00:01
In this question we have the 3 density functions.
00:04
The first one is row is equals to k.
00:07
The second one is row is equals to k y and the third one is row is equals to kx.
00:18
We are required to find the mass and center of mass of the square lamina whose vertices are 0 .0, a, 0, 0 comma a and a comma a.
00:35
So let's see how to solve this question.
00:40
The expression to calculate mass can be written as integration 0 to a, integration 0 to a, function row, dy, dx.
00:57
Now substitute all the values, so we will have mass is equals to integration 0 to a, integration 02a, integration 02a, row that means k, dy, d x.
01:09
Now let's integrate this function with respect to y, so we will have integration 0 to a, ky, and here we have the limits 0 to a d x.
01:21
Now substitute these limits, so we will have integration 0 to a, k, a, dx.
01:33
Now let's integrate this with respect to x, so we will have k -a -x and here we have the limit 0 to a.
01:43
So when we further calculate this, we get mass of the lamina m is equal to k -a -square.
01:52
And now let's find the center of mass...