# Thread: How do you do this inverse trig question?

1. ## How do you do this inverse trig question?

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I cannot figure out how to do part b

2. ## Re: How do you do this inverse trig question?

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.

3. ## Re: How do you do this inverse trig question?

Originally Posted by 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.
Is that correct?

4. ## Re: How do you do this inverse trig question?

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

5. ## Re: How do you do this inverse trig question?

Originally Posted by Sp3ctre
When x = 1/2 I get y = 2/sqrt(3)? (Just saw Drongoski's question so I tried doing it myself)

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