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

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

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

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  • $\begingroup$ $(n^{2}\sqrt {(1+n^{3})},-n)$ lies on this curve for every positive integer $n$. The region is actually unbounded and the areas is infinite. $\endgroup$ – Kavi Rama Murthy Feb 5 at 23:13
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HINT: If you're interested in finding the area inside the "loop" portion of the graph, you will want to integrate with respect to $y$.

Graph

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.

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  • $\begingroup$ I think the 'negative part$ is unbounded. $\endgroup$ – Kavi Rama Murthy Feb 5 at 23:16
  • $\begingroup$ @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 $\endgroup$ – Decaf-Math Feb 5 at 23:21
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    $\begingroup$ The correct wording of the question should be: find the area of the bounded region determined by ... @jackattack825 $\endgroup$ – Kavi Rama Murthy Feb 5 at 23:28
  • $\begingroup$ @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. $\endgroup$ – Decaf-Math Feb 5 at 23:32

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