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Let $x=2\cos^3\theta$ and $y=2\sin^3\theta$ known as the astroid.

enter image description here

In this case, radius $r=2$. and gray part's $x$ range is $1/\sqrt{2}\leq x\leq 2$. this deal with $0\leq\theta\leq \pi/4$.

Question. How can I calculate area of gray part in this picture?

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2 Answers 2

up vote 3 down vote accepted

The parametric representation of that asteroid is $x=2\cos^3\theta$, $y=2\sin^3\theta$. We hit the point $(2,0)$ when $\theta=0$, and the point $(1/\sqrt2,1/\sqrt2)$ when $\theta=\pi/4$. The area is between the curve and the $x$-axis, so equals the sum of infinitesimal vertical strips of width $|dx|=|x'(\theta)\,d\theta|$ and height $|y(\theta)|$. Therefore the area is $$ A=-\int_{\theta=0}^{\pi/4}x'(\theta) y(\theta)\,d\theta. $$ The minus sign comes from the fact that we are moving from right to left as the parameter $\theta$ grows (IOW $x'(\theta)<0$ in this interval).

I leave the calculation of that integral to you.

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The equation is $(\frac{x}{2})^{\frac{2}{3}}+(\frac{y}{2})^{\frac{2}{3}}=1$. You can solve this for $y$ to get $y=2\left(1-(\frac{x}{2})^{\frac{2}{3}}\right)^{\frac{3}{2}}$and integrate. I don't see how the lower limit is related to $r$

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