Surface under the 'love formula'

Question

To plot the love formula, you can plot two functions in one diagram:

$$y_1(x)=\sqrt[3]{x^2}+\sqrt{1-x^2}+1$$ $$y_2(x)=\sqrt[3]{x^2}-\sqrt{1-x^2}+1$$

I calculated the surface inside the heart as, am I right of that:

$$2\int_{0}^{1}\left(\sqrt[3]{x^2}+\sqrt{1-x^2}+1\right)\space\text{d}x-2\int_{0}^{1}2\space\text{d}x+2\int_{0}^{1}2\space\text{d}x-2\int_{0}^{1}\left(\sqrt[3]{x^2}-\sqrt{1-x^2}+1\right)\space\text{d}x=\pi$$

Your last equations are not clear! Please give some comments on them! In my point of view $$A=\int_{-1}^{1}[y_1(x)-y_2(x)]dx$$ — Hosein Rahnama, Dec 27 '15 at 14:16

score 0 · Accepted Answer · answered Dec 27 '15 at 13:55

0

The result is correct but I have no idea where all those integrals come from. For me the area would be just one integral of the difference between the two functions, which yields the same formula as the area of the unit circle.

answered Dec 27 '15 at 13:55

Justpassingby

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Surface under the 'love formula'

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