r^2=6-3\cdot\frac25=\frac{144}{25}$).
Step 2 · Identify the outer surface
At $z=0$: the cone gives $r^2=2(2)=4$, the ellipsoid gives $r^2=6$. So on $\left[-\frac{\sqrt2}5,\sqrt2\right]$ the ellipsoid is the outer boundary and the cone is the inner boundary.
Step 3 · Set up the volume integral
$$ V = 2\pi\int_{-\sqrt2/5}^{\sqrt2}\int_{\sqrt2(\sqrt2-z)}^{\sqrt{6-3z^2}}r\,dr\,dz = \pi\int_{-\sqrt2/5}^{\sqrt2}\left[(6-3z^2)-2(\sqrt2-z)^2\right]dz. $$
Expanding: $(6-3z^2)-2(2-2\sqrt2z+z^2)=2+4\sqrt2z-5z^2$.
Step 4 · Evaluate
$$ V = \pi\left[2z+2\sqrt2z^2-\frac{5z^3}3\right]_{-\sqrt2/5}^{\sqrt2}. $$
At $z=\sqrt2$: the bracket equals $\dfrac{8\sqrt2}3$. At $z=-\dfrac{\sqrt2}5$: the bracket equals $-\dfrac{16\sqrt2}{75}$. So:
$$ V = \pi\left(\frac{8\sqrt2}3+\frac{16\sqrt2}{75}\right) = \pi\sqrt2\cdot\frac{200+16}{75} = \frac{216\pi\sqrt2}{75} = \frac{72\pi\sqrt2}{25}. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically: direct evaluation of $\pi\int_{-\sqrt2/5}^{\sqrt2}\left[2+4\sqrt2z-5z^2\right]dz$ gives exactly $\dfrac{72\pi\sqrt2}{25}$.
Result
The volume is: