00:01
Hi, today we are solving the question in which given vertices of the tetrahedron as are 0 ,0, then 3 ,0, then 0 ,0, then 0, 3 ,0, and 0, 0, 0, 0, and 0, 0, 0, 0.
00:32
So from here, x varies from x varies from x such did 0 to 3.
00:45
Similarly, y varies from y such that 0 to 3.
00:55
Similarly, z varies from z such that 0 to 3.
01:06
So from here, let integral.
01:11
As i is equals to triple integral of z into dv over e so from here we can take it as i is equals to triple integral of z into d x d y into d z limits are from x is equal to zero 3, y is equal to 0 to 3 and z is equal to 0 to 3.
01:53
Now putting the limits and solving it further so we get it equals to double integral over here so x is equals to 0 to 3 and y is equals to 0 to 3.
02:07
Here we get it as z square by 2 and limits are from 0 to 3 into d .y into dz.
02:18
Now solving from here we get it as equal to, so from here it is equals to taking 9 by 2 outside integral of double integral of dy and dz dx and here also it will be dx...