Find the volumes of the solids generated by revolving the region between $y=5 \sqrt{x}$ and $y=\frac{x^{2}}{6}$ about a) the $x$-axis and $b$ ) the $y$-axis.
Answer
Final Answer: The volume of the solid generated by revolving the region between the functions \(y=5 \sqrt{x}\) and \(y=\frac{x^{2}}{6}\) about the x-axis is \(\boxed{225*30^{\frac{1}{3}}\pi}\) cubic units. The volume of the solid generated by revolving the region about the y-axis is \(\boxed{2\pi*(-283.794761322904 + 9.0*30^{\frac{2}{3}})}\) cubic units.
Step 1 ：Find the intersection points of the two functions \(y=5 \sqrt{x}\) and \(y=\frac{x^{2}}{6}\) to determine the limits of integration. The intersection points are at \(x=0\) and \(x=30^{\frac{2}{3}}\).
Step 2 ：Set up the integral for the volume of the solid generated by revolving the region between the two functions about the x-axis. The functions are \(f(x) = 5\sqrt{x}\) and \(g(x) = \frac{x^{2}}{6}\).
Step 3 ：Calculate the volume \(V_x\) by integrating from the lower limit to the upper limit. The volume \(V_x = 225*30^{\frac{1}{3}}\pi\) cubic units.
Step 4 ：Set up the integral for the volume of the solid generated by revolving the region between the two functions about the y-axis. The functions are \(f(y) = \frac{x^{2}}{25}\) and \(g(y) = 2.44948974278318*x^{0.5}\).
Step 5 ：Calculate the volume \(V_y\) by integrating from the lower limit to the upper limit. The volume \(V_y = 2\pi*(-283.794761322904 + 9.0*30^{\frac{2}{3}})\) cubic units.
Step 6 ：Final Answer: The volume of the solid generated by revolving the region between the functions \(y=5 \sqrt{x}\) and \(y=\frac{x^{2}}{6}\) about the x-axis is \(\boxed{225*30^{\frac{1}{3}}\pi}\) cubic units. The volume of the solid generated by revolving the region about the y-axis is \(\boxed{2\pi*(-283.794761322904 + 9.0*30^{\frac{2}{3}})}\) cubic units.
