I redid my shells because of a typo, and I got
2π∫[0,5] x(25-((x-2)^3-2)) dx
= 500π
How did you get 1250π? What was your integral?
We need to integrate over x, because the thickness of the shells is dx, not dy.
As for discs,
x = ∛(y+2) + 2
v = π∫[-10,25] πx^2 dy
= π∫[-10,25] (∛(y+2) + 1)^2 dy
= 500π
Find the volume of the solid generated by revolving the region bounded by the given lines and curves about the y-axis:
y=(x-2)^3-2, x=0, y=25
Solve by either the disk or washer method.
I calculated the volume using the shell method and got 1250pi. However, I can't figure out how to calculate it using the disk method. The answers should be the same unless I calculated the volume using the shell method incorrectly.
2 answers
I also checked my calculation in our previous post
http://www.jiskha.com/display.cgi?id=1385311297
I carelessly dropped the π in my last 3 lines, and should have used my calculator to add up the terms in my last line.
My last line should have been 500π
which then also agrees with Steve's new answer using shells
So you have your two methods,
mine using disks
Steve's using shells
both 500π
http://www.jiskha.com/display.cgi?id=1385311297
I carelessly dropped the π in my last 3 lines, and should have used my calculator to add up the terms in my last line.
My last line should have been 500π
which then also agrees with Steve's new answer using shells
So you have your two methods,
mine using disks
Steve's using shells
both 500π