Area of surface

Question

What is the area of the region that is bounded by the curve $$\vec{R}(t)=(\cos^3t, \sin^3t), 0\leq t<2\pi?$$ I have no idea how to start here or what i have to use.

Answer 1

The curve is given by $$x^{2/3} + y^{2/3} = 1$$ The graph of the curve looks like below.

Hence, the area of the graph is $$A = 4 \int_0^1 \left(1-x^{2/3}\right)^{3/2}dx$$ Setting $x=\sin^3(t)$, we obtain $dx=3\sin^2(t)\cos(t) dt$. Hence, we have the area to be \begin{align} A & = 4 \int_0^{\pi/2} \cos^3(t)\cdot 3\sin^2(t)\cos(t) dt = 12 \int_0^{\pi/2} \cos^4(t)(1-\cos^2(t))dt\\ & = 12 \left(\int_0^{\pi/2}\cos^4(t) dt - \int_0^{\pi/2}\cos^6(t) dt\right) \end{align} From here, we have $$\int_0^{\pi/2} \cos^{2m}(t)dt = \dfrac{\pi}{2^{2m+1}} \dbinom{2m}m$$ Hence, we obtain that the area is $$A = 12\left(\dfrac{\pi}{2^5} \dbinom{4}2 - \dfrac{\pi}{2^7} \dbinom{6}3\right) = 3\pi \left(\dfrac68 - \dfrac{20}{32}\right) = \dfrac{3\pi}8$$