00:02
So in this case we have to integrate over six separate surfaces corresponding to different sides of the cube.
00:12
These can be done in three pairs on each of which one of the coordinates is constant.
00:21
So for instance we can begin, we'd have the integral, all our integrals will be between zero and a, as all the side of the side of the of the cube we're integrating over our squares in side length a spanning from zero to a over different pairs of partition of universe.
01:07
So first we can look at the planes with constant to x that would be in the yz plane so we'll have one case with x equal to zero and we'll have one case with x equal to zero and we'll have one case with equal to a and in both cases we'll have plus y plus zd so we'll have altogether then an integral of y plus z and an integral of a plus y plus z which we can just combine to an integral of a plus 2 multiplied by y plus z and then this one of course is with respect to y and z and we can add our integrals in this way within the integral side while our integral grounds in this way within the integral sign because the region of integration is the same as each side has the same projection onto the wide plane namely this square with corners at the origin and a a and then in exactly precisely in analogous manner we would have an integral of a plus 2 times x plus z d x d x d z and an integral of a plus 2 times x plus y d x d y now the variables y and x are just dummy variables and we can interchange their names as they like as we like um so the for each pair of surfaces with one quarter and constant uh the integral will be the same so rather than writing out this expression once more with x and z and another time with x -y, we can realize that the resulting integrals of equal are equal.
03:34
Now, therefore, we can just pre -multiply this integral by three, as the three pairs will have the same integral value.
03:45
And then at this point, it's easy to proceed.
03:52
So our a term will integrate to ay, for instance, first of respect to y.
04:01
That domain from 0 to a, that's a squared less 0, is a squared.
04:06
Then integrating that with respect to z, from 0 to a, it is a squared z.
04:12
So a cubed less 0 is a cubed...