Answer: [tex]\dfrac{32\pi }{5}\ \text{unit}^3[/tex]
Step-by-step explanation:
Given
The area is bounded by [tex]y=x^2[/tex] and x-axis from [0,2]
The volume generated for [tex]y=f(x)[/tex] when rotated about the x-axis in the interval [a,b] is
[tex]V=\int_a^b\pi y^2dx[/tex]
Insert the values
[tex]\Rightarrow V=\int_0^2\pi x^4dx\\\\\Rightarrow V=\pi \int_0^2x^4dx\\\\\Rightarrow V=\pi \left[ \dfrac{x^5}{5}\right]_0^2\\\\\Rightarrow V=\dfrac{2^5\pi }{5}\\\\\Rightarrow V=\dfrac{32\pi }{5}\ \text{unit}^3[/tex]
