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
Best Answer
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.