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Evaluate the integral in Example 5 by integrating with respect to $x, y,$ and $z .$

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{'transcript': "were given an integral and were asked to evaluate this integral by integrating with respect to X, y and Z. So this is the integral from example five of this section. And so this is the integral from 0 to 3, integral from 0 to 4. Integral from why over 2 to 1. Plus why over to of the function two x minus y over to what z over three dx dy wide, easy. So in order to evaluate this integral, we will treat it as an iterated integral integrate first with respect to X, then with respect to why and finally with respect to Z. And then we'll compare this to the method used in example, five of this section so integrating with respect to X. This gives us integral from 0 to 3 into world from 0 to 4 and then we get X squared over two minus X y over to plus X Z over three from X equals y over to two. X equals one plus y over to D Y d Z and substituting. This simplifies to the integral from 0 to 3 integral from 0 to 4 of and after doing algebra, which I'm going to leave out. We get one half by plus one minus y over to plus Z over three d Y d Z And now integrating with respect to why we get the integral from 0 to 3 of why plus one squared over four minus y squared over four plus y Z over three from y equals zero toe y equals four. Dizzy and evaluating. We get the integral from 0 to 3 and I'm skipping some algebra here, but we get 9/4 plus four z over three minus 1/4 Dizzy. So taking the anti derivative With respect to Z, we get says to Z plus two Z squared over three from Z equals 0 to 3 and evaluating this gives us 12. And this is of course, the same is the answer from example five, however, an example. Five. We used the change of coordinates, which was much more convenient for evaluating this triple integral"}

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