Given the ellipsoid (x²)/(a²)+ (y²)/(b²)+ (z²)/(c²)=1
Given the ellipsoid $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$ and the paraboloid: $$ \frac{y^2}{b^2}+\frac{z^2}{c^2}=\frac{x-\lambda}{\mu-\lambda}\cdot\frac{a^2-\mu^2}{a^2}, $$ where $-a<\lambda<\mu<a$, show that the paraboloid divides the ellipsoid into two parts, and determine $\lambda$ (in terms of $a$ and $\mu$) such that the two parts have equal volume.
Explanation of Method
Both surfaces have elliptical cross-sections at each $x$, so the problem reduces to comparing two cross-sectional-area functions of $x$. At $x=\mu$, the paraboloid's cross-section exactly matches the ellipsoid's — the paraboloid is constructed to meet the ellipsoid precisely along the ellipse at $x=\mu$. For $x\in[\lambda,\mu]$ the paraboloid's (linear-in-$x$) cross-section lies inside the ellipsoid's (quadratic) cross-section, so the paraboloid surface together with the ellipsoid's own surface for $x\geqslant\mu$ encloses a cap-shaped sub-solid. Equating that cap's volume to half the ellipsoid's volume gives an equation for $\lambda$ in terms of $\mu$.
- Identify the cross-sectional area functions of both surfaces and confirm they agree at $x=\mu$.
- Set up the "cap" volume enclosed by the paraboloid ($x\in[\lambda,\mu]$) and the ellipsoid ($x\in[\mu,a]$).
- Solve for $\lambda$ by setting this cap volume equal to half the ellipsoid's volume.
Step 1 · Cross-sectional areas
The ellipsoid's cross-section at $x$ is an ellipse of area $A_e(x)=\pi bc\left(1-\dfrac{x^2}{a^2}\right)$. The paraboloid's cross-section (for $x\geqslant\lambda$) is an ellipse of area $A_p(x)=\pi bc\cdot\dfrac{x-\lambda}{\mu-\lambda}\cdot\dfrac{a^2-\mu^2}{a^2}$. At $x=\mu$: $A_p(\mu)=\pi bc\cdot\dfrac{a^2-\mu^2}{a^2}=A_e(\mu)$ — the two surfaces meet exactly at $x=\mu$, confirming the paraboloid genuinely divides the ellipsoid (it is nested inside for $x\in[\lambda,\mu]$ and coincides with it at $x=\mu$).
Step 2 · Set up the cap volume
The solid enclosed by the paraboloid (for $\lambda\leqslant x\leqslant\mu$) and the ellipsoid's far cap (for $\mu\leqslant x\leqslant a$) has volume:
$$ V_{\text{cap}} = \pi bc\left[\int_\lambda^\mu\frac{x-\lambda}{\mu-\lambda}\cdot\frac{a^2-\mu^2}{a^2}\,dx + \int_\mu^a\left(1-\frac{x^2}{a^2}\right)dx\right]. $$
Evaluating both integrals and simplifying:
$$ V_{\text{cap}} = \pi bc\left[\frac{2a}3-\frac\lambda2-\frac\mu2+\frac{\lambda\mu^2}{2a^2}-\frac{\mu^3}{6a^2}\right]. $$
Step 3 · Solve for $\lambda$
The full ellipsoid volume is $\dfrac43\pi abc$, so half is $\dfrac23\pi abc$. Setting $V_{\text{cap}}=\dfrac23\pi abc$ and solving the resulting linear equation for $\lambda$:
$$ \lambda = -\frac{\mu\big(3a^2+\mu^2\big)}{3\big(a^2-\mu^2\big)}. $$
Additional Notes
- A note on the source computation: the source's own derivation was internally inconsistent — its algebra broke down partway through Step 5 ("Through detailed calculation..."), and it then arbitrarily fixed $\mu=0$ (an unjustified restriction not implied by the problem, since $\mu$ is a free parameter satisfying $-a<\lambda<\mu<a$), producing first the impossible $\lambda=-\frac32a$ (violating $\lambda>-a$), then an unrelated final guess $\lambda=a\left(2^{-2/3}-1\right)\approx-0.37a$ with no valid derivation connecting it to the setup.
- The corrected formula above, $\lambda=-\dfrac{\mu(3a^2+\mu^2)}{3(a^2-\mu^2)}$, was verified by direct numerical integration for multiple values of $\mu$ (e.g. $\mu=0.5a$ and $\mu=0.2a$, with $a=1$): in both cases, the resulting cap volume matches exactly half the ellipsoid volume, and $\lambda$ correctly satisfies $-a<\lambda<\mu<a$.
Result
The value of $\lambda$ that splits the ellipsoid into two equal volumes is: