Having trouble solving question involving parametric equations

Question

I have been given the following:

$$y = a \cdot \cos^3t$$ $$x = a \cdot \sin^3t$$

$$0 \leqslant t \leqslant {\frac\pi2}$$

I am supposed to show that the mean value of $y$ over the interval $0\leqslant x \leqslant a$ is given by:

$$m = 3a\int^{\frac12\pi}_0(\cos^4t-\cos^6t)dt$$

I know that the mean fomula is given by:

$$m=\frac1{b-a}\int^b_aydx$$

But I'm having trouble finding an equation for $y$ in terms of $x$.

In addition, I am confused as to how the final answer will be given as an integral in terms of $t$.

I thought I should differentiate $y$, $x$ in terms of $t$ then maybe I would find a substitution:

$$\frac{dy}{dt} = -3a \cdot \cos^2t \cdot \sin t$$ $$\frac{dx}{dt} = 3a \cdot \cos t \cdot \sin^2t$$

$$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{-3a \cdot \cos^2t \cdot \sin t}{3a \cdot \cos t \cdot \sin^2t} = -\frac{\cos t}{\sin t} = -\cot t$$

Now I'm facing two problems:

1. I can't integrate both sides to get $y$ in terms of $x$ because the trigonometric function on R.H.S. is still in terms of $t$
2. Even if I did integrate this method seems too inefficient

What's the most appropriate method of solving this?

Answer 1

Think of the parametric equations as a change of variable in the integral defining $m$:

$$ m = \frac{1}{a} \int_0^a y \, dx = \frac{1}{a} \int_0^{\pi/2} y \frac{dx}{dt} \, dt $$

Now plug in the equations for $x$ and $y$ in terms of $t$ to get the desired result:

\begin{align} m &= \frac{1}{a} \int_0^{\pi/2} a \cos^3 t \left(3a \cos t (1 - \cos^2 t )\right) \, dt \\ &= 3a \int_0^{\pi/2} \left(\cos^4 t - \cos^6 t\right) \,dt \end{align}

Answer 2

$$y=a\cos^3t\implies \cos^2t=\left(\frac{y}{a}\right)^{2/3}$$ and $$x=a\sin^3t\implies \sin^2t=\left(\frac{x}{a}\right)^{2/3}$$

Since $$\cos^2t+\sin^2t=1$$ $$\implies \left(\frac{y}{a}\right)^{2/3}+\left(\frac{x}{a}\right)^{2/3}=1$$ $$\implies x^{2/3}+y^{2/3}=a^{2/3}$$ $$\implies y=(a^{2/3}-x^{2/3})^{3/2}$$

Now $$<y>=\frac{1}{a}\int_0^a(a^{2/3}-x^{2/3})^{3/2}dx$$

Now substitute $x=a\sin^3t$