Really ugly graph-volume problem

[Nikitin]

Homework Statement

Find the volume of the solid generated by revolving the region about the x-axis.
Region = the area separated by given function and lines:

x = sqrt(2y)/(y^2 +1)
y=0.

The Attempt at a Solution

i just need to solve the integral of ∫x dy... How? Is there any simple way to solve it?

 

[LCKurtz]

Homework Statement

Find the volume of the solid generated by revolving the region about the x-axis.
Region = the area separated by given function and lines:

x = sqrt(2y)/(y^2 +1)
y=0.

The Attempt at a Solution

i just need to solve the integral of ∫x dy... How? Is there any simple way to solve it?

Don't you want $\pi\int x^2\, dy$? Write out your integral, complete with limits so we know exactly what your problem is.

 

[Nikitin]

that would be the case if I wanted to revolve it around the Y-axis.

Here's a graph showing the problem:http://desmond.imageshack.us/Himg14/scaled.php?server=14&filename=63679970.jpg&res=landing

i need to find the area between the y=1 and the graph (which I require the integral of x to find). that way I can find the average radii of the solid, remake it into a cylinder with the same area and revolve it to find a volume identical to the 1 I'm supposed to find.

Duno if this will work, tho.

 

[LCKurtz]

that would be the case if I wanted to revolve it around the Y-axis.

Here's a graph showing the problem:http://desmond.imageshack.us/Himg14/scaled.php?server=14&filename=63679970.jpg&res=landing

i need to find the area between the y=1 and the graph (which I require the integral of x to find). that way I can find the average radii of the solid, remake it into a cylinder with the same area and revolve it to find a volume identical to the 1 I'm supposed to find.

Duno if this will work, tho.

Yes, I was thinking you were revolving about the Y axis. I see you have added the line y=1 to the problem. To revolve about the x-axis given you have x in terms of y, you want to use a dy integral and the shell method with an integral of this form:$$
V =2\pi\int_a^b yf(y)dy= 2\pi\int_0^1\frac{\sqrt 2 y^\frac 3 2}{1+y^2}\, dy$$which isn't quite the form in your original post. Whether that is an easy integral to work isn't obvious to me at the moment.

 

[Nikitin]

Yes, your method sounds much better than mine. I just learned the shell method yesterday so I don't have any problems with that.

But the integral of x needs to be solved anyway (and it's without thinking about 2*pi outside, imo).