 Hello everyone, myself, Mrs. Mayuri Kangae, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Vulture Institute of Technology, Solaapur. Today we are going to see multiple integrals, Part 1. The learning outcome is, at the end of this session, the students will be able to solve the multiple integrals over the region R. Previously, we studied the concept of double integrals and examined the tools needed to compute them. We learned the techniques and properties to integrate the function of two variables over the rectangular region, that is, with given limits of double integration. In this section, we consider the double integrals defined over the region R, which is a general bounded region. Most of the previous results holds in this section also. Double integrals of the functions defined over general bounded region R on a plane are expressed in three types. Type 1, the examples in Cartesian system, type 2, the examples in Polar system, type 3, the examples expressed in Cartesian system, but reducible to Polar system. In this video, we will see the examples expressed in Cartesian system. Let us see the steps to solve double integral over the region R. Step 1, draw the region R and trace out the region of integration. Step 2, find the limits of region R using vertical or horizontal strip. Here the region is expressed in Cartesian system, so to find the limits of region R, we use the vertical or horizontal strip and find the limits. Step 3, solve the integral. Now pause the video for a minute and give the answer of this question. What is the nature of following curves? y equals to x cube and x plus y equals to 2. I hope you all have written the solution. In the previous unit, curve tracing, we have seen how to trace a curve. Now here, y equals to x cube is symmetrical in opposite coordinates and passes through the origin. X plus y equals to 2 is a straight line with intercept 2 on both the axes. Let us see it graphically. Here y equals to x cube is a curve which is symmetrical in opposite coordinates and passing through the origin and the line x plus y equals to 2 is a straight line which cuts the 2 axes making an intercept 2, so it intersects x axis at the point 2 0 and it intersects the y axis at the point 0 2. Now let us go for examples, calculate the integral double integral over R x dx dy where R is bounded by y equals to x cube, x plus y equals to 2 and x equals to 0. It is given that the region R is bounded by the curves y equals to x cube, x plus y equals to 2 and x equals to 0. So first we will draw the region of integration, then we will find out the limits of integration and then we proceed to evaluate the integral. So let us draw the graph x axis which has the equation y equal to 0, y axis with the equation x equal to 0 which is one of the curve which bounds the region of integration. The x axis and y axis intersect at the point o that is origin. Now let us draw the curve y equals to x cube which passes through o. Now let us draw the line x plus y equal to 2 which intersects the x axis at the point 2 0 and y axis at the point 0 2. Just we have seen this, the two curves x plus y equals to 2 and y equals to x cube intersect at the point 1 1. This point of intersection can be obtained by solving the two equations y equals to x cube and x plus y equals to 2 simultaneously. The region of integration is between the line x plus y equals to 2, y equals to x cube and x equals to 0. So this region is the region of integration. Now let us find out the limits of integral. To find this we will draw a strip for the sake of simplicity. Let us draw a vertical strip. This strip will move within the region of integration which will give us the limits of outer integral and the ends of this strip will give us the limits of inner integral. As the strip is parallel to y axis the outer integral limits are the limits of x and the inner integral limits are the limits of y. We will draw a line which is parallel to y axis passing through the point 1 1. The equation of the line is x equals to 1. Now as the strip is vertical we can get the outer integral limit by moving this strip within the region of integration. When we move this strip within the region of integration it moves from x equals to 0 to x equals to 1. So the outer integral limits are x equals to 0 to x equals to 1. Now we can find the inner integral limits. To find the inner integral limits look at the end of this strip. Its lower end is on the curve y equals to x cube and upper end is on the straight line x plus y equals to 2. As these are the limits of y we will express both the limits in terms of y. So we get the inner integral limits as y equals to x cube to y equals to 2 minus x. So the given integral double integral over r x dx dy becomes integration from 0 to 1 integration from x cube to 2 minus x x dx dy. Now let us evaluate this integral. Here the integral is double integral over r x dx dy equals to integration from 0 to 1 integration from x cube to 2 minus x x dx dy. Observe the limits as the inner limits are expressed in terms of x. These are the limits of y. So first integration is with respect to y and then with respect to x. When we integrate with respect to y we treat x as constant. Therefore the given integral double integral over r x dx dy can be evaluated treating x as constant. So this integral can be written as integration from 0 to 1. This is taken outside the inner integral as the integration is with respect to y. So we get integration from 0 to 1 x integration from x cube to 2 minus x dy into dx. Now let us integrate dy. The integration of dy is y. So we get the integral as integration from 0 to 1 x y with the limits x cube to 2 minus x dx. Now let us put the limits. So we get integration from 0 to 1 x into the bracket, the upper limit 2 minus x minus the lower limit x cube into dx. Now we will multiply this x to the bracket. So we get the integral as integration from 0 to 1 2x minus x square minus x raise to 4 into dx. Now we will integrate this bracket with respect to x. So we get 2 as it is. The integration of x is x square by 2 minus sign as it is. The integration of x square is x cube by 3 again minus sign as it is. The integration of x raise to 4 is x raise to 5 upon 5. So we get the integration as 2x square upon 2 minus x cube by 3 minus x raise to 5 upon 5 with the limits 0 to 1. Here the 2 get cancelled and we get x square only. First we will put the upper limit. So we get 1 minus 1 upon 3 minus 1 upon 5 as we know that the square cube and raise to 5 of 1 is 1 itself minus when it is substituted 0 we get 0. Taking LCM we get 15 minus 5 minus 3 upon 15 that is equals to 7 upon 15. Therefore the value of double integral over r x dx dy is equals to 7 upon 15 where r is the region bounded by the curves y equals to x cube, x plus y equals to 2 and x equals to 0. Thank you.