Surface area of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ about $x,y$ axis.

Question

So I need to find the Surface area of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ about $x,y$ axis.

For $x-$axis, I found that

$$y=\frac{b}{a}\sqrt{a^2-x^2}$$

Then

$$y'=-\frac{bx}{a\sqrt{a^2-x^2}}.$$

Then

$$(y')^2=\frac{b^2x^2}{a^2(a^2-x^2)}.$$

Then $$1+(y')^2=\frac{x^2b^2+a^2(a^2-x^2)}{a^2(a^2-x^2)}$$

Then according to $S= 2\pi \int_a^b f(x)\sqrt{1+(f'(x))^2}dx$ we have

$$S = 2 \pi \int_{-a}^a \frac{b}{a^2}\sqrt{x^2b^2+a^2(a^2-x^2)} dx.$$

where am I going wrong? And for the $y-$axis do we just solve for $x$ and integrate $-b$ to $b$?

Answer 1

If $b \ge a$ we have $$S = 2 \pi \int_{-a}^a \frac{b}{a^2}\sqrt{x^2b^2+a^2(a^2-x^2)} dx = 2 \pi \frac{b}{a^2} \sqrt{ a^4+(b^2-a^2)x^2}dx =\\$$ $$ = 2 \pi \frac{b}{a^2\sqrt{b^2-a^2}}\int \sqrt{c^2+x^2}dx $$ where $c^2 = \frac{a^4}{b^2-a^2}$. If $b \le a$ we use $\int \sqrt{c^2-x^2}dx$ instead of $\int \sqrt{c^2+x^2}dx$.

If we swap $a$ and $b$ and swap $x$ and $y$ we will get an answer for the second question.

Answer 2

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is the equation of an ellipse

Not required but the area of ellipse is:

$$\begin{align} x&=a\cos t, y=b\sin t\\ A&=4\int_0^{\frac{\pi}{2}}{ydx}\\&=-ab\int_0^{\frac{\pi}{2}}{\sin^2t dt}\end{align}$$

About x-axis the area of ellipsoid is $$2\int_0^{\frac{\pi}{2}}{2\pi y dx}= \pi A$$

About y-axis the area of ellipsoid is $$2\int_0^{\frac{\pi}{2}}{2\pi x dy}= 4\pi \cdot (-ab)\int_0^{\frac{\pi}{2}}{\cos^2 t dt} $$