Calculate the surface area generated by rotating

Calculate the surface area generated by rotating the curve $y=\arcsin\sqrt{1-x^2}+\dfrac\pi2$ about the $x$-axis.

Step 1: Domain and symmetry

$x\in[-1,1]$; at $x=0$, $y=\pi$ (maximum); at $x=\pm1$, $y=\frac\pi2$. The curve is symmetric about $x=0$.

Step 2: Differentiate

\[\frac{dy}{dx}=\frac1{\sqrt{1-(1-x^2)}}\cdot\frac{-x}{\sqrt{1-x^2}} = -\frac{x}{|x|\sqrt{1-x^2}}\]

so $\left(\dfrac{dy}{dx}\right)^2=\dfrac1{1-x^2}$ regardless of sign, giving

\[1+(y')^2=\frac{2-x^2}{1-x^2}\]

Step 3: Substitute $x=\sin\theta$

Then $\sqrt{1-x^2}=\cos\theta$ and $\arcsin\sqrt{1-x^2}=\arcsin(\cos\theta)=\frac\pi2-\theta$ for $\theta\in[0,\pi/2]$, while $\sqrt{\frac{2-x^2}{1-x^2}}\,dx$ becomes $\sqrt{1+\cos^2\theta}\,d\theta$ (the $\cos\theta$ factors from $dx$ and the square root cancel).

Step 4: Rewrite the integral

By symmetry, integrate $\theta\in[0,\pi/2]$ (covering $x\in[0,1]$) and double:

\[S=2\cdot2\pi\int_0^{\pi/2}(\pi-\theta)\sqrt{1+\cos^2\theta}\,d\theta = 4\pi\left[\pi\int_0^{\pi/2}\sqrt{1+\cos^2\theta}\,d\theta - \int_0^{\pi/2}\theta\sqrt{1+\cos^2\theta}\,d\theta\right]\]

Step 5: Recognize the elliptic integral

Neither integral has an elementary closed form — the first is (a multiple of) the complete elliptic integral of the second kind, $E\!\left(\frac1{\sqrt2}\right)$. Both were evaluated by high-precision numerical integration.

Step 6: Numerical result

Numerically, $\displaystyle\int_0^{\pi/2}(\pi-\theta)\sqrt{1+\cos^2\theta}\,d\theta\approx4.6038$, so

\[S\approx4\pi(4.6038)\approx57.85\]

\[\boxed{S\approx57.85\text{ square units}}\]

(cross-checked by direct numerical integration of the original $x$-domain integral, which agrees to five significant figures.)