Find the area bounded by the arc of the curve y² = x(2

Find the area bounded by the arc of the curve $y^2 = x(2 - x)^3$.

We will use symmetry and a trigonometric substitution to reduce the integral to a Wallis form.

Step 1: Understand the curve

The equation $y^2=x(2-x)^3$ gives $y=\pm\sqrt{x(2-x)^3}$, real only where $x(2-x)^3\geq0$.

Step 2: Determine the domain

Since $(2-x)^3$ has the same sign as $2-x$, we need $x\geq0$ and $2-x\geq0$ together (the case $x\leq0,2-x\leq0$ is impossible). So the curve is defined for $x\in[0,2]$. At $x=0$ and $x=2$: $y=0$.

Step 3: Set up the integral

By symmetry about the $x$-axis:

\[A = 2\int_0^2 \sqrt{x}(2-x)^{3/2} \, dx\]

Step 4: Substitute $x = 2\sin^2 t$

Then $dx=4\sin t\cos t\,dt$; $x=0\Rightarrow t=0$, $x=2\Rightarrow t=\frac\pi2$; and $2-x=2\cos^2t$. So:

\[\sqrt{x}(2-x)^{3/2} = \sqrt{2}\sin t \cdot 2^{3/2}\cos^3 t = 2\sqrt2\sin t\cos^3 t\]

Step 5: Evaluate the integral

\[A = 2\int_0^{\pi/2} 2\sqrt2\sin t\cos^3t\cdot4\sin t\cos t\,dt = 16\sqrt2\int_0^{\pi/2}\sin^2t\cos^4t\,dt\]

Using the Wallis-integral formula for $\int_0^{\pi/2}\sin^mt\cos^nt\,dt$ with $m=2,n=4$ (both even):

\[\int_0^{\pi/2}\sin^2t\cos^4t\,dt = \frac{1!!\cdot3!!}{6!!}\cdot\frac\pi2 = \frac{1\cdot3}{6\cdot4\cdot2}\cdot\frac\pi2 = \frac{\pi}{32}\]

Step 6: Calculate the total area

\[A = 16\sqrt2\cdot\frac{\pi}{32} = \frac{\pi\sqrt2}{2}\]

\[\boxed{A = \frac{\pi\sqrt{2}}{2} \text{ square units}}\]

Step 7: Numerical value

\[A \approx 2.221 \text{ square units}\]

Verification via the Beta function

\[\int_0^{\pi/2}\sin^2t\cos^4t\,dt = \frac12B\!\left(\frac32,\frac52\right) = \frac12\cdot\frac{\Gamma(3/2)\Gamma(5/2)}{\Gamma(4)} = \frac{\pi}{32} \quad \checkmark\]

Summary

\[\boxed{A = \frac{\pi\sqrt{2}}{2} \approx 2.221 \text{ square units}}\]

Key insight: The curve forms a closed loop through the origin and $(2,0)$; the substitution $x=2\sin^2t$ converts the awkward power $(2-x)^{3/2}$ into a clean product of sine and cosine powers, matching a standard Wallis integral.