# Thread: Volume from Revolution

1. ## Volume from Revolution

Hey everyone. I have to revolve the surface bounded by $y=x\sqrt{x+1}$ and y = 0 around the y-axis and find the volume underneath it. I just have to set up the integral from [a,b]. The problem is, I don't know how to take the inverse of that function. From there I can just do
$\pi\int_a^b[ f^{-1}(x)]^2dx$
What should i do?

2. ## Re: Volume from Revolution

What are your $\displaystyle x$ bounds?

3. ## Re: Volume from Revolution

it wasn't given so i have to use constants, e.g., a for lower bound and b for upper bound. the surface i have to revolve is bounded by y = 0 and $y=x\sqrt{x+1}$

4. ## Re: Volume from Revolution

Originally Posted by lilaziz1
it wasn't given so i have to use constants, e.g., a for lower bound and b for upper bound. the surface i have to revolve is bounded by y = 0 and $y=x\sqrt{x+1}$
$x\sqrt{x+1} = 0$ at $x = 0$ and $x = -1$

the region between $y = x\sqrt{x+1}$ and $y = 0$ is in quad III

rotating this region about the y-axis using cylindrical shells ...

$V = 2\pi \int_{-1}^0 x^2\sqrt{x+1} \, dx$