Remember: Volume = pi. int[x^2 dy] (from y=a to y=b)
If you rearrange y = 1/sqrt(1-x^2) to make x^2 the subject, you will get: x^2 = 1 - 1/y^2
And borders:
x=0 y=1
x=1/2 y=sqrt(2)
Hence,
Volume = pi.int[ 1 - 1/y^2 dy ] (from y=1 to y=sqrt(2))
I think you should be able to do this question...