Derive ellipse area, ellipsoid volume and sphere surface area

On the upper half, $y=(b/a)\sqrt{a^2-x^2}$, so $A=4\int_0^ay\,dx=(4b/a)\int_0^a\sqrt{a^2-x^2}\,dx$. With $x=a\sin\theta$, $dx=a\cos\theta\,d\theta$ and $0\le\theta\le\pi/2$, the radical is $a\cos\theta$. Thus $A=4ab\int_0^{\pi/2}\cos^2\theta\,d\theta=2ab[\theta+\tfrac12\sin2\theta]_0^{\pi/2}=\pi ab$. The integration variable after substitution is $\theta$; one source line mistakenly retains $dx$.

The entire ellipse region has both positive and negative y-values. Rotating that whole region through $\pi$ already sweeps the complete circular cross-sections, so the solid is the full ellipsoid, not half its volume. $V=\pi\int_{-a}^ab^2(1-x^2/a^2)\,dx=(2\pi b^2/a^2)[a^2x-x^3/3]_0^a=4\pi ab^2/3$.

Use the upper semicircle $y\ge0$ as the radius. For $-r<x<r$, $y^{\prime}=-x/y$, so $2\pi y\sqrt{1+x^2/y^2}=2\pi\sqrt{x^2+y^2}=2\pi r$. Integrating from $-r$ to $r$ gives $S=2\pi r(2r)=4\pi r^2$. The endpoint derivatives are unbounded, but the combined integrand has the constant limiting value $2\pi r$.

$y=\sqrt{r^2-x^2}$ and $y^{\prime}=-x/\sqrt{r^2-x^2}$. Then $2\pi\sqrt{r^2-x^2}\sqrt{1+x^2/(r^2-x^2)}=2\pi r$ in the interior, and the same integral gives $4\pi r^2$. Rotating only this upper semicircle once generates the sphere surface without counting it twice.