 Hello everyone. Myself, Mrs. Mayuri Kangre, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Valchin Institute of Technology, Solapur. Today we are going to see multiple integrals, triple integration part 2. The learning outcome is at the end of this session the students will be able to solve the triple integrals. In the previous video lecture we have seen the definition, notation, geometric interpretation and uses of triple integration. Also, we have seen how to evaluate simple triple integrals that is with constant limits. The iterated integral is simple when capital V is a rectangular solid. For example, triple integration over V of f of x, y, z delta V is equals to integration from a to b, integration from c to d, integration from p to q, f of x, y, z, dz, dy, dx where a, b, c, d, p, q are the constants. Then such an integral is called as the simple integral and the order of integration can be decided using the differentials. Here, dz appears first, then dy and then dx, so the order of integration is first with respect to z, then with respect to y and then with respect to x and the limits of z are from p to q, the innermost integral, limits of y are from c to d, the middle integral and the limits of x are from a to b which can be illustrated here. For more complicated shapes, just remember these rules which are analogous to the rules we had for limits of double integral. First rule, the outer limits have to be constant, they cannot depend on any of the variables. Third rule, the middle limits can depend on the variables from the outer integral only, they cannot depend on the variable from the inner integral. Third rule, the inner limits can depend on variable from the outer integral and the variable from the middle integral. For example, triple integral over v f of x, y, z dv equals to integration from 2 to 3, integration from 1 minus z to 0, integration from minus y square minus z square to y square plus z square of f of x, y, z dx dy dz. Here if we observe the limits of inner integral, they are in terms of y and z, so these are the limits of x. So the first integration is with respect to x, if you observe the middle integral, it is expressed in terms of z, so this is the limit of y, so the next integration is with respect to y and the outer integral has the limits constant, so these are the limits of z which is to be evaluated at the end. Therefore, the order of integration is first with respect to x, then with respect to y and then with respect to z. Now, the following integral does not make any sense, see the integral, triple integral over v f of x, y, z dv is equals to integration from y to x, integration from 1 to 2 x, integration from 0 to 1 f of x, y, z dx dy dz. Why this does not make any sense, see its outer limit depend on x and y, middle limit depend on x and inner limits are constant, this integral does not obey the rule of triple integration. Therefore, we say that this integral does not make any sense. Now pause the video for a minute and give the answer of this question, what is the order of integration of the following triple integral? The integral is triple integration over v f of x, y, z dv equals to integration from 0 to 1, integration from 1 to 2 y, integration from 1 to x plus y, f of x, y, z, dz dx dy. I hope you all have written the order of integration, let us check the integration order. The order of integration is first with respect to z, then with respect to x and then with respect to y. See here, the given integral is of this form, here the inner integral has the limits 1 to x plus y, so these are the limits of z, you can find it by using the differential also. The middle integral has the limits 1 to 2 y, so these are the limits of x, see here the middle differential is dx and the outer integral has the limits of y. Therefore, the order of integration is first with respect to z, then with respect to x and then with respect to y. Now let us see the examples. First example, evaluate the integral, integration from 0 to 1, integration from 0 to x, integration from 0 to 1 plus x plus y, dz dy dx. Let us call the given integral as i, observe the integral, here the differentials are arranged in dz dy dx order, so the first integration is with respect to z, then with respect to y and then with respect to x. Irrespective of that, you can find the order of integration by observing the limits. Here the inner integral is expressed in terms of x and y, so these are the limits of z, the middle integral is expressed in terms of x, so this is the limit of y and outer integral will be the limit of x. So, here the order of integration is first with respect to z, then with respect to y and then with respect to x. Now evaluate this integral, i is equals to integration from 0 to 1, integration from 0 to x, integration from 0 to 1 plus x plus y, dz dy dx. First we will evaluate this inner integral, integration from 0 to 1 plus x plus y, dz, the integration of dz is z with the limits 0 to 1 plus x plus y, so the integral becomes integration from 0 to 1, integration from 0 to x, z with the limits 0 to 1 plus x plus y, dy into dx. We will put the limits, so we get the integral as integration from 0 to 1, integration from 0 to x, the upper limit 1 plus x plus y minus the lower limit 0 into dy into dx. We will separate the integral for all the terms, so we can write it as integration from 0 to 1, integration from 0 to x dy plus integration from 0 to x, x dy plus integration from 0 to x of y dy into dx. This integral is separated for all the terms. Now the integration of dy is y, the integration of x with respect to y is xy and the integration of y with respect to y is y square upon 2 with the limits 0 to x, so we can write the integral as integration from 0 to 1 into the bracket y with the limits 0 to x plus xy with the limits 0 to x plus y square upon 2 with the limits 0 to x into dx. Now let us substitute the limits, the upper limit is x and the lower limit is 0, so for the first bracket we can write it as x minus 0, for the second bracket x remains as it is, the limits are for y, so into the bracket x minus 0 plus the y is replaced by the upper limit x minus the lower limit 0 bracket square upon 2. So the integral becomes integration from 0 to 1 into the bracket x minus 0 plus x into x minus 0 plus x minus 0 bracket square upon 2 into dx. Here the integral becomes i equals to integration from 0 to 1, x minus 0 is x, x into x minus 0 is x square and x minus 0 bracket square upon 2 is x square upon 2, so we get the integral as integration from 0 to 1 x plus x square plus x square upon 2 into dx. Now we will integrate this with respect to x, the integration of x is x square by 2, the integration of x square is x cube upon 3, the integration of x square is x cube by 3 into this 1 by 2 which is a constant is written as it is. So we get the integration as x square upon 2 plus x cube upon 3 plus x cube upon 2 into 3 with the limits 0 to 1. Now let us substitute the limits, the upper limit is 1 and the lower limit is 0. So we get 1 square upon 2 plus 1 cube upon 3 plus 1 cube upon 2 into 3 is 6 minus 0 the lower limit which is not written here. Now 1 square upon 2 is 1 by 2, 1 cube upon 3 is 1 by 3 and 1 cube upon 6 is 1 by 6. So we get the value of the integral as 1 by 2 plus 1 by 3 plus 1 by 6. Now we will take the LCM which is 6, so in the numerator we get 3 plus 2 plus 1 which is nothing but 6. So we get 6 upon 6 which cancels each other and we get the value of integral i as 1. Therefore integration from 0 to 1, integration from 0 to x, integration from 0 to 1 plus x plus y dz dy dx is equal to 1. Thank you.