# 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?

Oct 8, 2015

$V o l u m e \approx 9.8 \pi$

#### Explanation:

Integrate by Method of Rings

Solution:
(1) Determine the plot of $y = \frac{10}{x} ^ 2$.
(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 $x = \sqrt{\frac{10}{y}}$
(6) Hence the area of the ring is, $A \left(y\right) = \pi {\left(\sqrt{\frac{10}{y}} - 1\right)}^{2}$
(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 $y = \frac{10}{x} ^ 2$. The limits are from $y = \frac{2}{5} \to y = 10$.

(8) Determining the volume, $V \left(y\right) = {\int}_{\frac{2}{5}}^{10} \pi {\left(\sqrt{\frac{10}{y}} - 1\right)}^{2} \mathrm{dy}$
(9) $V o l u m e \approx 9.8 \pi$