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December 20, 2014

December 20, 2014

Posted by **Tiffany** on Friday, April 12, 2013 at 10:33am.

The region in the second quadrant bounded above by the curve y = 16 - x2, below by the x-axis, and on the right by the y-axis, about the line x = 1

I have gathered, that washer method is to be used - (-4,0) is the shaded area.

- calculus -
**Steve**, Friday, April 12, 2013 at 11:09amwashers is a good way:

v = ∫[0,16] π(R^2-r^2) dy

where R=1+|x|, r=1

v = π∫[0,16] (1+√(16-y))^2 - 1 dy

= -π(4/3 (16-y)^(3/2) + 1/2 (16-y)^2) [0,16]

= 640π/3

Or, using shells,

v = ∫[-4,0] 2πrh dx

where r=1-x and h=y

v = 2π∫[-4,0] (x+1)(16-x^2) dx

= 2πx (x^3/4 - x^2/3 - 8x + 16)[-4,0]

= 640π/3

Hmm! shells is less complicated this time.

- calculus -
**Tiffany**, Friday, April 12, 2013 at 3:17pmWhere does the 4/3 and 3/2 come from?

Thank you

- calculus -
**Steve**, Friday, April 12, 2013 at 3:54pm∫(1+√(16-y))^2 - 1 dy

let u = 16-y

du = -dy

∫(1+√u)^2 - 1 du

∫1 + 2√u + u - 1

∫2√u + u

2(2/3)u^(3/2) + 1/2 u^2

because

∫u^n = 1/(n+1) u^(n+1) where n=1/2

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