How do you find the volume of the solid generated by revolving the graph of a function #f(x)# around a point on the x-axis?

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Answer 1 · 2016-07-26T14:45:28

#V = pi int (f(x))^2 d x#, between the given limits for x, around the point

#V = pi int (f(x))^2 d x#, between the given limits for x, around the

point.

This formula is based on

# triangle x to 0# of #sum F(x) triangle x=int F(x) d x,# over the

given range for x.

Here, an elementary area , in the form of a rectangle of length

f(x) and width #triangle x#, is revolved about its base on the x-axis,

to generate an elementary solid of revolution that is in the form of a

circular disc of radius f(x) and thickness #triangle x#. This elementary

volume for summation is

#triangle V=pi (f(x))^2 triangle x#.

Then, it is summation of the infinite series for V, in the limit.

#V = lim triangle V to 0# of #sum triangle V#

#= lim triangle x to 0# of #sum pi (f(x))^2 triangle x#

#=pi int (f(x))^2 d x#, over the given range for x.