# Calculus - Volume by Integration

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

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1. 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

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2. What happened to π in your calculations? Also, could you explain part (b)?

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3. 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.

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4. 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.

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5. Usine shells, you calculated it with respect to x, but it should be with respect to y. Maybe that's why your answers aren't the same? Could you look over again for me? I'm still not sure about it.

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6. Reiny's answer isn't right at all. I looked over her math.

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7. I was able to calculate the volume for the shell method. I got 1250pi on integral [0,5]. However, I still need help with the disk method. I wasn't able to figure it out and I'm not sure which integral to use.

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