so the limit as u approaches infinity from 4 to 0 of 2pi times the integral from 0 to 4 of (4-x)* -(6(u)^(-1/3) du since we have to rewrite the limits of integration as well because of the u-substitution?
How can I tell that I have to use the shell method?
Would the formula for this be 2pi times the integral from 0 to 4 of x* (6(4-x)^(-1/3) dx ?
So then, how would I take the limit of this? What test should I use?
I'm having some trouble with this problem:
Find the volume of the solid of revolution, or state that it does not exist. The region bounded by f(x)= 6(4-x)^(-1/3) and the x-axis on the interval [0,4) is revolved avout the y-axis.
How would I be able to tell whether to use the shell, disk, or...
Thanks so much! I got 5/2!
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I have one more question though. Why wasn't it an indeterminate form? It looked like it would be infinity over infinity.
Also, I tried proceeding from where you left off, applying the FTOC and I got pi * (lim as b approches infinity of (b^-2 +3b^-3) - 4. So then wouldn't that equal pi * ( infinity + 4) so it would be infinity so it would diverge?
Thank you so much for replying! I was just wondering, would it be possible to use lhopital's rule to find the limit since V=πlimt→∞(∫t1x+3x3dx) would be infinity over infinity? I tried that and I got 1/(3x^2) and then tried to apply the fundamental theorem of calculus, but I got the wrong...
Find the volume of the solid of revolution, or state that it does not exist. The region bounded by f(x)= the square root of ((x+3)/(x^3)) and the x-axis on the interval [1,infinity) is revolved around the x-axis.
I tried using the disk method: pi* (sqrt(((x+3)/(x^3)))^2
Then I think I have to...