 Hello everyone! Myself, Mrs. Mayuri Kangai, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Valjan Institute of Technology, Solapur. Today, we are going to see multiple integrals. The learning outcome is, at the end of this session, the students will be able to compute the double integrals by using change of order of integration. In the previous videos, you have learned how to change the order of integration. So, pause the video for a minute and answer this question. Change the order of integration. The integral is integration from 0 to 2, integration from 0 to x, f of x, y, dy, dx. Yes, the solution of this example is, the given integral i is integration from 0 to 2, integration from 0 to x, f of x, y, dy, dx. Here, the region of integration is bounded by the curves, tend by the limits of this double integral. Here, the limits of double integrals are from 0 to 2 and from 0 to x. The inner integral is expressed as a function of x. So, the inner integral limits are for the variable y and the outer integral limits are for x. So, the curves which bounds the region of integration are x equal to 0, x equals to 2, y equal to 0 and y equals to x. See the graph. The region of integration is bounded by x equal to 0 and x equals to 2. We know that x equals to 0 is nothing but the y axis and x equals to 2 is a straight line which is parallel to y axis intersects the x axis at the point A whose coordinates are 2, 0. So, the region of integration lies between these two lines and the region of integration for the variable y between y equal to 0 to y equal to x. So, y equal to 0 is nothing but the x axis and y equal to x is the line which passes through the origin. So, the region of integration is the triangle OAB and initial strip is parallel to y axis as the y is expressed as the function of x. To change the order of integration, we will reverse the strip so it becomes parallel to x axis. Now, to find out the limits of this grill, we will move this strip within the region of integration OAB. To find out the limits, we will draw a perpendicular from this point B on y axis which intersects the y axis at the point C02 and the equation of this straight line will be y equals to 2. So, this strip moves within this region of integration which gives us the outer limit as y equal to 0 and y equals to 2. Now, to find out the inner limits, look at the ends of this strip. The lower end which is nearer to y axis is on the line x equals to y and upper end is on the line x equals to 2. So, the inner limits are x equals to y and x equals to 2. So, the given integral i that is integration from 0 to 2, integration from 0 to x, f of xy dy dx. When we are going to change the order of integration, it becomes integration from 0 to 2, integration from y to 2, f of xy dy dx. Now, we will solve the examples on evaluation of double integral by changing the order of integration. The first example is evaluate i equals to integration from 0 to 1, integration from root x to 1, y square upon under root y raise to 4 minus x square dx dy. Observe the limits, the limits are 0 to 1, root x to 1, here the inner integral is having the limit as function of x. So, these are the limits of y and the outer integral is having the limits of x. So, the given order of integration is first with respect to y and then with respect to x. When we are going to integrate first with respect to y, see the integrant, the integration is not possible. But if we change the order of integration, that is the first will be with respect to x and then with respect to y, the integration will be convenient. So, we will change the order of integration. In the given example, the limits of x are 0, 1 and the limits of y are root x to 1, we will change the order of integration. Let us assign the given integral as i, that is integration from 0 to 1, integration from root x to 1, y square upon under root y raise to 4 minus x square dx dy. See the limits which gives us the region of integration. The region of integration is bounded by the curves x equal to 0, x equals to 1, y equal to root x and y equals to 1. Just we have seen that the inner integrals are having the limits of y. So, these are expressed as y equal to root x and y equal to 1. The outer integral are the limits of x, so it can be written as x equal to 0 and x equals to 1. Rewriting these equations, so we can write it as x equal to 0, x equals to 1. Squaring the equation gives us y square equal to x and y equal to 1. Now, we will draw the graph x axis, y axis, equation is x equal to 0 which is nothing but the first curve which we want. Then we will draw the next line x equals to 1 which is parallel to y axis. Now, we will draw the curve y square equal to x which is a parabola and we will draw the line y equals to 1 which is parallel to x axis. This y axis intersects the line y equal to 1 at the point A and line intersects the curves as well as the x equals to 1 at the point B. The coordinates of A are 0, 1 which can be obtained by solving the equations of the curves and the coordinates of B are 1, 1. Again, it can be obtained by solving the curve equations. See the limits x equal to 0 to x equals to 1. So, x equal to 0 is here, x equals to 1 is here. So, the region of integration is between these two lines. Next, y square equal to x to y equals to 1 as y is expressed as a function of x. The initial strip is parallel to y axis whose lower end is on y square equal to x and upper end is on y equals to 1. So, this region OAB is the region of integration. To change the order of integration, we will reverse the strip. So, it becomes parallel to x axis. Now, we will find out the limits of integral. To find out the limits, we will find out the outer limits. So, we will move this strip within the region of integration. It varies between x axis and the line y equals to 1. So, y equal to 0 to y equals to 1 are the outer limits. Now, to find out the inner limits, look at the ends of this strip. Its lower end is on y axis and upper end is on the curve x equals to y square. So, the limits of this integral is y equal to 0, y equal to 1, x equal to 0 and x equals to y square. Now, we will evaluate the integral by this changed order. So, we can write down the given integral i as integration from 0 to 1, integration from 0 to y square, y square upon under root y raise to 4 minus x square dx dy. Now, let us evaluate this integral. First integration is with respect to x. So, we will take the terms of y as constant and outside the integral. So, we can write integration from 0 to 1, y square dy, integration from 0 to y square, 1 upon under root y raise to 4 minus x square dx. Now, to solve this integral, let us assign y square equals to a constant a for inner integral only. Therefore, we can write it as integration from 0 to 1, y square dy as it is. Here, the inner integral is having the limit 0 to y square at the place of y square, a will be there. So, we can write it as integration from 0 to a, 1 upon under root. Now, y raise to 4 is there. So, at the place of y raise to 4, we will write down it as a square minus x square dx. Now, we know that the integration of 1 upon under root a square minus x square is sin inverse of x upon a. So, this integral is integration from 0 to 1, y square dy, sin inverse of x upon a with the limits 0 to a. Now, we will substitute the limits. So, the integral becomes integration from 0 to 1, y square dy, sin inverse of a upon a minus sin inverse of 0. As a upon a is 1 and the sin inverse of 1 is pi by 2, sin inverse of 0 is 0. We can write the given integral as integration from 0 to 1, y square dy, pi by 2 minus 0. As pi by 2 is constant, can be taken outside the integral and we know that the integration of y square is y cube by 3. So, we can write it as pi by 2 into the bracket y cube by 3 with the limits 0 to 1. Again, substitute the limits pi by 2 as it is 1 by 3 minus 0. It gives us the value of integral as pi by 6. Thank you.