Find the surface area of the body bounded by the sphere
Find the surface area of the body bounded by the sphere $x^2+y^2+z^2=3a^2$ and the paraboloid $x^2+y^2=2az$ ($z>0$).
Explanation of Method
The two surfaces intersect in a circle at $z=a$, splitting the body's boundary into a lower paraboloid cap and an upper spherical cap, both projecting onto the same disk $x^2+y^2\le2a^2$. Each piece's surface-area integrand is radially symmetric, so each reduces to a single elementary radial integral, and the two areas are added.
- Find the intersection circle.
- Compute the paraboloid cap's area $S_1$.
- Compute the spherical cap's area $S_2$ and add.
Step 1 · Intersection
Substituting $x^2+y^2=2az$ into the sphere equation: $2az+z^2=3a^2\iff(z+3a)(z-a)=0$, so (for $z>0$) $z=a$, with $x^2+y^2=2a^2$ there — a circle of radius $a\sqrt2$.
Step 2 · Paraboloid cap $S_1$
With $z=\dfrac{x^2+y^2}{2a}$: $\sqrt{1+z_x^2+z_y^2}=\sqrt{1+r^2/a^2}$, over the disk $r\le a\sqrt2$:
$$ S_1 = 2\pi\int_0^{a\sqrt2}r\sqrt{1+\frac{r^2}{a^2}}\,dr = \pi a^2\left[\frac23u^{3/2}\right]_1^3 = \frac{2\pi a^2}3\big(3\sqrt3-1\big), $$
(with $u=1+r^2/a^2$).
Step 3 · Spherical cap $S_2$
With $z=\sqrt{3a^2-x^2-y^2}$: $\sqrt{1+z_x^2+z_y^2}=\dfrac{a\sqrt3}{\sqrt{3a^2-r^2}}$, over the same disk $r\le a\sqrt2$:
$$ S_2 = 2\pi a\sqrt3\int_0^{a\sqrt2}\frac{r\,dr}{\sqrt{3a^2-r^2}} = \pi a\sqrt3\Big[2\sqrt v\Big]_{a^2}^{3a^2} = 2\pi a^2\big(3-\sqrt3\big), $$
(with $v=3a^2-r^2$).
Step 4 · Total
$$ S = S_1+S_2 = \frac{2\pi a^2}3\big(3\sqrt3-1\big) + 2\pi a^2\big(3-\sqrt3\big) = \frac{2\pi a^2}3\Big[(3\sqrt3-1)+(9-3\sqrt3)\Big] = \frac{16\pi a^2}3. $$
Additional Notes
- This derivation matches the source exactly.
- Verified numerically (e.g. $a=1.3$): $S_1\approx14.8524$, $S_2\approx13.4638$, total $\approx28.3162$, matching $\dfrac{16\pi a^2}3$ exactly.
Result
The total surface area is: