Area of My Valentine

Since the heart curve is even, we need only consider the positive $x$ half of the curve and then multiply the result by 2. Then we have:

$\begin{aligned} x^2 + \left(\frac {5y}4-\sqrt{\color{#3D99F6}|x|}\right)^2 & = 1 & \small \color{#3D99F6} \text{For }x \ge 0 |x| = x \\ x^2 + \left(\frac {5y}4-\sqrt{\color{#3D99F6}x}\right)^2 & = 1 \\ \frac {5y}4-\sqrt{x} & = \pm \sqrt{1 - x^2} \\ y & = \frac 45 \left(\sqrt x \pm \sqrt{1-x^2} \right) \end{aligned}$

We note that $y$ has two parts $y = \dfrac 45 \left(\sqrt x + \sqrt{1-x^2} \right)$ , the blue curve, and $y = \dfrac 45 \left(\sqrt x - \sqrt{1-x^2} \right)$ , the red curve. The area of half the heart $A = \displaystyle \int_0^1 y\ dx$ also has two parts $A_1 = \displaystyle \int_0^1 \frac 45 \left(\sqrt x + \sqrt{1-x^2} \right) dx$ , the blue $+$ purple regions, and $A_2 = \displaystyle \int_0^1 \frac 45 \left(\sqrt x - \sqrt{1-x^2} \right) dx$ , the pink $+$ purple regions. And that:

$\begin{aligned} A & = A_1 - A_2 \\ & = \int_0^1 \frac 45 \left(\sqrt x + \sqrt{1-x^2} \right) dx - \int_0^1 \frac 45 \left(\sqrt x - \sqrt{1-x^2} \right) dx \\ & = \frac 85 \int_0^1 \sqrt{1-x^2} \ dx & \small \color{#3D99F6} \text{Let }x = \sin \theta \implies dx = \cos \theta \ d\theta \\ & = \frac 85 \int_0^\frac \pi 2 \cos^2 \theta \ d\theta \\ & = \frac 85 \int_0^\frac \pi 2 \frac {1+\cos 2\theta}2 d\theta \\ & = \frac 45 \left[ \theta + \frac {\sin 2 \theta}2\right]_0^\frac \pi 2 \\ & = \frac 25 \pi \end{aligned}$

Therefore, the area of the heart $= 2A = \dfrac 45 \pi \approx \boxed{2.513274}$ .

$A_{1} = \dfrac{4}{5}\int_{0}^{1} (\sqrt{1 - x^2} + \sqrt{x} - 1) dx = \dfrac{4}{5}(\int_{0}^{\frac{\pi}{2}} \cos^{2} d\theta + \int_{0}^{1} \sqrt{x} - 1 dx) =$ $\dfrac{4}{5} (\dfrac{1}{2}(\theta + \dfrac{1}{2} \sin(2\theta))|_{0}^{\frac{\pi}{2}} +(\dfrac{2}{3} x^{\frac{3}{2} } - x)|_{0}^{1}) = \dfrac{4}{5} (\dfrac{\pi}{4} - \dfrac{1}{3})$

Similarly, $A_{2} = \dfrac{4}{5}\int_{0}^{1} (1 + \sqrt{1 - x^2} - \sqrt{x}) dx = \dfrac{4}{5}(\int_{0}^{\frac{\pi}{2}} \cos^{2} d\theta + \int_{0}^{1} 1 - \sqrt{x} dx) =$ $\dfrac{4}{5} (\dfrac{1}{2}(\theta + \dfrac{1}{2} \sin(2\theta))|_{0}^{\frac{\pi}{2}} +(x - \dfrac{2}{3} x^{\frac{3}{2} })|_{0}^{1}) = \dfrac{4}{5} (\dfrac{\pi}{4} + \dfrac{1}{3})$

$\implies A_{1} + A_{2} = \dfrac{2\pi}{5} \implies$ the area of the heart $A = 2(A_{1} + A_{2}) = \dfrac{4\pi}{5} \approx \boxed{2.513274}$ .

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