We know, that the area of the surface generated by revolving the curve y about the x-axis is given by:
[tex]\boxed{A=2\pi\cdot\int\limits_a^by\sqrt{1+\left(y'\right)^2}\, dx}[/tex]
In this case a = 0, b = 15, [tex]y=\dfrac{x^3}{15}[/tex] and:
[tex]y'=\left(\dfrac{x^3}{15}\right)'=\dfrac{3x^2}{15}=\boxed{\dfrac{x^2}{5}}[/tex]
So there will be:
[tex]A=2\pi\cdot\int\limits_0^{15}\dfrac{x^3}{15}\cdot\sqrt{1+\left(\dfrac{x^2}{5}\right)^2}\, dx=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\left(\star\right)\\\\-------------------------------\\\\
\int x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\,dx=\int\sqrt{1+\dfrac{x^4}{25}}\cdot x^3\,dx=\left|\begin{array}{c}t=1+\dfrac{x^4}{25}\\\\dt=\dfrac{4x^3}{25}\,dx\\\\\dfrac{25}{4}\,dt=x^3\,dx\end{array}\right|=\\\\\\[/tex]
[tex]=\int\sqrt{t}\cdot\dfrac{25}{4}\,dt=\dfrac{25}{4}\int\sqrt{t}\,dt=\dfrac{25}{4}\int t^\frac{1}{2}\,dt=\dfrac{25}{4}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}= \dfrac{25}{4}\cdot\dfrac{t^{\frac{3}{2}}}{\frac{3}{2}}=\\\\\\=\dfrac{25\cdot2}{4\cdot3}\,t^\frac{3}{2}=\boxed{\dfrac{25}{6}\,\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}}\\\\-------------------------------\\\\ [/tex]
[tex]\left(\star\right)=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\dfrac{2\pi}{15}\cdot\dfrac{25}{6}\cdot\left[\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}\right]_0^{15}=\\\\\\=
\dfrac{5\pi}{9}\left[\left(1+\dfrac{15^4}{25}\right)^\frac{3}{2}-\left(1+\dfrac{0^4}{25}\right)^\frac{3}{2}\right]=\dfrac{5\pi}{9}\left[2026^\frac{3}{2}-1^\frac{3}{2}\right]=\\\\\\=
\boxed{\dfrac{5\Big(2026^\frac{3}{2}-1\Big)}{9}\pi}[/tex]
Answer C.
