Posted by **Beth** on Friday, December 2, 2011 at 11:43am.

Find the colume, in cubis units, generated by revolving the region bounded by y=x^3 x=0 and y=8 around the line x=7

- Calculus -
**Steve**, Friday, December 2, 2011 at 12:23pm
Best to use discs here, I think, integrating along y from 0 to 8

V = Int(pi*(R^2 - r^2) dy) [0,8]

where r = 7-x and R=7 and x = y^1/3

V = Int(pi*(49 - (7-y^1/3)^2)dy)[0,8]

= pi*Int(49 - (49 - 14y^1/3 + y^2/3) dx)[0,8]

= pi*Int(14y^1/3 - y^2/3)[0,8]

= pi*(21/2 y^4/3 - 3/5 y^5/3)[0,8]

= pi*(21/2 * 16 - 3/5 * 32)

= pi(168 - 96/5)

= 148.8pi

= 467.46

What the heck. Let's try shells:

V = Int(2pi*r*h dx)[0,2]

= 2pi Int((7-x)*(8-x^3) dx)[0,2]

= 2pi Int(x^4 - 7x^3 - 8x + 56 dx)[0,2]

= 2pi (1/5 x^5 - 7/4 x^4 - 4x^2 + 56x)[0,2]

= 2pi (1/5 * 32 - 7/4 * 16 - 4*4 + 112)

= 2pi (74.4)

= 148.8 pi

= 467.46

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