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 from here.

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