# Find the volume V of the solid obtained by rotating the region bounded by the given curves…?

0

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?

Also Check This  LA Hand Symbol???????????

• 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.

• 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 )

• I got 27pi/4…if that’s right I can send solution…

Pain to type into my phone…lol…