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.
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
5 answers
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.
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)
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.
but, after all, I am on third cup of coffee.