1 Answer | Add Yours
The solid formed by graph of y = 1/(x^2) between x=1 and x=2 rotated about the y-axis, can be taken to be a series of cylinders of height dy and a radius given by the value of x.
As y = 1/x^2, we first write this as x^2 = 1/y or x = sqrt (1/y)
Also the limits x = 1 and x = 2 are substituted with y = 1 and y = 1/4.
The volume now is: Int [ pi* x^2 dy], y = 1 to y = 1/4
=> Int[ pi* (sqrt (1/y) )^2 dy] , y = 1 to y = 1/4
=> Int[ pi* (1/y) dy] , y = 1 to y = 1/4
=> Int[ pi* y^(-1) dy] , y = 1 to y = 1/4
=> (-1)*[ pi* y^-2] , y = 1 to y = 1/4
=> (-pi)*[ 1^-2 - (1/4)^-2]
=> (-pi)*[ 1 - 16]
The required volume of the solid is 15*pi.
We’ve answered 318,913 questions. We can answer yours, too.Ask a question