Using Integration to calculate volume

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

Now we will start with a function:

$f(x)={\sqrt {x}}$

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

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

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

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

So the area of the disk will be like this $A=\pi r^{2}$

$r={\sqrt {x}}$

$r^{2}=x$

so

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

$A(x)=\pi x$

so what we will get this

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

so $\pi$ is a constant so we can pull it out

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