Calculate the surface area generated by rotating

Calculate the surface area generated by rotating the curve $y=4-\dfrac{\sqrt[8]{9x^2}}2$ about the $x$-axis for $x>0$ (the arc between the coordinate axes).

Step 1: Simplify and find the domain

With $k=\dfrac{9^{1/8}}2\approx0.6580$: $y=4-kx^{1/4}$. At $x=0$, $y=4$; setting $y=0$ gives $x^{1/4}=\dfrac4k$, so the arc runs from $x=0$ to

\[x_{\max}=\left(\frac4k\right)^4=\frac{4096}3\approx1365.3\]

Step 2: Differentiate

\[y'=-\frac k4x^{-3/4} \implies 1+(y')^2=1+\frac{k^2}{16x^{3/2}}\]

Step 3: Substitute $u=x^{1/4}$

Then $x=u^4$, $dx=4u^3\,du$, $x^{3/4}=u^3$, $x^{3/2}=u^6$, $y=4-ku$, and the upper limit becomes $u_{\max}=\dfrac8{\sqrt[4]3}\approx6.079$. Substituting into $S=2\pi\int y\sqrt{1+(y')^2}\,dx$:

\[S=2\pi\int_0^{u_{\max}}(4-ku)\sqrt{16u^6+k^2}\,du\]

Step 4: Evaluate numerically

With $k^2=\dfrac{\sqrt3}4\approx0.4330$, high-precision numerical integration of $\displaystyle\int_0^{6.079}(4-ku)\sqrt{16u^6+k^2}\,du$ gives $\approx1093.68$, so

\[S\approx2\pi(1093.68)\approx6871.8\]

\[\boxed{S\approx6871.8\text{ square units}}\]

Step 5: Sanity check on the scale

This large value is expected: the domain spans $x\in[0,\approx1365]$, and $y$ stays of order $1$–$4$ over most of that enormous range (since $y$ depends only on the slowly growing $x^{1/4}$), so the swept surface — roughly a long, gently tapering horn — has a correspondingly large area, consistent with a rough order-of-magnitude estimate of $2\pi\cdot(\text{typical radius})\cdot(\text{length})\sim2\pi\cdot2\cdot1365\approx17000$.