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December 22, 2014

December 22, 2014

Posted by **Andrew** on Tuesday, December 16, 2008 at 7:36pm.

Find The volume of y=x^3 , y=1 , x = 2

revolving around the X axis

- Calculus -
**bobpursley**, Tuesday, December 16, 2008 at 8:01pmV=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 -
**Reiny**, Tuesday, December 16, 2008 at 8:03pmI 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 -
**Damon**, Tuesday, December 16, 2008 at 8:10pmMaybe 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 -
**Damon**, Tuesday, December 16, 2008 at 8:11pmWell, we check this time :)

- Calculus -
**Reiny**, Tuesday, December 16, 2008 at 8:16pmnoticed that too, lol

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

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