 Hello everyone, welcome to the session of application of multiple integrals part 3. This is Swati Nikam, assistant professor, Department of Humanities and Sciences, Vulturen Institute of Technology, Solapur. At the end of this session, student will be able to find area of region bounded by curves by using double integration. Hence, please pause your video for a minute and write the formula for finding area of curve in Cartesian coordinates. I hope you have written your answer. So area of curve in Cartesian coordinate region over a region r is a is equal to double integration over r dx into dy. Now let us see area by double integration in polar coordinates. Let us consider area enclosed by two plane curves r is equal to r1 is equal to f1 theta and r is equal to r2 is equal to f2 theta intersecting in point A with coordinates r1 alpha and b with r2 beta. Then the area of elementary rectangle is r into dr into d theta. In Cartesian coordinate system, we have seen that area is dx into dy. So after replacing in polar form x equal to r cos theta, y is equal to r sin theta, we get the replacement of dx dy is equal to r dr d theta. In polar coordinate system, the area A is equal to integration alpha to beta, inner integration f1 theta to f2 theta of r dr d theta. Let us solve an example. Find the area of cardioid r is equal to A into bracket 1 minus cos theta by using double integration. Answer, we know that the cardioid r is equal to A in bracket 1 minus cos theta is symmetrical about initial line and passing through pole which is drawn as this one. So initial line and perpendicular line. To find area of cardioid, it is convenient to find area of cardioid above the initial line and multiply it by 2. Because of symmetry, we can use this technique. For the cardioid r varies from, so the strip is always considered from pole 0 to boundary of cardioid r is equal to A times bracket 1 minus cos theta and theta varies from. So to cover this region, r theta will start with theta is equal to 0 to theta is equal to pi above the initial line. Therefore, area A is equal to twice integration 0 to pi integration 0 to A times bracket 1 minus cos theta r dr d theta which is equal to 2 times integration 0 to pi. Now limits are in terms of theta, so we will integrate with respect to r first. Integration of r is r square by 2 limits 0 to A times 1 minus cos theta into d theta which is equal to integration 0 to pi this 2 and 2 get cancelled in place of r upper limit A square in bracket 1 minus cos theta bracket square into d theta. Therefore, area A is equal to integration 0 to pi A square in bracket 1 minus cos theta whole square d theta which is equal to A square into integration 0 to pi. Let us expand this bracket by using A minus B whole square, so 1 minus twice cos theta plus cos square theta d theta. In order to integrate this trigonometric term should have degree 1, hence reduce cos square theta by using trigonometric formula 2 1 plus cos 2 theta divided by 2. Therefore integration is A square into bracket 0 to pi bracket 1 minus 2 cos theta plus 1 plus cos 2 theta divided by 2 into d theta which is equal to A square into integration 0 to pi this 1 plus half is 3 by 2 minus 2 cos theta plus cos 2 theta divided by 2 into d theta which is equal to integrate it now A square in bracket 3 by 2 into theta minus 2 times integration of cos theta is sin theta plus integration of cos 2 theta is sin 2 theta by 4 limits 0 to pi. Therefore required area A is equal to A square into bracket 3 by 2 theta minus 2 sin theta plus sin 2 theta by 4 limits 0 to pi which is equal to A square into bracket 3 by 2 into pi in place of theta substitute upper limit minus 2 times sin pi plus sin 2 pi by 4 minus now in place of each theta lower limit 0 0 plus 2 sin 0 minus sin 0 by 4. Now sin pi is 0 sin 2 pi is 0 sin 0 is 0 and therefore from this bracket we will left only with the first term 3 by 2 pi hence our final answer of area A is equal to 3 by 2 A square into pi example number 2. Find the total area of the curve R is equal to A into cos 2 theta now the curve is 4 leaves rows as shown in the diagram. In curve tracing we have learned that how to get these type of rows curves. So the curve is symmetrical to both initial line as well as a line perpendicular to initial line. So that to find area of rows curve it is convenient to find area of only one loop and multiplied by 4. So we are going to consider only one loop here. Now area of one loop is equal to again twice area of half loop because loop is again symmetric above the initial line and below the initial line and on the half loop above the initial line R varies from pole 0 to boundary R is equal to A cos 2 theta and theta varies from theta is equal to 0 to theta is equal to pi by 4. Therefore area has formula 4 times 4 loops so 4 times bracket twice area of half loop therefore 2 times integration 0 to pi by 4 inner integration 0 to A cos 2 theta R dr d theta which is equal to 8 times integration 0 to pi by 4 into inner integration is with respect to R. So R square by 2 limits 0 to A cos 2 theta d theta. Therefore area is equal to 4 times integration 0 to pi by 4 A square cos 2 theta d theta which is equal to 4 times integration 0 to pi by 4 A square outside in bracket reduce this cos square by using trigonometric formula to degree 1 as 1 plus cos 4 theta divided by 2 into d theta which is equal to 4 by 2 is 2 into A square constant and now integration of 1 is theta plus integration of cos 4 theta is sin 4 theta divided by 4 limits 0 to pi by 4 which is equal to 2 A square inside the big bracket in place of theta put upper limit pi by 4 plus sin 4 into bracket pi by 4 upon 4 minus lower limit 0 minus sin 4 times 0 by 4 so that now 4 4 get cancelled and it is sin pi. Sin pi by has value 0 and this is sin 0 has value 0. So from this bracket again we left only with the first term pi by 4 and hence area of the rose curve is pi into A square by 2. I have referred the textbook higher engineering mathematics by Dr. B. S. Grebar. Thank you so much for watching this video.