Question
The area enclosed by the asteroid $\left(\frac{x}{a}\right)^{2 / 3}+\left(\frac{y}{a}\right)^{2 / 3}=1$ is(a) $\frac{3}{4} a^{2} \pi$(b) $\frac{3}{18} \pi a^{2}$(c) $\frac{3}{8} \pi a^{2}$(d) $\frac{3}{4} a \pi$
Step 1
We can write $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$ where $\theta$ ranges from $0$ to $2\pi$. Show more…
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