What is the volume of the solid made by rotating the region bounded by $y=x^2+1$ and $y=-2x+3$ about the $x$ - axis.
Please show the process using:
(i) disks/washers
(ii) cylindrical shells
What is the volume of the solid made by rotating the region bounded by $y=x^2+1$ and $y=-2x+3$ about the $x$ - axis.
Please show the process using:
(i) disks/washers
(ii) cylindrical shells
Copyright © 2021 JogjaFile Inc.
The photo shows the region to be rotated. The first step is to find the limits on the $x$-axis, namely the solution to $x^2+1=-2x+3$. This can be written $x^2+2x=2$. Complete the square to get $(x+1)^2=3$, so the limits are $x=-1-\sqrt{3}$ and $-1+\sqrt{3}$.
Now imagine a vertical slice. It will have a CD shape (or washer) where the outer radius is the line $-2x+3$ and the inner radius is the curve $x^2+1$. The width will just be $dx$.
Thus our integral will be $\int\limits_{-1-\sqrt{3}}^{-1+\sqrt{3}}\pi((-2x+3)^2-(x^2+1)^2)dx$.
As for the second method, you will need to use the volume of a cylindrical shell, $2\pi rh\Delta r$. I won't do the whole problem, but you have the $x$ limits, and it shouldn't be too different from the first part.