# What is the volume of the solid produced by revolving f(x)=sqrt(1-x), x in [0,1] around the x-axis?

Here, $x \le 1$. i have changed the interval of integration to [0, 1] from the inadmissible [0, 4].
The volume =$\pi \int {y}^{2} \mathrm{dx} = \pi \int \left(1 - x\right) \mathrm{dx} = \pi \left[x - {x}^{2} / 2\right]$, between the limits [0, 1].
answer = $\frac{\pi}{2}$, cubic units.