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Calculus

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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

  • Calculus -

    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.

  • Calculus -

    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.

  • Calculus -

    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)

  • Calculus -

    Well, we check this time :)

  • Calculus -

    noticed that too, lol

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

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