Find the volume of revolution formed by revolving the region bounded by the graph of y=1-x and the x-axis and the y axis revolved around the x-axis

Question

Please Help, I'm in Calculus 2 and my professor didn't explain how to solve these type of questions

Answer 1

The solid of revolution obtained in this way is a cone with base a disc centred at the origin of radius 1 contained in the plane $x=0$ and with height the segment $[0,1]$ along the $x$ axis. It's volume is given by the formula: $$\pi\int_0^1 f^2(x) dx=\pi\int_0^1 (1-x)^2 dx=\pi\int_1^0 t^2 d((1-t))=\pi\int_0^1 t^2 dt=\left[\frac{t^3}{3}\right]_0^1=\frac{\pi}{3}$$ The same result can be obtained in a more traditional way. In fact the volume of a cone is one third the product of the base area by the height size: $\frac{1}{3}\cdot (\pi 1^2) \cdot 1=\frac{\pi}{3}.$