How do you find the volume of the solid generated by revolving the region bounded by the curves y = 10 / x², y = 0, x = 1, x = 5 rotated about the y-axis?

1 Answer
Oct 8, 2015

#Volume approx 9.8pi#


Integrate by Method of Rings

(1) Determine the plot of #y=10/x^2#.
(2) In this solution, the positive side of the curve was used (so as not to deal with negative signs ^_^).
(3) Since the curve is to be rotated about the y-axis, the cross section of the solid should be perpendicular to the y-axis and has its area a function of y.
(4) Since the curve is to be bounded from x=1, the inner radius of the ring or the distance of the line x=1 to the axis of rotation (which is x=0) is equal to 1 .
(5) As for the outer radius of the ring, the distance of the curve to the axis of rotation is expressed as #x=sqrt(10/y)#
(6) Hence the area of the ring is, #A(y)=pi(sqrt(10/y)-1)^2#
(7) If we take a differential element, dy, multiply it to the cross sectional area, then integrate it, we get the volume of the solid. As for the limits of integration, find the values of y at x = 1 and x = 5 based on the curve #y=10/x^2#. The limits are from #y=2/5 to y=10#.

(8) Determining the volume, #V(y)=int_(2/5)^10 pi(sqrt(10/y)-1)^2dy#
(9) #Volume approx 9.8pi#