# integral 1

• March 19th 2013, 11:00 AM
mathe26
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...Attachment 27599
• March 19th 2013, 11:33 AM
TheEmptySet
Re: integral 1
Quote:

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...Attachment 27599

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

$\int_{-1}^1 \pi r^2dx=\int_{-1}^1 \pi (\sqrt{x+1}+1)^2dx$
• March 19th 2013, 11:39 AM
mathe26
Re: integral 1
sorry but i do not know to continue,can u help a little bit more :)