Respuesta :

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]

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