Question

How do you find the volume bounded by `y=sqrt(x + 1)`, x = 0, x = 3, and y = 0 revolved about the x-axis?

Answer 1

We will use the disk method. Where `r=y`

Explanation:

The volume is `V=int_0^3pir^2dx=int_0^3pi(sqrt(x+1))^2dx=`
`piint_0^3(x+1)dx=pi[x^2/2+x]_0^3=pi(9/2+3)=(15pi)/2`