Calculate the following integral: double integral over D

Calculate the following integral: $$ \iint_D y^2\,dx\,dy, $$ where the region $D$ is bounded by the $x$-axis and the first arch of the cycloid $x=a(t-\sin t)$, $y=a(1-\cos t)$.

Step 1 · Understand the cycloid

$$ x=a(t-\sin t), \qquad y=a(1-\cos t), \qquad 0\leqslant t\leqslant2\pi. $$

At $t=0$: $(0,0)$. At $t=2\pi$: $(2\pi a,0)$. This is the first arch.

Step 2 · Reduce to a single integral in $x$

Since $D$ is bounded below by $y=0$ and above by the cycloid:

$$ I = \int_0^{2\pi a}\int_0^{y(x)} y^2\,dy\,dx = \int_0^{2\pi a} \frac{[y(x)]^3}{3}\,dx. $$

Step 3 · Convert to the parameter $t$

With $x=x(t)$, $dx=x'(t)\,dt=a(1-\cos t)\,dt$, and $y(t)=a(1-\cos t)$:

$$ I = \int_0^{2\pi} \frac{[a(1-\cos t)]^3}{3}\cdot a(1-\cos t)\,dt = \frac{a^4}{3}\int_0^{2\pi} (1-\cos t)^4\,dt. $$

Step 4 · Expand $(1-\cos t)^4$

By the binomial theorem:

$$ (1-\cos t)^4 = 1-4\cos t+6\cos^2t-4\cos^3t+\cos^4t. $$

Step 5 · Integrate term by term

$$ \int_0^{2\pi}1\,dt=2\pi, \qquad \int_0^{2\pi}\cos t\,dt=0, \qquad \int_0^{2\pi}\cos^2t\,dt=\pi, $$

$$ \int_0^{2\pi}\cos^3t\,dt=0 \quad(\text{odd power, symmetric}), \qquad \int_0^{2\pi}\cos^4t\,dt=\frac{3\pi}{4}. $$

Step 6 · Combine results

$$ \int_0^{2\pi}(1-\cos t)^4\,dt = 2\pi - 4(0) + 6\pi - 4(0) + \frac{3\pi}{4} = 8\pi+\frac{3\pi}{4} = \frac{35\pi}{4}. $$

Step 7 · Final answer

$$ I = \frac{a^4}{3}\cdot\frac{35\pi}{4} = \frac{35\pi a^4}{12}. $$

Additional Notes

• This technique — reducing $\iint y^n\,dx\,dy$ to $\int\frac{[y(x)]^{n+1}}{n+1}\,dx$ and then substituting a parametrization — works for any region bounded below by an axis and above by a parametrically-given curve.
• $\int_0^{2\pi}\cos^4t\,dt=\frac{3\pi}{4}$ follows from the reduction $\cos^4t=\left(\frac{1+\cos2t}{2}\right)^2$, whose average value over a period is $\frac38$.