Use the shell method to find the volume of the solid of revolution obtained by rotating the region bounded by y = 1 and y = tan x about the x-axis from x = 0 to x = pi/4
Can someone check this please, i'm kind of rusty on the shell method.
so the shell method uses the formula:
V = int{2pi*x*h}dx where x is the radius and h is the height. since with this problem we are rotating about the x-axis, we will do our integral with respect to y.
y = tan(x)
=> x = arctan(y) ...........this is the height, see diagram
so
V = int{2pi*y*h}dy evaluated between 0 and 1
=> V = int{2pi*yarctan(y)}dy
=> V = 2pi*int{yarctan(y)}dy
=> V = 2pi [(y^2 + 1)/2 * arctan(y) - y/2] .......i used a formula forthis, but the integral can be done by parts i believe.
now we evaluate:
=> V = 2pi(arctan(1) - 1/2)
=> V = 2pi(pi/4 - 1/2)
=> V = (pi^2)/2 - pi
=> V = (pi^2 - 2pi)/2