Using the Disk Method or washer method (im not sure which one)
Find The volume of y=x^3 , y=1 , x = 2
revolving around the X axis
V=PI*INT y^2 dx
=PI INT x^6 dx from x=1 to x=2
PI*y^2 is the area of the disk
dx is its thickness.
I will assume you want to rotate the region bounded by your three curves.
vol = π(integral) (x^6 - 1)dx
= π[(x^7)/7 - x]from 1 to 2
= pi[128/7-2 - (1/7 - 1)]
= π[127/7 - 1) cubic units
check my arithmetic, I have been making some silly typing errors lately.
Maybe you mean the area between the horizontal line y = 1 and the curve y = x^3 from their intersection at (1,1) to the vertical line x = 2 spun around the x axis?
That would be washers.
inner radius = 1
outer radius = x^3
area of washer = pi (x^3)^2 - pi
= pi (x^6-1)
integrate that times dx from x = 1 to x = 2
pi ( x^7/7 - x) at x = 2 = pi 128/7 -14/7) = pi (114/7)
pi (x^7/7 - x) at x = 1 = pi(1/7 - 1)= pi (-6/7)
so
pi (120/7)
Well, we check this time :)
noticed that too, lol
but, after all, I am on third cup of coffee.
To find the volume of the solid formed by revolving the region between the curve y = x^3, the line y = 1, and the x-axis from x = 0 to x = 2 about the x-axis, we can use the disk method.
The disk method involves using infinitesimally small disks to approximate the volume of the solid. We integrate the cross-sectional area of each disk over the interval [0, 2] to find the total volume.
First, let's find the equation of the region bounded between y = x^3, y = 1, and the x-axis. The curve y = x^3 intersects the line y = 1 at x = 1, since 1^3 = 1. So, the region can be defined by the inequalities x^3 ≤ y ≤ 1.
The radius of each disk will be equal to the x-coordinate of the function y = x^3, which is x. The area of each disk is given by A = π * radius^2 = π * x^2. Integrating this area over the interval [0, 1], we get the volume V as follows:
V = ∫[0, 1] π * x^2 dx
To evaluate this integral, we can rewrite it as:
V = π * ∫[0, 1] x^2 dx
Integrating x^2 with respect to x gives:
V = π * (x^3/3) | from 0 to 1
V = π * (1^3/3 - 0^3/3)
V = π * 1/3
Hence, the volume of the solid generated by revolving the region bounded between y = x^3, y = 1, and the x-axis from x = 0 to x = 1 about the x-axis is π/3 cubic units.