 Hello everyone, welcome to the session of application of multiple integrals part 4. This is Swati Nikam, assistant professor, Department of Humanities and Sciences, Valchan Institute of Technology, Solapur. At the end of this session, student will be able to find mass of lamina by using double integration. Now, let us see mass of lamina in cartation coordinate system. If the surface density rho of a plane varies from point to point of the lamina and if it can be expressed as a function of the coordinates of point, then the mass of elementary area dA is rho into dA. If rho is represented as f of xy and since dA is equal to dx into dy, so formula for mass of lamina is capital M is equal to double integration over f of xy dx in friends. Before we proceed further, please pause your video for a minute and answer this simple question, what is mean by lamina? I hope you have written your answer. So lamina is a thin layer plate or scale of sedimentary rock, organic tissue or other material. In this video, we are going to find mass of these type of lamina. Let us have an example. 1. A lamina is bounded by y is equal to x square and y is equal to 2x. If the density at any point is given by x into y, find the mass of lamina. Now let us know the region of this lamina, so we will plot these two curves. Here the curve y is equal to x square represents parabola symmetric to positive y axis and y is equal to 2x is the line passing through origin. So y is equal to x square is parabola and this is the line passing through origin. Let us consider region a, o, a and a strip parallel to y axis say pq. On this strip y varies from at bottom y is equal to x square and at top y is equal to 2x. On this strip x varies from x has constant limits, left side x is equal to 0 and right side x is equal to thus the required mass of lamina of the region a, o, a is given by capital M is equal to double integration over r of rho of xy into dx dy which is equal to double integration over r. In the example it is given that the density at any point is given by xy and hence we have to substitute value of as xy here dx into dy which is equal to outer integration 0 to 1, inner integration x square to x over x into y into dx into dy. Now as the limits are in terms of x, so let us integrate with respect to y first, hence it is equal to integration 0 to 1, inner integration x square to x, y, dy outside the bracket x dx. Therefore, M is equal to integration 0 to 1, integration x square to x, y dy, x dx is equal to integration 0 to 1. Integration of y is y square by 2 limits of this integration x square to x into x dx that is equal to integration 0 to 1 in place of y now substitute upper limit and lower limit which is x square by 2 minus x raise to 4 by 2 outside the bracket x dx that is equal to integration 0 to 1 x square into x is x cube by 2 minus x raise to 4 into x is x raise to 5 by 2 into dx. So M is equal to integration 0 to 1 bracket x cube by 2 minus x raise to 5 by 2 into dx is equal to integration of x cube is x raise to 4 upon 4 into 2 minus integration of x raise to 5 is x raise to 6 upon 6 into 2 outside the bracket 0 to 1 which is equal to put upper and lower limit in place of x 1 so 1 by 8 minus 1 by 12 minus lower limit 0 plus 0 which is equal to 1 by 24. So mass of lamina in this example is 1 by 24. Let us have one more example, example number 2 a lamina is bounded by 2 y is equal to x square and 2 x is equal to y square if the density at any point varies as the distance of the point from x axis find the mass of lamina here the curve 2 y is equal to x square represents parabola symmetric to positive y axis and 2 x equal to y square represents parabola symmetric to positive x axis these curves bounds the lamina it is also given that the density is directly proportional to the distance of the point from x axis which is equal to y actually therefore density can be written as rho of x y is equal to k into y where k is proportionality constant. Now the required mass of lamina of the region AOA which is bounded by the 2 curves x square is equal to 2 y and y square is equal to 2 x is capital M equal to double integration over r of x y into dx dy. Therefore let us consider a strip parallel to x axis say Mn which is representation of this region AOA. Now on this strip x y is from x has variable limits at the left boundary x represents y square by 2 2 on the right boundary x is represented as under root of 2 y on this strip y varies from y has constant limits from bottom to top y is equal to 0 2 y is equal to 2 therefore M is equal to integration 0 to 2 inner integration y square by 2 to under root of 2 y k y dx into dy as we are writing value of density is equal to k into y. Therefore it is equal to let us have k outside constant into integration 0 to 2 we will integrate with respect to x integration of dx is x outside the bracket limits y square by 2 to under root of 2 y into y dy. So M is equal to k into integration 0 to 2 x limits y square by 2 to under root of 2 y y dy is equal to k into integration 0 to 2 bracket under root of 2 y minus y square by 2 outside the bracket y dy. We have substituted upper and lower limit in place of x here which is equal to k into integration 0 to 2 bracket root 2 into this is y raise to half into y which is y raise to 3 by 2 minus y square into y is y cube by 2 into dy that is equal to k into bracket let us have integration now root 2 into integration of y raise to 3 by 2 is y raise to 5 by 2 upon 5 by 2 minus integration of y cube is y raise to 4 upon 4 into 2 is 8 outside the bracket limits 0 to 2. Therefore M is equal to k into bracket root 2 into y raise to 5 by 2 upon 5 by 2 minus y raise to 4 upon 8 outside the bracket 0 to 2 that is equal to k into bracket. Now this denominator 5 by 2 shifted into numerator as 2 by 5 into root 2 into in place of y substitute upper limit 2 raise to 5 by 2 minus 2 raise to 4 upon 8 minus lower limit 0 now plus 0 that is equal to k into bracket 2 times 2 raise to 6 by 2 upon 5 minus 16 upon 8. So M is equal to k into bracket 2 into 2 raise to 6 by 2 upon 5 minus 16 by 8 is equal to k into bracket 2 times 2 cube by 5 minus 2 which is equal to k into let us have 2 common bracket 8 by 5 minus 1 that is k into 2 into 3 by 5. So mass of lamina of this region is 6 into k by 5. I have referred advanced engineering mathematics by H. K. Das for creation of this video. Thank you so much for watching this video.