Calculate surface area of flat figure by using double integral and polar coordinates

Question

Check me please. I tried check it via WolframAlpha, but I don't trust in it 100%.

Task:

Calculate surface area of flat figure by using double integral in polar coordinates. Figure confined by line:

$$\left ( x^{2} + y^{2} \right )^{2} = a^{2}\left ( 2x^{2} + 3y^{2} \right )$$

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My steps:

1. $x = \rho \cos\varphi , \: y = \rho \sin\varphi $

2. $\left ( x^{2} + y^{2} \right )^{2} = a^{2}\left ( 2x^{2} + 3y^{2} \right ) \Rightarrow \\ \Rightarrow \rho^{4} = a^{2}\left ( 2\rho^{2}\cos^{2}\varphi + 3\rho^{2}\sin^{2}\varphi \right ) \Rightarrow 0\leq \rho \leq a\sqrt{2+ \sin^{2}\varphi} = A$

3. $\int_{0}^{2\pi} d\varphi \int_{0}^{A}\rho d\rho = \frac{1}{2}\int_{0}^{2\pi}A^{2}d\varphi = \frac{a^{2}}{2}\int_{0}^{2\pi} (2+\sin^{2}\varphi))d\varphi = \frac{a^{2}}{2}\left ( 4\pi + \frac{1}{2}\int_{0}^{2\pi} \frac{1-\cos2\varphi }{2}d2 \varphi\right ) = 3\pi a^{2}$

Last integral in WolframAlpha. I do not understand where is true :)

Answer 1

HINT:

The shape is like an ellipse, only slightly bigger at $ \varphi = \pi/4 $ points. A lower bound for area is $ \pi \sqrt 6 a^2. $

To me your working & results seem ok, origin isolation is some bother.