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Mathematics 13 Online
OpenStudy (anonymous):

what is surface of revolution of y=(x^2+1)^(1/2), [0

OpenStudy (anonymous):

I think they are trying to say find the integral over 0 to 3.

OpenStudy (anonymous):

your ksaa

OpenStudy (anonymous):

there is no such thing as the integral of (2x^2+1)^(1/2)

OpenStudy (anonymous):

am i right?

OpenStudy (anonymous):

I am not sure what you are working on in class. Usually when I see questions like this they something like what is the volume of the solid revolved around a certain x or y. But you can integrate this, let u=2x^2+1 then when you get dx you can solve for u and substitute for x.

OpenStudy (anonymous):

yes that is ture but, the du doesnt cancel out the x that is there cause of the equation

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