Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = 27x^3, y = 0, x = 1; about x = 2
I tried but got 42pi/45, which is wrong…
I am supposed to take the integral with respect to y correct? The part that messes me up is the fact that there is a space in between the line or rotation and the object. I know how to use the washer method, just probably messed up somewhere along the way. any help?
3 Answers
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We’ve got y in terms of x and a vertical axis, so let’s use shells.
Each shell has radius r = 2 – x
and height h = y = 27x³, so
V = ∫[a,b] 2πrh dx = 2π∫[0,1] (2-x)27x³ dx = 2π((27/2)x^4 – (27/5)x^5) |[0,1]
V = 2π(27/2 – 27/5) = (π/5)(135 – 54) = 81π/5
Money.
If you must use washers, you need dy slices, so
x = (³√y)/3, and the limits are 0 ≤ y ≤ 27.
Each washer has outer radius R = 2 – x = 2 – (³√y)/3
and inner radius r = 2 – 1 = 1. Then
V = ∫[a,b] π(R² – r²) dy = π∫[0,27] ( (2 – (³√y)/3)² – 1) dy
Plug and chug. Barring error, you’ll get 81π/5.
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i got 2511π / 5 .. is it correct ?
V = ∫ πx^2 dy … ( y from a → b )
y = 27x^3 => x^2 = 1/9 y ^(2/3)
x = 1 => y = 27
x = 2 => y = 216
=>
V = ∫ π ( 1/9 y^2/3 ) dy … ( y from 27 → 216 )
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I got 27pi/4…if that’s right I can send solution…
Pain to type into my phone…lol…
