00:01
In this question, we are given with z equals to 1 minus y square and x plus 6 z equals to 6 and x equals to 0 z equals to 0.
00:16
And the density function x, y, z equals to 5.
00:23
Now, mass is given by mass equals to integration from minus 1 to plus 1, integration from 0 to 1 minus y square and integration from 0 to 6 minus 6z.
00:41
Z and the density function that is 5 dx dz dy.
00:47
So, on integrating this will be integration from minus 1 to plus 1, integration from 0 to 1 minus y square and taking 5 outside of integration.
01:00
So, this will be x and the limit will be from 0 to 6 minus 6 z dz dy.
01:09
Now, now on substituting upper limit because on substituting lower limit it will be 0.
01:14
So taking 5 out of integration.
01:17
So integration from 0 to 1 minus y.
01:21
On substituting limit this will be 6 minus 6z dz dy.
01:27
Then on further integrating this will be plus 5 integration from minus 1 to plus 1.
01:34
This will be 6z minus 6z square upon 2 and the limit is again from 0 to 1 minus y square dz.
01:44
Now, d1.
01:45
Now, on substituting lower limit, it will be 0...