How do you do this inverse trig question? (1 Viewer)

[emprxss]

I cannot figure out how to do part b

 

[fluffchuck]

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.

 

[Drongoski]

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.

Is that correct?

 

[Sp3ctre]

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.

When x = 1/2 I get y = 2/sqrt(3)? (Just saw Drongoski's question so I tried doing it myself)

 

[fluffchuck]

When x = 1/2 I get y = 2/sqrt(3)? (Just saw Drongoski's question so I tried doing it myself)

Yes, sorry, my bad.