Answer: [tex]9.6\pi[/tex]
Step-by-step explanation:
Given
Curve is [tex]y=3-3x^2[/tex]
boundary is y=0 i.e.
[tex]\Rightarrow 0=3-3x^2\\\Rightarrow x^2=1\\\Rightarrow x=\pm 1[/tex]
The volume of solid generated when rotated about the x-axis is
[tex]\Rightarrow V=\int_a^b\pi y^2dx[/tex]
Putting values we get
[tex]\Rightarrow V=\int_{-1}^{1}\pi (3-3x^2)^2dx\\\\\Rightarrow V=\int_{-1}^{1}\pi(9+9x^4-18x^2)dx\\\\\Rightarrow V=\pi \left [ \frac{9x^5}{5} - 6 x^3 + 9 x\right ]_{-1}^{1}\\\\\Rightarrow V=9.6\pi[/tex]