# Math Help - integral 1

1. ## integral 1

y=sqrt(x+1),x=1,y=-1
i have to find a volume(certain integral) of geometric body which is showed (produced) by y=sqrt(x+1) rotation about y=-1...

2. ## Re: integral 1

Originally Posted by mathe26
y=sqrt(x+1),x=1,y=-1
i have to find a volume(certain integral) of geometric body which is showed (produced) by y=sqrt(x+1) rotation about y=-1...
The radius (or distance) between the functions and the line y=-1 is

$r=\sqrt{x+1}-(-1) \implies r=\sqrt{x+1}+1$

Now to find the volume of the solid of rotation use the disk method

$\int_{-1}^1 \pi r^2dx=\int_{-1}^1 \pi (\sqrt{x+1}+1)^2dx$

3. ## Re: integral 1

sorry but i do not know to continue,can u help a little bit more