Bogus 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)
A donut is generated by rotating the circle, S, defined by (x-3)^2 +y^2 =4 about the y axis. If the anagement of dunkin donuts wants to know how to price these donuts use intergral calculus to help them
(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!
1 answer
1 answer