<s>[INST] Surface area of an ellipse 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. 
$$\left(\frac{x}{3}\right)^{2}+\left(\frac{y}{2}\right)^{2}=1.$$

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