Question

Find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis.

Answer 1

Answer: $9.6\pi$

Step-by-step explanation:

Given

Curve is $y=3-3x^2$

boundary is y=0 i.e.

$\Rightarrow 0=3-3x^2\\\Rightarrow x^2=1\\\Rightarrow x=\pm 1$

The volume of solid generated when rotated about the x-axis is

$\Rightarrow V=\int_a^b\pi y^2dx$

Putting values we get

$\Rightarrow V=\int_{-1}^{1}\pi (3-3x^2)^2dx\\\\\Rightarrow V=\int_{-1}^{1}\pi(9+9x^4-18x^2)dx\\\\\Rightarrow V=\pi \left [ \frac{9x^5}{5} - 6 x^3 + 9 x\right ]_{-1}^{1}\\\\\Rightarrow V=9.6\pi$