00:01
So, we have given here function of p x ,y that is equal to y square.
00:12
So implies that from here we need to find p y differentiation with respect to y that is equal to 2y.
00:24
And we have given here q x y that is equal to x square.
00:33
So from here we need to find that our q x differentiation with respect to x that is equal to 2x.
00:46
So from here our dq upon dx minus del p upon del y is equal to 2x minus 2y that is equal to implies that 2 into x minus y.
01:08
Now we need to plot the graph for this condition for given vertices.
01:14
So let's say this is our y axis and let's say this is our x axis.
01:22
So we have given some point that is 0 ,0.
01:26
Let's say this is point which is 4 ,0 and this is our third vertices which is 0 ,4.
01:36
So this is we have here our triangle which this path is c1, this path is c2 and this path is c3.
01:46
Now let r be the reason closed reason enclosed by c1, c2, c3.
01:54
So this will be our reason.
01:56
So from here we need to write that is double integration of reason r of del q minus del q upon del x minus del p upon del y with respect to dx dy.
02:16
So that is equal to then we have integration 0 to 4, integration 0 to 4 and value of dq upon dx minus dp upon dy is and that is we need to put here into 2.
02:38
Taking 2 outside so it become the value of this is x minus y and now we need to integrate this with respect to dx dy.
02:49
So first we need to integrate this with respect to dx so we will get from here that is 2.
02:57
First we need to integrate this with respect to y then we will get xy minus y square by 2 and limit we have here 0 to 4 and with respect to now integrate dx.
03:15
Then after putting the limit we will get 2 into integrating 0 to 4 it become 4x minus 4 square means 16 and 16 by 2 is 8 and lower limit becomes 0 so it whole value become 0 and now we need to integrate this with respect to dx.
03:42
So after integrating with respect to dx we will get 4 into x square by 2 minus 8x and limit we have here 0 to 4...