Calculus Evaluate volume of the curve y=ln(x),y=1,y=2,x=0 about the y axis?

V = pi ∫ x^2 dy (disks method)

x = e^y

V = pi ∫ e^2y dy

y = 1 to 2

pi (1/2 e^2y)|

pi/2 ( e^4 - e^2)

y=e^-x, y=1, x=2, about y=2

e^-x = 1 at x = 0.

V = pi ∫ (y - 2)^2-1 dx

x = 0 to 2

y - 2 because we are rotating about y = 2

the -1 comes because we are bounded above by y = 1

pi ∫ (e^-x - 2)^2 - 1 dx

pi ∫ e^-2x - 2e^-x + 3 dx

pi (-1/2 e^-2x + 2 e^-x + 3x)| [0,2]

pi (-1/2 e^-4 + 1/2 + 2 e^-2 - 2 + 6)

pi ( 2 e^-2- 1/2 e^-4 + 4.5)

This kind of difficult question but not impossible as it helps if graph this

To make this massively simply you can just type

Evaluate volume of the curve x = e^y from x = 1 to x = 2

around the y - axis

use the disk integration method as

pi * int_a^b r(y) dy as

r(y) = e^y a = 1 & b = 2

pi * int_1^2 e^y dy

pi*(e^2 - e)

e*pi(e - 1)