Re: help with some calc 2
jay shark did this incorrectly, and the fact that he obtained -1/3 as a volume should have made it obvious.
To solve the problem, find the cross sectional area and integrate to get volume.
If you use shell's method for 1, you get the height of the solid is (sqrt(x)-x^2). You still need the radius which is equal to sqrt(x).
This makes the volume=2*pi*integral(sqrt(x)*(sqrt(x)-x^2)) evaulated from 0 to 1.
volume=2*pi*[1/2*x^2-2/7*x^(7/2)] evaluated at x=0 to x=1
After doing the algebra, you obtain V=3/7*pi.
2)You can also use shell's method here. Except the axis of revolution is translated to the left by 1, so radius would be (1+x). You should be able to find the height. From there, you set up the integral and evaulate it from 0 to 3.
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