I'm currently stuck with this question and would appreciate any tips on how I can approach this question.
Right now, my intuition is to split the area into 2 portions, the left and the right, and focus on finding one of them. So within each portion: I'll find the area with respect to y-axis for the region bounded by y = [-a, a]. But even for this, I'm having difficulties manipulating the equation such that I only have y terms on one side of the equation.
But from y=a to y=(max point), I believe I need to decompose the graph into 2 separate equations. This is where I'm having difficulties.
Would appreciate if anyone can advise if I'm in the right direction and any other tips on how to decompose the graph
$\endgroup$ 11 Answer
$\begingroup$Manipulating the given equation, you get
$$\begin{equation}\begin{aligned} \left(y - x^{\frac{2}{3}}\right)^2 & = a^2 - x^2 \\ y - x^{\frac{2}{3}} & = \pm \sqrt{a^2 - x^2} \\ y & = \pm \sqrt{a^2 - x^2} + x^{\frac{2}{3}} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus, the lower bound for $y$ would be $-\sqrt{a^2 - x^2} + x^{\frac{2}{3}}$ and the upper bound would be $\sqrt{a^2 - x^2} + x^{\frac{2}{3}}$. Thus, the height to integrate would be their difference, i.e., $\left(\sqrt{a^2 - x^2} + x^{\frac{2}{3}}\right) - \left(-\sqrt{a^2 - x^2} + x^{\frac{2}{3}}\right) = 2\sqrt{a^2 - x^2}$. Also, the range for $x$ would be from $-a$ to $a$. Alternatively, as you suggest, due to the symmetry (since replacing $x$ with $-x$ in \eqref{eq1A} doesn't change the equation), you can you just find the area for one half (e.g., for $x = 0$ to $x = a$) and double the result to get the total area. In that case the area would be
$$2\int_{0}^{a} 2\sqrt{a^2 - x^2}dx \tag{2}\label{eq2A}$$
I'll leave it to you to solve this integral.
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