Volume of a solid of revolution

[Guest2]

Consider the solid in three dimensions that is formed when the graph of a function $f(x)$, with $f(x) \ge  0$ for all $x \in [a, b]$, is revolved around the $x$-axis on the segment $x \in [a, b]$. Derive the following formula for the volume $V$ of this solid: $V = \pi\int_a^b f^2(x)dx$. Use the formula to establish that the volume of a sphere with radius $R$ equals $V = \frac{4}{3}\pi R^3$.

I don't know how to start this.

 

[MarkFL]

In order to obtain a sphere by revolving a function about the $x$-axis, we need to revolve a semi-circle. Now, a circle having radius $R$, and centered at the origin is:

\(\displaystyle x^2+y^2=R^2\)

If we solve this for $y=f(x)\ge0$ we obtain:

\(\displaystyle f(x)=\sqrt{R^2-x^2}\implies f^2(x)=R^2-x^2\) where $-R\le x\le R$. Because $f^2$ is an even function, and the limits are symmetrical about the $y$-axis, we may use the even function rule to then state:

\(\displaystyle V=2\int_0^R R^2-x^2\,dx\)

Now all that is left is to carry out the integration. :)

 

[Guest2]

Thank you.

How do I prove the first part of the question? Deriving the formula: [tex]\displaystyle V = \pi\int_a^b f^2(x)\;{dx}[/tex]

 

[MarkFL]

Thank you.

How do I prove the first part of the question? Deriving the formula: [tex]\displaystyle V = \pi\int_a^b f^2(x)\;{dx}[/tex]

Think of the sphere as being composed of a bunch of disks, having thickness $dx$. What would then be the radius of an arbitrary disk, and hence its volume $dV$?