# How do I calculate surface of ellipse rotating around x axel?

How to calculate surface of ellipse rotating around x axel when ellipse equation is :

$y^{2}+4x^{2} = 36$

I started with this formula for calculating surface of rotating object around x axel

$\int_{a}^{b}f(x) \sqrt{1 + f'(x)^{2}} dx$

but i always end up with complicated solution, I believe there is easier way of solving this but I am missing it. Any hint and help is welcome.

• Why? Can you give me little bit more explanation i will be grateful. – Zvnoky Brown Dec 12 '16 at 23:55
• @JackD'Aurizio Twice the volume of a sphere with radius $3$ is $72\pi$. But in any case, shouldn't it be related to the surface area? I can possibly understand stretching a radius $3$ sphere up the $y-$axis to give $2\cdot 4\pi r^2 = 72\pi$. Or am I completely missing what you're hinting at? – John Dec 13 '16 at 0:01
• Sorry, I got it entirely wrong. The surface area is given by an elliptic integral. – Jack D'Aurizio Dec 13 '16 at 0:13

The surface area is given by

\begin{equation} A=2\pi\int_{a}^{b}y\sqrt{1+(y^\prime)^2}\,dx \end{equation}

with $y=2\sqrt{9-x^2}$, $a=-3,\,b=3$. Because of symmetry we can double the integral from $0$ to $3$.

Since $\sqrt{1+(y^\prime)^2}=\sqrt{\dfrac{9-3x^3}{9-x^2}}$ we have

\begin{eqnarray} A&=&4\pi\int_{0}^{3}2\sqrt{9-x^2}\sqrt{\dfrac{9+3x^2}{9-x^2}}\,dx\\ &=&8\sqrt{3}\pi\int_{0}^{3}\sqrt{3+x^2}\,dx \end{eqnarray}

which can be done by trigonometric substitution or by using the formula

\begin{equation} \int\sqrt{a^2+x^2}\,dx=\dfrac{1}{2}\left[x\sqrt{a^2+x^2}+a^2\ln\vert x+\sqrt{a^2+x^2}\vert\right]+c \end{equation}

If I substituted correctly, this gives $A=\left[72+12\sqrt{3}\ln\left(1+\frac{2\sqrt{3}}{3}\right)\right]\pi$

• I think there is an error in the second term but I am being called to dinner. – John Wayland Bales Dec 13 '16 at 0:49