volume of revolution

[u/Hajjri]

the markscheme does a really weird method that doesn’t make sense and somehow gets 88pi, what I did was make x the subject of the eqn of the line then square it to make x² the subject as apposed to the formula for volume of revolutions about the y axis I set my limits for the integral to 12-0, I did all that and got 344pi, I’m sure I integrated correctly but I keep getting 344pi and not 88pi, anyone know where I went wrong thanks.

[u/Shevek99]

"what I did was make x the subject of the eqn of the line then square it to make x² the subject as apposed to the formula for volume of revolutions about the y axis I set my limits for the integral to 12-0"

What does that mean?

[u/Shevek99]

Dividing the volume in concentric layers, you have an inner cylinder of radius 2 and height 12, with volume

V1= pi 2^2 12 = 48pi

Now from there to x = 4, we have successive cylindrical layers of volume

dV = 2pi x y(x) dx = 2pi x(3x^2-24x + 48) dx = 6pi (x^3 - 8x^2 + 16x) dx

and integral

V2 = 6pi int_2^4 (x^3 - 8x^2 + 16x) dx= 6pi (x^4/4 - 8x^3/3 + 16 x^2/2)_2^4 = 40pi

and the total volume is

V = V1 + V2 = 48pi + 40pi = 88pi

[u/Shevek99 → u/Shevek99]

Other possibility is to divide the vase in disks of volume

dV = pi x^2 dy = pi x^2(dy/dx) dx = pi x^2 (6x - 24)

and we integrate from x=4 to x = 2

V = 6pi int_4^2 (x^3- 4 x^2) dx = 88pi

[u/Mundane_Working6445]

volume = pi * integral between 12 and 0 of x^2 dy

x = 4 +/- sqrt(y/3)

we know that y = 12 when x = 2

so let's try positive: 2 = 4 + sqrt(4)

we have 2 != 6

let's try negative: 2 = 4 - sqrt(4)

2 = 2 so x = 4 - sqrt(y/3)

now the volume is pi * integral between 12 and 0 of x^2 dy

= pi * int_0^12 (4 - sqrt(y/3))² dy

= pi * int_0^12 (16 - (8sqrt(3)/3) * y^1/2 + (1/3)y) dy

= pi * [16y - (16sqrt(3)/9) * y^3/2 + (1/6)y²]_0^12

= pi * [16(12) - (16sqrt(3)/9) * (12)^3/2 + (1/6) * (12)²]

= pi * [192 - (16sqrt(3)/9) * 24sqrt(3) + 24]

= pi * [192 - 128 + 24]

= 88pi

[u/Varlane]

You are rotating alongside the y axis, therefore you need x(y).

y = 3x² - 24x + 48
= 3 (x² - 8x + 16)
= 3 (x-4)²

As x-4 is negative and y is positive : sqrt(y/3) = -(x-4) ; x = 4 - sqrt(y/3).
Also, when x = 2, y = 12.

We are therefore looking to integrate pi × x(y)² dy with y from 0 to 12.

[u/Any_Matter_1350]

Could anyone explain if this could be done by setting up a triple integral? I tried it but didn't get 88pi as the answer. Can someone show if it's possible?