Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer

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 $r$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.