Using Integration to calculate volume

This time we will start to learn how to use integration to calculate the volume.

Now we will start to calculate if the function rotating 360 degrees on the x-axis, and the function will never leave the x-axis, and then it will form a 3-dimensional volume. We will call this solid of revolution because we obtain it by revolving a region about a line.

if we say the function is $f(x)={\sqrt {x}}$ $\int _{0}^{1}A(x)dx$

The radius of a circle will be r and $r={\sqrt {x}}$

And we can understand it as the volume is made of many disks and it will form this volume.

The area of the disk will be A and$A=\pi r^{2}$

$r={\sqrt {x}}$

$r^{2}=x$

So replaced r with ${\sqrt {x}}$

$A=\pi {({\sqrt {x}})}^{2}$

$A(x)=\pi x$

so what we will get this

$\int _{0}^{1}\pi xdx$

Because $\pi$ is a constant so we can pull it out of the intergerl.

Then we will get.

$\pi \int _{0}^{1}xdx$