Question

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?

Answer 1

`Volume approx 9.8pi`

Explanation:

Integrate by Method of Rings

Solution:
(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`