Surface area of an ellipse

Question

I have this problem that I have worked out. Will someone check it for me? I feel like it is not correct. Thank you!

Rotate the graph of the ellipse about the $x$-axis to form an ellipsoid. Calculate the precise surface area of the ellipsoid.

$$\left(\frac{x}{3}\right)^{2}+\left(\frac{y}{2}\right)^{2}=1.$$

Answer 1

$\frac {x^2}{9} + \frac {y^2}{4} = 1\\ \frac {d}{dx}(\frac {x^2}{9} + \frac {y^2}{4} = 1)\\ \frac {2x}{9} + \frac {2y}{4}\frac {dy}{dx} = 0\\ \frac {dy}{dx} = - \frac {4 x}{9 y}\\ 2\pi \int_{-3}^{3} y\sqrt {1+(\frac {dy}{dx})^2}\ dx\\ 2\pi \int_{-3}^{3} \sqrt {y^2+ \frac {16}{81}x^2}\ dx$

Note you have the wrong limits of integration. Now you can substitute $y^2 = 4 - \frac 49 x^2$

$2\pi \int_{-3}^{3} \sqrt {4 - \frac {20}{81}x^2}\ dx\\ 2\pi \int_{-3}^{3} \frac {2}{9}\sqrt {81 - 5 x^2}\ dx$

Which gets you back where you were. Now you need to make a trig substitution $x = \frac 9{\sqrt 5} \sin \theta$ $ 2\pi \int_{-\arcsin \frac{\sqrt 5}3}^{\arcsin \frac{\sqrt 5}3} \frac {2}{9}\sqrt {81 - 81\sin^2\theta} (\frac 9{\sqrt 5} \cos \theta\ d\theta)\\ 2\pi \int_{-\arcsin \frac{\sqrt 5}3}^{\arcsin \frac{\sqrt 5}3} \frac {18}{\sqrt 5} \cos^2\theta\ d\theta\\ 2\pi \frac {9}{\sqrt 5}(\theta + \sin\theta\cos\theta)|_{-\arcsin \frac{\sqrt 5}3}^{\arcsin \frac{\sqrt 5}3}\\ \frac {36\pi}{\sqrt 5}\arcsin \frac{\sqrt 5}3 + 8\pi $

Answer 2

I recognize that you have a correct solution by @DougM, but I wanted to point out a few things. First, Pappus's $(1^{st})$ Centroid Theorem states that the surface area $A$ of a surface of revolution generated by rotating a plane curve $C$ about an axis external to $C$and on the same plane is equal to the product of the arc length $s$ of $C$ and the distance d traveled by its geometric centroid ([Pappus's centroid theorem][1]). Simply put, $S=2\pi RL$, where $R$ is the normal distance of the centroid to the axis of revolution and $L$ is curve length. The centroid of a curve is given by

$$\mathbf{R}=\frac{\int \mathbf{r}ds}{\int ds}=\frac{1}{L} \int \mathbf{r}ds$$

Second, I'll demonstrate how to solve this in the complex plane. And finally, I'll carry out the solution for an arbitrary ellipse given by

$$\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1$$

In the complex plane, the surface area of a curve rotated about the $x$-axis is given by

$$S=2\pi\int \Im\{z\}|\dot z| du,\quad z=z(u)$$

Thus we have

$$ \begin{align} &z=a\cos\theta+ib\sin\theta,\quad \theta\in[0,\pi]\\ &\dot z=-a\sin\theta+ib\cos\theta\\ &|\dot z|=\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}=b\sqrt{\left(\frac{a^2-b^2}{b^2}\right)\sin^2\theta+1}\\ \end{align} $$

So that finally

$$ \begin{align} S &=2\pi\int_0^{\pi}b^2\sin\theta\sqrt{\left(\frac{a^2-b^2}{b^2}\right)\sin^2\theta+1}\\ &=2\pi b^2\left[1+\frac{(1+A)\tan^{-1}\sqrt{A}}{\sqrt{A}}\right],\quad A=\left(\frac{a^2-b^2}{b^2}\right) \end{align} $$

This has been tested numerically for arbitrary $a$ and $b$, and, of course, agrees with @DougM for $a=3$ and $b=2$.