Find the surface area of the smaller body bounded
Find the surface area of the smaller body bounded by the surfaces $x^2+y^2+z^2=2$, $x^2+y^2+z^2=6$, and $y^2+z^2=x$.
Explanation of Method
The paraboloid meets the inner sphere at $x=1$ (circle of radius $1$) and the outer sphere at $x=2$ (circle of radius $\sqrt2$). The bounded solid these three surfaces enclose is a "corked paraboloid segment": a spherical cap of the inner sphere near its far pole, a spherical cap of the outer sphere near its far pole, and the paraboloid's lateral wall joining them for $1\le x\le2$. Each of the three pieces reduces to a single elementary integral — the two caps via the standard spherical-cap area formula, the paraboloid strip via a polar radial integral.
- Find the two intersection circles.
- Compute the two spherical-cap areas.
- Compute the paraboloid strip's area and sum all three.
Step 1 · Intersections
Substituting $y^2+z^2=x$ into each sphere: $x^2+x=2\Rightarrow x=1$ (inner sphere, circle of radius $1$), and $x^2+x=6\Rightarrow x=2$ (outer sphere, circle of radius $\sqrt2$). The bounded solid is the part of the region between the two spheres that lies on the "inside" of the paraboloid, i.e. its boundary consists of the cap of the inner sphere with $x\ge1$, the cap of the outer sphere with $x\ge2$, and the paraboloid wall for $1\le x\le2$.
Step 2 · Spherical caps
On the inner sphere ($R_1=\sqrt2$), using $x=R_1\cos\varphi$: at $x=1$, $\cos\varphi_1=\frac1{\sqrt2}\Rightarrow\varphi_1=\frac\pi4$, so with $A=2\pi R^2(1-\cos\varphi)$:
$$ A_1 = 2\pi(2)\left(1-\frac1{\sqrt2}\right) = 2\pi\big(2-\sqrt2\big). $$
On the outer sphere ($R_2=\sqrt6$), at $x=2$: $\cos\varphi_2=\dfrac2{\sqrt6}$, so:
$$ A_2 = 2\pi(6)\left(1-\frac2{\sqrt6}\right) = 2\pi\big(6-2\sqrt6\big). $$
Step 3 · Paraboloid strip and total
With $x=\rho^2$ ($\rho=\sqrt{y^2+z^2}$), the surface element is $\sqrt{1+4\rho^2}\,\rho\,d\rho\,d\theta$, over $\rho\in[1,\sqrt2]$:
$$ A_3 = 2\pi\int_1^{\sqrt2}\rho\sqrt{1+4\rho^2}\,d\rho = \frac\pi6\big(27-5\sqrt5\big) $$
(via $u=1+4\rho^2$). Summing all three pieces:
$$ P = A_1+A_2+A_3 = \frac{\pi\big(123-12\sqrt2-24\sqrt6-5\sqrt5\big)}6. $$
Additional Notes
- This derivation matches the source's computation exactly. (The source's own prose description of which cap is "inside" vs. "outside" the paraboloid was self-contradictory between its Step 1 and Step 2, but the actual $\varphi$-ranges used in the formulas — $\varphi\in[0,\pi/4]$ for the inner cap, $\varphi\in[0,\varphi_2]$ for the outer cap — correctly identify the caps needed to close the bounded solid, as confirmed numerically below.)
- Verified numerically: $A_1\approx3.6806$, $A_2\approx6.9179$, $A_3\approx8.2832$, total $\approx18.8817$, matching the boxed closed form exactly.
Result
The total surface area is: