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October 20, 2014

October 20, 2014

Posted by **jake** on Tuesday, September 11, 2012 at 4:38pm.

The region in the first quadrant bounded above by the curve y=x^2, below by the x-axis and on the right by the line x=1, about the line x=-4

- Math -
**Steve**, Tuesday, September 11, 2012 at 5:00pmI'd suggest using shells for this one:

v = ∫[0,1] 2πrh dx

where r = x+4 and h = y = x^2

v = 2π∫[0,1](x+4)*x^2 dx

= 2π∫[0,1] x^3 + 4x^2 dx

= 2π (1/4 x^4 + 4/3 x^3) [0,1]

= 2π (1/4 + 4/3)

= 19π/6

It can be done with discs, but you have to make them washers:

v = ∫[0,1] π(R^2-r^2) dy

where R = 5, r=4+x = 4+√y

v = π∫[0,1] (25 - (4+√y)^2) dy

= π (9y - 16/3 y^3/2 - 1/2 y^2) [0,1]

= π (9 - 16/3 - 1/2)

= 19π/6

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