Answer:
[tex]\frac{1250\pi}{81}[/tex]
Step-by-step explanation:
Volume of solid of revolution between the curves y=5x and y=3x^2 around x-axis on interval [0, 5/3] can be found by using integral as follows:
[tex]Volume = \int\limits^{\frac{5}{3}}_0 \pi ((5x)^2-(3x^2)^2)dx=\pi\int\limits^{\frac{5}{3}}_0(25x^2-9x^4)dx=\\\\=\pi(\frac{25}{3}x^3-\frac{9}{5}x^5)|^{\frac{5}{3}}_0=\frac{1250\pi}{81}[/tex]