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November 27, 2014

November 27, 2014

Posted by **Anonymous** on Monday, February 25, 2013 at 6:50am.

(i) set up a definite intergral to determine the volume,V1, of a donut having no hole by rotating the right hand semi circle about the y axis.

Please help!

- Calculus -
**Steve**, Monday, February 25, 2013 at 11:09amBogus problem!

The right semicircle is 3 units from the y axis. If you rotate that piece around the y-axis, it will have a 6-unit diameter hole in the middle.

However, the integral as described above can take advantage of the symmetry about the x-axis, and is thus found using washers:

2∫[0,2] π (R^2-r^2) dy

where R = x = 3+√(4-y^2) and r=3

= 2π ∫[0,2] (3+√(4-y^2))^2 - 9 dy

= 4π (3π + 8/3)

Or, integrating over x using shells,

2∫[3,5] 2πrh dx

where r = x and h = y = √(4-(x-3)^2)

= 4π ∫[3,5] x√(4-(x-3)^2) dx

= 4π(3π + 8/3)

Or, using the Theorem of Pappus,

area of semi-circle is 1/2 π * 2^2 = 2π

The centroid is at 3+(4r/3π) = 3+(8/3π)

So, the volume is

2π(2π*(3+(8/3π)))

= 2π(6π + 16/3)

= 4π(3π + 8/3)

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