 Hello everyone, myself, Mrs. Mayuri Kangre, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Valchan Institute of Technology, Sallapur. 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 coordinate system of integration. In earlier videos, we have seen how to change the order of integration, but now we will see how to change the coordinate system of integration. Before that, we will see what is the need of change of coordinate system of integration. We go for change to polar if the one of the curve which will bound the region of integration is circle or the integrand is complicated function of x square plus y square. Before going to see the steps to change the coordinate system, we will see some basics related to it. The first point is the limits of outer integral are always the limits of theta. Second point, the limits of inner integral are always the limits of R. The order of integration of polar form examples is first with respect to R and then with respect to theta. This is because the outer integral is having the limits of theta and inner integral is having the limits of R. So, first integration will be with respect to R and then with respect to theta. Fourth point, within the region of integration, we take the strip along the radius vector which is called as radial strip passing through the pole. Now we will see the steps to change coordinate system of integration. Step one, with the limits, write four curves which will bound the region of integration. Trace out the region of integration. Step two, change the equation of boundaries of region in the polar form by putting x equals to r cos theta, y equals to r sin theta, r equals to under root x square plus y square and dx dy equals to r dr d theta. Draw the radial strip and move. When we move the radial strip, we get the limits. To obtain the outer integral limit, rotate the strip in anti-clockwise direction within the region of integration and write the limits of theta. To obtain the inner integral limit, look at the ends of the strip sliding on the curves. The equation of curves are the limits. And write the limits of r. After changing the coordinate system, we will go for evaluating the integral. Now we will see the examples. First example, evaluate integration from 0 to a, integration from 0 to under root a square minus x square e raised to minus of x square plus y square dx dy. Zern the limits. The limits are from 0 to a and from 0 to under root a square minus x square. The inner integral is having the limits which are expressed as the function of x. So these are the limits of y and outer integral is having the limits of x. So the region of integration is bounded by the curves x equals to 0, x equals to a, y equals to 0 and y equals to under root a square minus x square. Squaring this equation gives us x square plus y square equals to a square which is an equation of a circle whose center is at origin and the radius is a. Also if you observe the integral involves e raised to minus of x square plus y square. So we have to change the coordinate system. Remember that we will draw the graph x axis that is nothing but y equal to 0, y axis that is x equal to 0 which is the first curve, y equals to a, a straight line parallel to y axis, circle x square plus y square equals to a square whose center is at origin and the radius is a. Let the given integral be i. So the region r is the region of integration which is between the lines. y axis and x equals to a and from y equal to 0 to circle y equals to under root a square minus x square. Now to change the coordinate system we will put x equals to r cos theta, y equals to r sin theta, r equals to under root x square plus y square and dx dy equals to r dr d theta. Therefore y equals to under root a square minus x square which is nothing but r square equals to a square. So gives us r equals to a. Equation of circle becomes r equals to a. We take the radial strip which will rotate in the region of integration in anticlockwise direction for the values of theta. Then the x axis becomes the initial line theta equal to 0 and the y axis becomes theta equals to pi by 2. Now to find out the outer integral limit we will rotate this radial strip within the region of integration from theta equal to 0 to theta equals to pi by 2. So the outer integral limits are theta equal to 0 to theta equal to pi by 2. Now to obtain the inner integral limit look at the ends of the strip its lower end is on the pole that is on the origin and upper end is on the circle r equals to a. Therefore r varies from 0 to a. As we know that the outer integral will have the limits of theta and the inner integral will have the limits of r. So the given integral i can be written as integration from 0 to pi by 2 integration from 0 to a e raised to minus x square plus y square is there. So at the place of this x square plus y square we will write down r square and dx dy is replaced by r dr d theta. Now let us evaluate this integral. The first integration is with respect to r theta is treated as constant. So we can write integration from 0 to pi by 2 d theta integration from 0 to a e raised to minus r square r dr. To evaluate this let us substitute r square equals to t which gives us r dr equals to dt by 2 when r is 0 t is 0 and when r is a t is a square. So the given integral i can be written as integration from 0 to pi by 2 d theta at the place of r square we will write down t and at the place of r dr we will write down dt by 2 so it becomes integration from 0 to a square e raised to minus t dt upon 2. The integration of e raised to minus t is minus of e raised to t 1 by 2 is a constant can be taken outside. So we will get it as integration from 0 to pi by 2 d theta 1 by 2 into minus of e raised to minus t with the limits 0 to a square. Substituting the limits we will get it as integration from 0 to pi by 2 d theta 1 by 2 minus of e raised to minus a square plus 1. This bracket is a constant so can be taken outside the integral so 1 by 2 1 minus e raised to minus of a square into the integration of d theta is a theta with the limits 0 to pi by 2. After substituting the limits we will get i as pi by 4 into bracket 1 minus e raised to minus a square. Now we will go for the next example change it to polar form and evaluate integration from 0 to n integration from under root x minus x square to under root 1 minus x square dx dy upon under root 1 minus x square minus y square. The region of integration r is bounded by the curves x equals to 0, x equals to 1, y equals to under root x minus x square and y equals to under root 1 minus x square. Look at the limits the inner integral is having the limits of x so these are the limits of y the outer integral is having the limits of x. Now rewriting these two equations the first equation gives us y square equals to x minus x square so we can write y square plus x square equals to x which can be written as x minus 1 by 2 bracket square plus y square equals to 1 which is a circle with center at 1 by 2 0 and radius 1 and the equation y equals to under root 1 minus x square is nothing but x square plus y square equals to 1 which is a circle center at the origin and radius 1. Now we will draw the graph x axis that is y equal to 0, y axis x equal to 0 the first curve. Now we will draw the line x equals to 1 which is parallel to y axis. Now we will draw the circle x square plus y square equals to 1 and then we have drawn the circle x minus 1 by 2 bracket square plus y square equals to 1 whose the center is at the point 1 by 2 0. We will draw the strip initially the strip is parallel to y axis as y is expressed as the function of x whose lower end is on the circle under root x minus x square the smaller circle and upper end is on the circle y equals to under root 1 minus x square so the region between y axis and x equals to 1 and between the two circles is the region of integration. We change the given example to the polar form by using x equals to r cos theta y equals to r sin theta r equals to under root x square plus y square and dx dy equals to r dr d theta for this we will take a radial strip which will rotate in anticlockwise direction for the values of theta the x axis changes to initial line theta equal to 0 and y axis changes to the line theta equals to pi by 2 when we move this strip within the region of integration it will move from theta equal to 0 to theta equal to pi by 2 so the outer integral limits are theta equal to 0 to theta equals to pi by 2. Now pause the video for a minute and think over it what will be the inner integral limit yes look at the end points of the strip its lower end is on the smaller circle that is x square plus y square equals to x now x square plus y square is r square and x is r cos theta so we can write the equation of the circle as r equals to cos theta which is the lower limit of inner integral and upper end is on the circle y equals to under root 1 minus x square gives us r equals to 1 now let us evaluate the integral by substituting these limits by the substitution we can write down this integration from 0 to pi by 2 integration from cos theta to 1 r dr d theta upon under root 1 minus r square let us substitute 1 minus r square as t so we can write r dr as minus dt upon 2 when r is equals to cos theta t which is equals to sin square theta and when r is equals to 1 t 0 so t varies from sin square theta to 0 so the given integral i can be written as integration from 0 to pi by 2 d theta integration from sin square theta to 0 minus dt upon 2 root t the first integration is with respect to t so theta is considered as constant and taken outside this 1 upon root of t can be written as t raised to minus 1 by 2 dt we will integrate this gives us integration from 0 to pi by 2 d theta minus 1 by 2 t raised to 1 by 2 upon 1 by 2 with the limits sin square theta to 0 substitute the limits gives us integration from 0 to pi by 2 d theta minus 1 this 2 in the denominator get cancelled and we will get the bracket as 0 minus sin square theta bracket raised to 1 by 2 sin square theta raised to 1 by 2 is sin theta so we can write it as integration from 0 to pi by 2 sin theta d theta the integration of sin theta is minus cos theta with the limits 0 to pi by 2 which gives us minus cos of pi by 2 minus minus cos 0 which gives the value of integral as 1 thank you