Question

Find the area of the surface obtained by rotating the curve \displaystyle y = \sqrt[3]{x} about y-axis for 1 \le y \le 4.

Answer 1

\[y=\sqrt[3]{x}~~\iff~~x=y^3~~\implies~~\frac{dx}{dy}=3y^2\]
\[\begin{align*}A&=2\pi\int_1^4x(y)\sqrt{1+\left(\frac{dx}{dy}\right)^2}~dy\\\\
&=2\pi\int_1^4y^3\sqrt{1+9y^4}~dy\end{align*}\]
Let \(u=1+9y^4\), then \(du=36y^3~dy~~\iff~~\dfrac{du}{36}=y^3~dy\).
\[\begin{align*}A&=\frac{1}{18}\pi\int_{10}^{2305}\sqrt u~du\end{align*}\]