# How can I determine the area of the region bounded by the curve $x^2=y^4(1-y^3)$ [closed]

$$x^2=y^4(1-y^3)$$

On a graph it looks like a diamond. This is for my integral calculus class.

## closed as off-topic by max_zorn, Leucippus, Cesareo, Lee David Chung Lin, mrtaurhoFeb 6 at 9:12

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• $(n^{2}\sqrt {(1+n^{3})},-n)$ lies on this curve for every positive integer $n$. The region is actually unbounded and the areas is infinite. – Kavi Rama Murthy Feb 5 at 23:13

HINT: If you're interested in finding the area inside the "loop" portion of the graph, you will want to integrate with respect to $$y$$.
The form you're given the equation in makes things "simpler" in some respects. In particular, $$x^2=(y^4)(1-y^3)$$ \begin{align}\implies x & =\pm\sqrt{y^4(1-y^3)}.\\ &= \pm y^2\sqrt{1-y^3}\end{align} From here, notice the graph is symmetric about the $$y$$-axis. You can choose the positive portion, $$x = y^2\sqrt{1-y^3},$$ and find the integral here and double it to find the total area.
• I think the 'negative part$is unbounded. – Kavi Rama Murthy Feb 5 at 23:16 • @KaviRamaMurthy I'm not sure what you mean.$(0,0)$is considered a root of this equation, because$0\stackrel{?}{=} 0(1-0) \quad \checkmark$. Hence the loop is bounded – Decaf-Math Feb 5 at 23:21 • The correct wording of the question should be: find the area of the bounded region determined by ... @jackattack825 – Kavi Rama Murthy Feb 5 at 23:28 • @KaviRamaMurthy if I correctly understand what your concern is, the graph of$x^2 = y^4(1-y^3)$makes no shape with the$x$-axis other than the small loop I put in the picture. Indeed, I think the OP meant to imply they want to find the area of the small loop section more than anything, as everything else below the graph is simply infinite. EDIT: Pedagogically speaking, finding the area in the loop is the only thing that makes sense, as it seems like the intention is to integrate$\int_0^1y^2\sqrt{1-y^3}\,dy$, which is a simple$u\$-substitution. – Decaf-Math Feb 5 at 23:32