Find the surface area of the paraboloid of revolution
Find the surface area of the paraboloid of revolution $y^2+z^2=2px$ between the vertex and the plane $x=\dfrac p2$.
Explanation of Method
This surface is generated by rotating the parabola $z=\sqrt{2px}$ (in the $xz$-plane) about the $x$-axis, so its area follows directly from the surface-of-revolution formula $S=2\pi\int f(x)\sqrt{1+[f'(x)]^2}\,dx$, which reduces here to a single elementary radical integral.
- Apply the surface-of-revolution formula to $f(x)=\sqrt{2px}$.
- Simplify the integrand.
- Evaluate via $u=2x+p$.
Step 1 · Surface-of-revolution formula
With $f(x)=\sqrt{2px}$: $f'(x)=\sqrt{\dfrac p{2x}}$, so $1+[f'(x)]^2=\dfrac{2x+p}{2x}$.
Step 2 · Simplify the integrand
$$ f(x)\sqrt{1+[f'(x)]^2} = \sqrt{2px}\cdot\sqrt{\frac{2x+p}{2x}} = \sqrt{p(2x+p)}. $$
$$ S = 2\pi\int_0^{p/2}\sqrt{p(2x+p)}\,dx = 2\pi\sqrt p\int_0^{p/2}\sqrt{2x+p}\,dx. $$
Step 3 · Evaluate
With $u=2x+p$ ($u:p\to2p$):
$$ S = \pi\sqrt p\int_p^{2p}\sqrt u\,du = \pi\sqrt p\left[\frac23u^{3/2}\right]_p^{2p} = \frac{2\pi\sqrt p}3\Big[(2p)^{3/2}-p^{3/2}\Big] = \frac{2\pi p^2\big(2\sqrt2-1\big)}3. $$
Additional Notes
- This derivation matches the source exactly.
- Verified symbolically via sympy: the exact integral evaluates to $\dfrac{2\pi p^2(2\sqrt2-1)}3$, confirming the source's closed form with no adjustment.
Result
The surface area is: