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
(a)solve by either the disk or washer method
(b)solve by the shell method
(c)state which method is easiest to apply
A quick look at the region bounded by
y=(x-2)^3-2, x=0, y=25
http://www.wolframalpha.com/input/?i=plot+y%3D%28x-2%29%5E3-2%2C+x%3D0%2C+y%3D25
shows me that I should scan horizontally from y = -10 to y = 25
using V = π∫ x^2 dy from y = -10 to 25
I need x^2
y = (x-2)^3 - 2
(x-2)^3 = y+2
x-2 = (y+2)^(1/3)
x = (y+2)^(1/3) + 2
x^2 = [ (y+2)^(1/3) + 2 ]^2
= (y+2)^(2/3) + 4(y+2)^(1/3) + 4
V = π∫( (y+2)^(2/3) + 4(y+2)^(1/3) + 4) dy from -10 to 25
= π( (3/5)(y+2)^(5/3) + 3(y+2)^(4/3) + 4y) | from -10 to 25
= (3/5)(27)^(5/3) + 3(27)^(4/3) + 100 - (3/5(-8)^(5/3) + 3(-8)^(4/3) - 40)
= 729/5 + 243 + 100 -( -96/5 + 48 - 40)
= 516
check my arithmetic
What happened to π in your calculations? Also, could you explain part (b)?
recall that the volume of a shell is its circumference times its height (since the shell is very thin).
Since y(3) = 25, our limits of integration on x are [0,3]
So, using shells,
v = ∫[0,3] 2πrh dx
where r=x and h=25-y
v = 2π∫[0,3] x(25-(x-2)^3-2) dx
= 2π∫[0,3] -x^4+6x^3-12x^2+31x dx
= 1044/5 π
I agree that Reiny's answer should contain pi somewhere.
I think that the disc method ought to consider the function y+10, so that all the y values are positive
y = (x-2)^3 + 8
x = ∛(y-8) + 2
y = 8 to 35
v = π∫[8,35] (∛(y-8) + 2)^2 dy
= 2484/5 pi
Hmmm. Not the same as shells. Better check my math as well.