 Hello everyone, welcome to the session of application of multiple integrals, part one. This is Swati Nikam, assistant professor, department of humanities and sciences, Valchan Institute of Technology, Sulapur. At the end of this session, student will be able to find area of region bounded by curves by using double integration. Now let us see area by double integration, we will go for Cartesian coordinate system first. Consider the area enclosed by two plane curves x is equal to phi 1y this and x is equal to phi 2y the other curve intersecting in point A with coordinate AC and point B with coordinate BD. Now consider a strip parallel to x axis, while studying double integration we have seen that to know the limits of integration in the given region we have to consider a strip either parallel to x axis or parallel to y axis. So here we have considered a strip parallel to x axis. On this strip y varies from y has constant limits y is equal to c bottom to y is equal to d top and x has variable limit since strip is parallel to x axis. So x is equal to phi 1y to x is equal to phi 2y and therefore the formula for area is capital A is equal to outer integration c to d inner integration phi 1y to phi 2y of dx dy. Now on the other way consider the area enclosed by two plane curves y is equal to f 1x this curve and y is equal to f 2x this upper curve intersecting in point A and B. Consider a strip parallel to y axis so that on this strip y varies from bottom first which indicates lower limit y is equal to f 1x to top side indicates upper limit y is equal to f 2x and x varies from x limits are from left to right so x has constant limit x is equal to A to x is equal to B and therefore the whole area is capital A is equal to integration A to B inner integration f 1x to f 2x dx dy. Friends before we proceed further pause your video for a minute and answer this simple question evaluate the limits of double integration over the positive quadrant of an ellipse x square by A square plus y square by B square is equal to 1. I hope you have written your answer so this is the required ellipse the limits of double integration over the positive or first quadrant of an ellipse is obtained as follows. This is x y coordinate system I have drawn an ellipse let us consider region A O B because we are interested in positive quadrant or first quadrant so region O A B is the required region of integration. Now to indicate this region let us consider a strip parallel to y axis say P Q while its motion one end is on lower boundary of this strip is y equal to 0 and other end is on an ellipse on this strip y varies from so y has variable limits lower side is x axis that is y equal to 0 to upper side is an ellipse so let us rewrite this equation in terms of y as y is equal to B into under root of 1 minus x square upon A square and on this strip x varies from so we have to write down constant limits for x from left x equal to 0 to write x is equal to A. Now let us solve an example for finding area of the circle x square plus y square is equal to A square to find area of this circle it is convenient to find area of the circle in first quadrant and multiply it by 4 thus the total area of the circle is given by A is equal to 4 times double integration over region R dx into dy. Let us consider the region A O B so in first quadrant the region is A O B and now in order to get its limits let us consider a strip parallel to y axis say P Q while its motion one end is on y equal to 0 so see the lower end is on the x axis which represents y equal to 0 and the other end is on the circle x square plus y square is equal to A square so on this strip y varies from y is equal to 0 to upper end indicates upper limit y is equal to square root of A square minus x square at actually it is plus or minus under root of A square minus x square but we are in positive quadrant so let us consider the positive value only and now x has constant limits x varies from left side x equal to 0 and right side x is equal to A therefore the formula for area becomes now A equal to 4 times integration 0 to A integration 0 to square root of A square minus x square dx into dy which is equal to 4 times integration 0 to A inside the bracket we will integrate with respect to y first because limits of inner integration are in terms of x and outside the bracket dx integration of dy is y simply and substitute the limits upper and lower limits so that it is equal to 4 into integration 0 to 1 inside the big bracket y writing lower limit 0 upper limit square root of A square minus x square into dx. Now in place of y substitute upper limit minus lower limit into dx therefore A equal to 4 times now we will write down formula for this square root of A square minus x square integration as x by 2 into under root of A square minus x square plus A square by 2 into sin inverse of x by A outside the bracket limits 0 to A which is equal to 4 times inside the bracket wherever we have x substitute upper limit A so A square minus A square becomes 0 plus A square by 2 into sin inverse of it is A upon A which is 1 and sin inverse of 1 is pi by 2 minus lower limit in place of x 0 minus sin inverse of 0 is 0 so it is equal to 4 into bracket A square into pi divided by 4 4 get cancelled and therefore area of the standard circle A is equal to A square into pi. Let us have one more example for practice find the area bounded by x equal to 0 y equal to 2x and y is equal to 4 so let us consider a region A o v this is a region bounded by all the three curves and a strip parallel to y axis say pq the point of the intersection of the curve y is equal to 4 and y is equal to 2x is b and which is obtained simply as with coordinate 2 comma 4 thus the required area of region oab is given by A equal to double integration over r dx into dy on the strip y varies from so let us write down the limits see strip is parallel to y axis so limits from bottom at bottom y is equal to 2x and at top y is equal to 4 so y is equal to 2x 2 y is equal to 4 and the constant limits are left side x is equal to 0 to right side x is equal to 2 hence area A is equal to integration 0 to 2 constant limit outside and variable limits integration 2x to 4 dx into dy which is equal to integration 0 to 2 inner integration 2x to 4 dy into dx as the limits are in terms of x so integrate with respect to y first which is equal to integration 0 to 2 integration of dy is y limits 2x to 4 into dx so now in place of y put upper limit minus lower limit which is equal to integration 0 to 2 4 minus 2x in outside dx is equal to integration integration of 4 is 4x minus 2 times integration of x is x square by 2 outside the bracket limits 0 to 2 which is equal to in place of x substitute upper limit 2 so 4 into 2 8 minus this to get cancelled 2 square is 4 minus lower limit in place of x 0 minus minus plus square in place of x square it is 0 minus 0 square 0 which is equal to 8 minus 4 is equal to 4 so required area A is equal to 4 I have referred a textbook named higher engineering mathematics by Dr. B.