# 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?

Jul 26, 2016

$V = \pi \int {\left(f \left(x\right)\right)}^{2} d x$, between the given limits for x, around the point

#### Explanation:

$V = \pi \int {\left(f \left(x\right)\right)}^{2} d x$, between the given limits for x, around the

point.

This formula is based on

$\triangle x \to 0$ of $\sum F \left(x\right) \triangle x = \int F \left(x\right) d x ,$ 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

$\triangle V = \pi {\left(f \left(x\right)\right)}^{2} \triangle x$.

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 $\sum \pi {\left(f \left(x\right)\right)}^{2} \triangle x$

$= \pi \int {\left(f \left(x\right)\right)}^{2} d x$, over the given range for x.