Calculate the volume of the solid generated by rotating

Calculate the volume of the solid generated by rotating the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ about: 1° the $x$-axis, 2° the $y$-axis.

We apply the disk method for each axis and confirm consistency with the sphere formula when $a=b$.

Part 1: Rotation about the $x$-axis

From the ellipse, $y^2=\dfrac{b^2}{a^2}(a^2-x^2)$, so the disk at position $x\in[-a,a]$ has radius $r(x)=\dfrac ba\sqrt{a^2-x^2}$ and area $A(x)=\pi\dfrac{b^2}{a^2}(a^2-x^2)$.

\[V_x = \frac{\pi b^2}{a^2}\int_{-a}^{a}(a^2-x^2)\,dx = \frac{\pi b^2}{a^2}\cdot\frac{4a^3}{3} = \frac{4\pi ab^2}{3}\]

\[\boxed{V_x = \frac{4\pi ab^2}{3} \text{ cubic units}}\]

Part 2: Rotation about the $y$-axis

By the symmetric role of $x$ and $y$: $x^2=\dfrac{a^2}{b^2}(b^2-y^2)$, so the disk at $y\in[-b,b]$ has area $A(y)=\pi\dfrac{a^2}{b^2}(b^2-y^2)$.

\[V_y = \frac{\pi a^2}{b^2}\int_{-b}^{b}(b^2-y^2)\,dy = \frac{\pi a^2}{b^2}\cdot\frac{4b^3}{3} = \frac{4\pi a^2b}{3}\]

\[\boxed{V_y = \frac{4\pi a^2b}{3} \text{ cubic units}}\]

Step 3: Consistency check

When $a=b=r$ (the ellipse degenerates to a circle), both formulas give $V_x=V_y=\dfrac{4\pi r^3}{3}$, exactly the sphere volume. ✓

\[\frac{V_x}{V_y} = \frac{ab^2}{a^2b} = \frac ba\]

Summary

\[\boxed{V_x = \frac{4\pi ab^2}{3}\ (\text{about the } x\text{-axis}), \qquad V_y = \frac{4\pi a^2b}{3}\ (\text{about the } y\text{-axis})}\]

Key insight: Both formulas are special cases of the general ellipsoid volume $V=\frac{4\pi abc}{3}$ (semi-axes $a,b,c$), with the axis of revolution's own semi-axis appearing once and the perpendicular semi-axis appearing twice (squared).