Sketch the curve 9y² = x(3-x)² and find the surface area

Sketch the curve $9y^2 = x(3-x)^2$ and find the surface area generated by rotating the loop about the $x$-axis.

We first describe the loop, then compute the surface generated by revolving it about the $x$-axis.

Step 1: Analyze the curve

\[y=\pm\frac{\sqrt x(3-x)}3,\qquad x\ge0\]

The loop runs from $(0,0)$ to $(3,0)$, symmetric about the $x$-axis, with maximum height $\frac23$ at $x=1$ (found by setting $y'=0$ on the upper branch).

Step 2: Differentiate the upper branch

\[y=\frac{\sqrt x(3-x)}3 \implies y'=\frac{3-3x}{6\sqrt x}=\frac{1-x}{2\sqrt x}\]

Step 3: Simplify $1+(y')^2$

\[1+\left(\frac{1-x}{2\sqrt x}\right)^2 = \frac{4x+(1-x)^2}{4x}=\frac{x^2+2x+1}{4x}=\frac{(x+1)^2}{4x}\]

\[\sqrt{1+(y')^2}=\frac{x+1}{2\sqrt x}\]

Step 4: Set up the surface integral

Rotating the loop about the $x$-axis, the upper branch alone generates the entire surface (its mirror image below the axis produces the same surface in 3D), so

\[S=2\pi\int_0^3 y\sqrt{1+(y')^2}\,dx = 2\pi\int_0^3\frac{\sqrt x(3-x)}3\cdot\frac{x+1}{2\sqrt x}\,dx = \frac\pi3\int_0^3(3-x)(x+1)\,dx\]

Step 5: Expand and integrate

\[(3-x)(x+1)=-x^2+2x+3\]

\[S=\frac\pi3\int_0^3(-x^2+2x+3)\,dx = \frac\pi3\left[-\frac{x^3}3+x^2+3x\right]_0^3 = \frac\pi3(-9+9+9)=\frac\pi3\cdot9=3\pi\]

\[\boxed{S=3\pi\approx9.42}\]