00:01
All right, so for problem four, we have to evaluate this triple integral.
00:05
As you can see, we're integrating with respect to x first, which means we're going to treat y and the z as constants, which means the z over y plus 1, all in itself is going to be treated as a constant.
00:19
So we can move it outside of the integral and treat it as a constant multiple.
00:25
So it's going to be z over y plus 1 times the integral from 0 to root 1 minus z squared of 1 dx we integrate this with respect to x with respect to x so it's just going to be x from 0 to root 1 minus z squared and we're just going to be left with root 1 minus z squared and we multiply this with a constant multiple that is the constant with respect to x and we continue integrating so if we rewrite the integral is just going to be the integral from 0 to 1, the integral from 0 to 1, z times 1 root 1 minus z squared all over y plus 1 d z d y.
01:17
And now we can split apart this double integral into two separate integrals, and i'll show you how.
01:23
So for the left integral will just be the integral for y.
01:28
So the integral for y is the outside integral.
01:31
So it's the integral from 0 to 1.
01:35
And we're going to like factor out a part of the integrant that has the y variable.
01:42
In this case, it's 1 over y plus 1, d .y...