View source for 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 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 
<math>f(x)= \sqrt x</math>
<math>\int_{0}^{1} A(x)dx</math>

so the radius of a circle will be r and <math> r = \sqrt x</math> 

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 <math> A = \pi r^2</math> 

<math> r = \sqrt x</math> 

<math> r^2 = x</math> 

so 

<math> A = \pi {(\sqrt x)}^2</math> 

<math> A(x) = \pi x</math> 

so what we will get this 

<math>\int_{0}^{1}\pi x dx</math>

so <math> \pi </math> is a constant so we can pull it out 

<math>\pi \int_{0}^{1} x dx</math>