YACT Calc Question

[SXMP]

I've spent the last hour on this:

Find the volume bounded by:
x=y^2
y=x^2
revolved around x=3

The book says the answer is 17pi/10. No matter what I try, both Washer method or Shell, nothing seems to give me that answer. Someone please tell me that 17pi/10 is not the answer, that this must be a misprint.

Thanks

 

[FelixDeCat]

I dont really have an idea since I didnt study it, but it probably can be answered somewhere. Math forums somewhere perhaps? Its probably something like x raised to they power of y divided by the arctangent of pi subtracted from the inverse of Fermats last theorem, plus 1.

 

[SXMP]

Originally posted by: FelixDeKat
I dont really have an idea since I didnt study it, but it probably can be answered somewhere. Math forums somewhere perhaps? Its probably something like x raised to they power of y divided by the arctangent of pi subtracted from the inverse of Fermats last theorem, plus 1.

I really dont think that is it. I think the answer is 3pi/10. But I just need someone to confirm.

 

[crt1530]

I got 13pi/15.

 

[SXMP]

Originally posted by: crt1530
I got 13pi/15.

Could you put what your Integral looked like before you evaluated?

Thanks

 

[crt1530]

I computed pi times the integral of (3-y^2)^2 - (3-y)^2 evaluated from 0 to 1.

 

[SXMP]

Originally posted by: crt1530
I computed pi times the integral of (3-y^2)^2 - (3-y)^2 evaluated from 0 to 1.

Shouldn't that be square root of y?

 

[crt1530]

Yes. Good catch.

 

[crt1530]

That'll leave you with the book's answer.

 

[SXMP]

Originally posted by: crt1530
That'll leave you with the book's answer.

Sure did. Thanks!