 Hello everyone, myself, Mrs. Mayuri Kangal, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Valchan Institute of Technology, Solapur. Today, we are going to see multiple integrals, triple integration, part one. The learning outcome is at the end of this session, the students will be able to solve the triple integrals. The concept of double integral of a function f of xy over a given region in xy plane can be extended further to define triple integral. Consider a function f of xyz defined over a finite region v of three-dimensional space let the region be subdivided into n subintervals delta vn delta v2 up to delta vn. Let the point p with the coordinates xr, yr, zr be a point in rth subinterval. If limit n tends to infinity delta v tends to 0 summation r equal to 1 to n f of xr, yr, zr into delta vr exist, then it is called as triple integral of f of xyz over the region v. Let us see the notations, the triple integral of f of xyz over the region v is denoted as triple integral over the region v f of xyz delta v. Therefore, triple integral over v f of xyz delta v is equals to limit of n tends to infinity delta v tends to 0 summation r equal to 1 to n f of xr, yr, zr into delta vr. Now let us see the uses of triple integral, triple integrals can be used to find first volume just like the double integral and second mass when the volume of the region we are interested in has variable density. Here we can see the geometric interpretation of triple integral. Now let us see the evaluation of triple integral. There are some rules to evaluate a triple integral. Let us see it, first rule, the triple integral can be evaluated by successive single integrals. Rule number 2, the integration always proceeds from inside to outside. Rule number 3, the integration order may coincides with the order that the differentials for example dx or dy appear. So we can specify the integration order by a list of differentials. For example, triple integral over v f of xyz delta v is equals to integration from a to b integration from c to d integration from p to q f of xyz dz dy dx where a, b, c, d and pq are the constants. Then to evaluate this triple integral we will follow the successive single integrals and these integrals will proceed from inner integral to outer integral that is from right to left. Now to decide the order of integration we will see the differentials here the differentials are dz dy dx. So the first integration is with respect to z then with respect to y and then with respect to x. The triple integral can be evaluated by successive single integrals as shown see here from inner bracket to outer where the integration with respect to z is performed first by treating x and y as constant then with respect to y treating x as constant and finally with respect to x is performed. Now pause the video for a minute and solve this example integration from 0 to 1 integration from 0 to 3 x plus a y dx dy. I hope you all have solved this example let us check the solution let the given integral be i equal to integration from 0 to 1 integration from 0 to 3 x plus a y dx dy the first integration is with respect to x. So we can write it as integration from 0 to 1 integration from 0 to 3 x plus a y dx into dy let us integrate with respect to x treating y as constant the integration of x is x square upon 2 y is treated as constant so the integration of dx is x. So we get the integral as integration from 0 to 1 x square upon 2 plus x y with the limits is 0 to 3 into dy. Now let us substitute the limits x is replaced by the limits so we can write it as integration from 0 to 1 3 square upon 2 plus 3 y minus the lower limit is 0 minus 0 into dy so we get integration from 0 to 1 9 upon 2 plus 3 y into dy. Now here we will integrate with respect to y so the integration is 9 y upon 2 plus 3 into y square upon 2 with the limits is 0 to 1 now here again we will substitute the limits so we get 9 upon 2 plus 3 upon 2 minus 0 minus 0 the denominator is same so the addition of numerator gives us 12 upon 2 that is 6 therefore the value of the integral is 6. Now let us go for the examples evaluate the integral integration from 0 to 1 integration from 0 to 2 integration from 1 to 2 x y z square dx dz dy let i equals to integration from 0 to 1 integration from 0 to 2 integration from 1 to 2 x y z square dx dz dy. Let us decide the order of integration here the differentials are arranged as dx dz dy therefore the order of integration is first with respect to x then with respect to z and then with respect to y so we can rewrite the i as integration from 0 to 1 integration from 0 to 2 integration from 1 to 2 x y z square dx dz dy. Let us evaluate the integral we will evaluate this integral from the innermost integral the first integration is with respect to x so we will treat y and z as constant therefore we can take y and z square outside the innermost integral therefore we can write the given integral as integration from 0 to 1 integration from 0 to 2 y z square integration from 1 to 2 x dx dz dy the integration is with respect to x and the integration of x is x square upon 2 therefore we can write the integral as integration from 0 to 1 integration from 0 to 2 y z square into the bracket x square upon 2 with the limits 1 to 2 dz dy now let us substitute the limits upper limit is 2 so 2 square minus the lower limit is 1 1 square upon 2 we get the integral as integration from 0 to 1 integration from 0 to 2 y z square 2 square minus 1 square upon 2 dz dy 2 square is 4 minus 1 is 3 upon 2 which is a constant which can be taken outside the both integrals so we can write it as 3 upon 2 integration from 0 to 1 integration from 0 to 2 y z square dz dy the inner integration is with respect to z so we will treat y as constant with respect to z which can be taken outside the inner integral so we can write it equals to 3 upon 2 integration from 0 to 1 y integration from 0 to 2 z square dz into dy the integration of z square is z cube upon 3 so we get 3 upon 2 integration from 0 to 1 y z cube upon 3 with the limits 0 to 2 into dy now let us substitute the limits so we get 3 upon 2 integration from 0 to 1 y into the bracket the upper limit is 2 so 2 cube minus lower limit 0 upon 3 into dy 2 cube is 8 so we get here this bracket value as 8 upon 3 8 upon 3 is a constant which can be taken outside the integral so we get 3 upon 2 into 8 upon 3 into integration from 0 to 1 y dy here 3 get cancelled from numerator and denominator and 2 4 z 8 so we get it equal to 4 and the integration of y is y square upon 2 with the limits 0 to 1 so we get 4 y square upon 2 with the limits 0 to 1 now substitute the limits 4 into bracket 1 square minus 0 upon 2 which gives us 4 into 1 upon 2 2 2 z 4 so we get the value of integral as 2 therefore i equal to integration from 0 to 1 integration from 0 to 2 integration from 1 to 2 x y z square dx dz dy is equal to 2 thank you.