Problem 126
Medium Physical Applications

Calculate the surface area generated by rotating

Calculate the surface area generated by rotating the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ (with $a>b$) about: a) the $x$-axis (prolate spheroid), b) the $y$-axis (oblate spheroid).

Explanation of Method

Differentiate the upper branch for each axis, express $1+(y')^2$ (or $1+(x')^2$) in terms of the eccentricity $e=\sqrt{1-b^2/a^2}$, and integrate using a trigonometric substitution for part (a) and a direct antiderivative for part (b), producing an $\arcsin$ term in one case and a logarithm in the other.

Part a: About the $x$-axis (prolate spheroid)

Step 1: Set up

$y=\frac ba\sqrt{a^2-x^2}$, so $y'=-\dfrac{bx}{a\sqrt{a^2-x^2}}$, and with $e^2=\dfrac{a^2-b^2}{a^2}$:

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

Step 2: Set up the integral

By symmetry, \[S_x=4\pi\int_0^a\frac{b}{a}\sqrt{a^2-x^2}\cdot\frac{\sqrt{a^2-e^2x^2}}{\sqrt{a^2-x^2}}\,dx = \frac{4\pi b}a\int_0^a\sqrt{a^2-e^2x^2}\,dx\]

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

\[\int_0^a\sqrt{a^2-e^2x^2}\,dx=\frac{a^2}{2e}\left[\arcsin e+e\sqrt{1-e^2}\right]=\frac{a^2}{2e}\left[\arcsin e+\frac{eb}a\right]\]

Step 4: Combine

\[S_x=\frac{4\pi b}a\cdot\frac{a^2}{2e}\left[\arcsin e+\frac{eb}a\right]=2\pi b^2+\frac{2\pi ab}e\arcsin e\]

\[\boxed{S_x=2\pi b^2+\frac{2\pi ab}e\arcsin e}\]

Part b: About the $y$-axis (oblate spheroid)

Step 1: Set up

$x=\frac ab\sqrt{b^2-y^2}$, so $x'=-\dfrac{ay}{b\sqrt{b^2-y^2}}$, and using $a^2-b^2=a^2e^2$:

\[1+(x')^2=\frac{b^4+a^2e^2y^2}{b^2(b^2-y^2)} \implies \sqrt{1+(x')^2}=\frac{\sqrt{b^4+a^2e^2y^2}}{b\sqrt{b^2-y^2}}\]

Step 2: Set up the integral

\[S_y=\frac{4\pi a}{b^2}\int_0^b\sqrt{b^4+a^2e^2y^2}\,dy\]

Step 3: Evaluate with the standard formula

Using $\int\sqrt{A^2+B^2y^2}\,dy=\frac y2\sqrt{A^2+B^2y^2}+\frac{A^2}{2B}\ln\left(By+\sqrt{A^2+B^2y^2}\right)$ with $A=b^2,\,B=ae$, and simplifying $\sqrt{b^4+a^2e^2b^2}=ab$ (since $b^2+a^2e^2=a^2$):

\[\int_0^b\sqrt{b^4+a^2e^2y^2}\,dy = \frac{ab^2}2+\frac{b^4}{2ae}\ln\frac{a(1+e)}b\]

Step 4: Combine and simplify

\[S_y=\frac{4\pi a}{b^2}\left[\frac{ab^2}2+\frac{b^4}{2ae}\ln\frac{a(1+e)}b\right] = 2\pi a^2+\frac{2\pi b^2}e\ln\frac{a(1+e)}b\]

Since $\left[\frac{a(1+e)}b\right]^2=\frac{1+e}{1-e}$ (because $a^2(1-e^2)=b^2$), this equals

\[\boxed{S_y=2\pi a^2+\frac{\pi b^2}e\ln\frac{1+e}{1-e}}\]

Special case

When $a=b=r$ (a circle), $e\to0$ and both formulas reduce, in the limit, to the sphere's $S=4\pi r^2$.


Result

Final answer:

\[\boxed{\begin{array}{ll}\text{a) About } x\text{-axis:} & S_x=2\pi b^2+\dfrac{2\pi ab}e\arcsin e\\[0.5em]\text{b) About } y\text{-axis:} & S_y=2\pi a^2+\dfrac{\pi b^2}e\ln\dfrac{1+e}{1-e}\end{array}}\]