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

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

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

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

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 :)

• The Oval looks almost like an ellipse with semi-axes √ 2 a, √3 a but is different. The origin is an isolated point, needing to be isolated. Only a quadrant needs to be evaluated. Nov 1, 2015 at 18:35

The shape is like an ellipse, only slightly bigger at $\varphi = \pi/4$ points. A lower bound for area is $\pi \sqrt 6 a^2.$
• Did you mean $4 \int_{0}^{\pi / 2} d\varphi \int_{0}^{ a \sqrt{2+ \sin^{2}\varphi }} \rho d \rho$ ? Nov 1, 2015 at 19:26