Calculate the surface area generated by rotating

Calculate the surface area generated by rotating the parabola $y^2=2x$ about the line $y=2x$, over the arc between their two intersection points.

Step 1: Find the intersection points

Substituting $y=2x$ into $y^2=2x$: $4x^2=2x\implies x=0$ or $x=\frac12$. The intersections are $(0,0)$ and $\left(\frac12,1\right)$, so the relevant arc has $y\in[0,1]$.

Step 2: Distance from the curve to the line

With $x=\frac{y^2}2$, the (signed) distance to $2x-y=0$ is $\dfrac{2x-y}{\sqrt5}=\dfrac{y^2-y}{\sqrt5}$. For $y\in(0,1)$ this is negative (the parabola lies above the line there), so the distance is

\[d(y)=\frac{y(1-y)}{\sqrt5}\]

Step 3: Arc-length element

\[\frac{dx}{dy}=y \implies ds=\sqrt{1+y^2}\,dy\]

Step 4: Set up the integral

\[S=2\pi\int_0^1\frac{y(1-y)}{\sqrt5}\sqrt{1+y^2}\,dy = \frac{2\pi}{\sqrt5}\left[\int_0^1y\sqrt{1+y^2}\,dy-\int_0^1y^2\sqrt{1+y^2}\,dy\right]\]

Step 5: First integral

\[I_1=\int_0^1y\sqrt{1+y^2}\,dy=\left[\frac13(1+y^2)^{3/2}\right]_0^1=\frac{2\sqrt2-1}3\]

Step 6: Second integral

Using the standard reduction $\displaystyle\int y^2\sqrt{1+y^2}\,dy=\frac{y(2y^2+1)\sqrt{1+y^2}}8-\frac18\ln\left(y+\sqrt{1+y^2}\right)+C$:

\[I_2=\int_0^1y^2\sqrt{1+y^2}\,dy = \frac{3\sqrt2}8-\frac18\ln(1+\sqrt2)\]

Step 7: Combine

\[I_1-I_2=\frac{2\sqrt2-1}3-\frac{3\sqrt2}8+\frac18\ln(1+\sqrt2)\]

Putting over a common denominator of 24: $I_1-I_2=\dfrac{7\sqrt2-8}{24}+\dfrac18\ln(1+\sqrt2)$, so

\[S=\frac{2\pi}{\sqrt5}\left[\frac{7\sqrt2-8}{24}+\frac18\ln(1+\sqrt2)\right] = \frac{\pi\sqrt5}{60}\Big(7\sqrt2+3\ln(1+\sqrt2)-8\Big)\]

\[\boxed{S=\frac{\pi\sqrt5}{60}\Big(7\sqrt2+3\ln(1+\sqrt2)-8\Big)}\]

Step 8: Numerical value

\[S\approx\frac{\pi\sqrt5}{60}(9.899+2.644-8)\approx\frac{\pi\sqrt5}{60}(4.544)\approx0.532\]