 Hello everyone, myself, Mrs. Mayuri Kangre, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Valchan 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, we have learned how to evaluate the double integrals by using change of order of integration. So pause the video for a minute and solve this question. The question is evaluate by changing the order of integration. The given integral is integration from 0 to pi, integration from x to pi, sine y upon y, dy dx. See the solution, the given integral i is integration from 0 to pi, integration from x to pi, sine y upon y, dy dx. Observe the limits, the inner integral is having the limits x to pi, here the limits are expressed in terms 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, but the evaluation becomes easier if we change the order of integration. The given limits of y are x and pi and those for x are 0 and pi, now we will change the order of integration, the given integral is i, limits of the double integral gives us the equations of the curves which bounds the region of integration, so the region of integration is bounded by the curves x equal to 0, x equals to pi, y equal to x, y equal to pi, the graph is shown here, x equal to 0 is nothing but the y axis, x equals to pi is the straight line parallel to y axis, the region of integration lies between these two lines, y is expressed as the function of x, so the initial strip is parallel to y axis whose lower end is on the curve y equal to x, so the strip will be here whose lower end is on the line y equal to x and upper end is on the line y equals to pi, the line y equals to pi is parallel to x axis, so the region OPQ is the region of integration where P is having the coordinates pi pi and Q is having the coordinates 0 pi, these coordinates can be obtained by solving the curve equations, reverse the strip so it becomes parallel to x axis, to find out the limits of the integral move this strip within the region of integration OPQ, when we move this strip within the region it moves from x axis to y equals to pi, so the outer limits are y equal to 0 to y equals to pi, to find out the inner limits look at the ends of the strip, lower end is on y axis that is x equal to 0 and upper end is on the straight line y equals to x, so we get the new limits as y equal to 0, y equals to pi, x equal to 0 and x equals to y, so the integral becomes integration from 0 to pi, integration from 0 to y, sine y upon y dy dx, here the inner integral is having the limits of x, so the first integration will be with respect to x, when we integrate with respect to x first we will treat y as constant, so this integration from 0 to pi, integration from x to pi, sine y upon y dy dx can be written as integration from 0 to pi, sine y upon y dy into integration from 0 to y dx, now we know that the integration of dx is x with the limits 0 to y which gives us the integral as integration from 0 to pi, sine y upon y into y dy, the y gate cancelled, so the integral becomes integration from 0 to pi, sine y dy, we know that the integration of sine y is minus cos y with the limits 0 to pi, it gives us minus cos pi, minus of minus cos 0 becomes minus cos pi plus cos 0, as cos pi is minus 1, minus of minus 1 becomes plus 1, plus cos 0 is 1, so we get the value of the integral as 2, now we will go for the examples, first example, change the order of integration and evaluate, integration from 0 to 1, integration from x square to 2 minus x, xy dx dy, in this integral look at the limits of the integral, the inner integral is having the limits which are expressed as the function of x, so these are the limits of y and outer limits are constants, these are the limits of x, which shows that the given order of integration is first with respect to y and then with respect to x, with the limits for x as 0, 1 and for y as x square to 2 minus x, now let us assign the given integral as i, the region of integration is bounded by the curves x equals to 0, x equals to 1, y equal to x square and y equals to 2 minus x, these equations are obtained from the limits of the double integral, now to trace the region of integration, we will draw the graph x axis that is y equal to 0, y axis x equal to 0 which is the first curve, now we will draw the line x equals to 1 which is parallel to y axis, now draw the graph y equals to x square a parabola and the line y equals to 2 minus x, which intersects the y axis at the point a, whose coordinates are 0 to and intersects the curve and the straight line at the point b, whose coordinates are 1, 1, these coordinates can be obtained by solving the curve equations, now here the y is expressed as a function of x, so initial strip is parallel to y axis, this strip moves between the lines y axis and x equals to 1 and whose lower end is on the curve y equal to x square and upper end is on the line y equals to 2 minus x, so o, b, a is the region of integration, now to change the order of integration reverse the strip, so it becomes parallel to x axis, when we move this strip within the region of integration, see the upper end of the strip first moves on the straight line and then on the parabola, so the region of integration here divides in two regions r1 and r2, now we will find out the limits of both the regions separately, in the region r1, when we move this strip within the region of integration, it moves from x axis to the line bc, where bc is a perpendicular drawn from the point b on a y axis, then the equation of this line becomes y equals to 1, so this strip moves from y equal to 0 to y equals to 1, which are the outer limits, to find out the inner limits look at the ends, its lower end is on y axis and upper end is on the curve y equals to x square, so the inner limits are x equal to 0 to x equals to root y, to find out the limits in region r2, let us draw a perpendicular from the point a on line x equals to 1, the equation of this line is y equals to 2, when this strip moves within the region of integration a, b, c, it moves from y equals to 1 to y equals to 2, so the outer limits are y equal to 1 to y equals to 2, look at the ends of this strip, its lower end is on the y axis and upper end is on the line y equals to 2 minus x, which can be written as x equals to 2 minus y, so the inner limits are x equals to 0 to x equals to 2 minus y, now let us evaluate this integral, i is given as integration from 0 to 1, integration from x square to 2 minus x, x, y, dx, dy, as the region splits into two regions r1 and r2, this integral also divides into two integrals, the first integral is integration from 0 to 1, integration from 0 to root y, x, y, dx, dy which is for region r1 and for r2 the integral from 1 to 2, integration from 0 to 2 minus y, x, y, dx, dy, the first integration is with respect to x, so we will treat y as constant, we can take y dy outside the integral for both the integrals, so it becomes integration from 0 to 1, y dy, integration from 0 to root y, x dx plus integration from 1 to 2, y dy, integration from 0 to 2 minus y, x dx, we know that the integration of x dx is x square by 2, so we can write it as integration from 0 to 1, y dy, x square by 2 with the limit 0 to root y, plus integration from 1 to 2, y dy, x square by 2 with the limit 0 to 2 minus y, substitute the limits, we can write y as it is, square of this root y becomes y, so we get y into y by 2 dy plus integration from 1 to 2, y into, this x is replaced by 2 minus y, we can get 2 minus y bracket square upon 2 dy, so first integral gives us integration from 0 to 1, y square upon 2 dy, plus second integral gives us integration from 1 to 2, y into bracket, squaring the bracket gives us 4 minus 4 y plus y square, whole divided by 2 into dy, 1 by 2 taken outside the integral for both the integrals as it is constant, now integration of y square is y cube by 3 with the limit 0 to 1, again in the second integral, first term is 4 y, its integration is 4 y square by 2, minus sign as it is, the second term is 4 y square, 4 as it is the integration of y square is y cube by 3 plus, third term is y into y square gives us y cube, whose integration is y raise to 4 upon 4 with the limits 1 to 2, now we will substitute the limits, for the first integral we get 1 by 2 into the bracket 1 by 3 plus for the second integral we get 1 by 2 into bracket, 4 into 2 square upon 2 minus 4 into 2 cube upon 3 plus 2 raise to 4 upon 4, the upper limit is substituted, now let us substitute the lower limit which gives us minus 4 upon 2 plus 4 upon 3 minus 1 by 4, we will evaluate this which gives us 1 by 6 plus 1 by 2 into bracket 8 minus 32 by 3 plus 4 minus 2 plus 4 by 3 minus 1 by 4, simplifying this integral gives us the value of given integral as i equals to 3 upon 8, thank you.