 Hello everyone, myself, Mrs. Mayuri Kangal, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Walshchen Institute of Technology, Solapur. Today we are going to see multiple integrals part 2. The learning outcome is at the end of this session the students will be able to solve the multiple integrals over the region R. In the previous video we have seen how to evaluate double integrals over the region R where the region R is a Cartesian region. Now in this video we will see how to evaluate double integrals over the region R where the region R is polar region. Let us see the steps to solve double integral over the region R where region R is a polar region. Step 1, draw the region R and trace out the region of integration. Step 2, find the limits of region R using radial strip. Step 3, solve the integral. Now pause the video for a minute and give the answer of this question. What is the nature of the following curve? The curve equation is R equals to 2A cos theta. I hope you all have written the solution. Let us check it. R equals to 2A cos theta. Multiply both the side by R gives us R square equals to 2A R cos theta. We know that R square is x square plus y square and R cos theta is x. So we can write the equation as x square plus y square equals to 2A x. Now we will shift this 2A x to the left side. So we can write it as x square minus 2A x plus y square equal to 0. For completing this square we add a square on both sides. So we get the equation as x square minus 2A x plus a square plus y square equal to a square. Now this equation can be rewritten as x minus a bracket square plus y minus 0 bracket square equal to a square which is an equation of a circle with center at a 0 and the other area is a. Now let us go for the next example. Calculate the integral double integral over R R square sin theta dr d theta where R is region of a circle R equal to 2A cos theta and lying above the initial line. Here the region R is the region of the circle R equal to 2A cos theta and lying above the initial line. Just now we have seen that R equal to 2A cos theta is a circle with center at a 0 and radius a. Therefore the region R is upper half of this circle. Let us see it point A0 is drawn, now we will draw a circle. The equation of the circle is R equal to 2A cos theta and the region of integration is upper half of this circle. Now here the given example is polar so the strip is radial. To find out the outer integral limits, let us move this strip within the region of integration. It moves from theta equal to 0 to theta equal to pi by 2. Therefore the outer integral limits are theta equal to 0 to theta equal to pi by 2. To find out the inner integral limits, look at the end of the strip. Its lower end is on the pole and upper end is on the circle R equal to 2A cos theta. Therefore the inner integral limits are R equal to 0 to R equal to 2A cos theta. Therefore the given integral becomes double integral over R, R square sin theta dr d theta is equals to integration from 0 to pi by 2, integration from 0 to 2A cos theta R square sin theta dr d theta. Now let us evaluate this integral, observe the limits we can get here. The order of integration is first with respect to R and then with respect to theta. So let us evaluate it, integration from 0 to pi by 2 to integration from 0 to 2A cos theta R square sin theta dr d theta is equals to. Now the first integration is with respect to R. So the terms of theta are considered to be constant with respect to R, so can be taken outside the inner integral. So we get the integral as integration from 0 to pi by 2 sin theta d theta into integration from 0 to 2A cos theta R square dr. Let us evaluate the inner integral, it is with respect to R and the integration of R square is R cube by 3. So we get the integral as integration from 0 to pi by 2 sin theta d theta R cube upon 3 with the limits 0 to 2A cos theta. Now let us put the limits, so we get the integral as integration from 0 to pi by 2 sin theta d theta 2A cos theta minus 0 bracket cube upon 3 here 2 cube into A cube upon 3 that is 8A cube upon 3 is a constant. So it can be taken outside the integral therefore we can write it as 8A cube upon 3 integration from 0 to pi by 2 sin theta into cos cube theta d theta. Now to evaluate this integral we will use the definition of beta function, we know that integration from 0 to pi by 2 sin rest to p theta cos rest to q theta d theta is equals to 1 by 2 into beta of p plus 1 by 2 comma cube plus 1 by 2 comparing this definition of beta function with this given integral we get p equal to 1 and q equal to 3. So by using this definition we can get the value of the integral as 8A cube upon 3 into bracket 1 by 2 beta of p plus 1 that is 1 plus 1 upon 2 comma cube plus 1 here cube is 3 therefore we get 3 plus 1 upon 2. So we can write it as integration from 0 to pi by 2 integration from 0 to 2A cos theta r square sin theta d r d theta equals to 8A cube upon 3 into the bracket 1 by 2 beta 1 plus 1 upon 2 comma 3 plus 1 by 2 which is nothing but 1 plus 1 is 2 upon 2 gives us 1 and 3 plus 1 is 4 upon 2 gives us 2. So we get the value of the integral as 8A cube upon 3 into the bracket 1 by 2 beta of 1 comma 2. Now let us evaluate this beta function we know that beta mn is equals to gamma m into gamma n upon gamma m plus n. So we will convert this beta function to the gamma function. So we get 8A cube upon 3 1 by 2 as it is. Here m is equals to 1 and n is equals to 2 therefore we can write gamma 1 into gamma 2 upon gamma 1 plus 2 we know that gamma 1 is 1 gamma 2 is 1 factorial and here we get gamma 3 which has the value 2 factorial. So we can write it as 8A cube upon 3 into the bracket 1 by 2 into 1 into 1 factorial upon 2 factorial. We know that 1 factorial is 1 and 2 factorial is 2 therefore we get the value of this bracket as 1 upon 4. So we can write 8A cube upon 3 into the bracket 1 by 4 4 to the 8 so we can write 2A cube upon 3 which is the value of the integral therefore the value of the given integral is 2A cube upon 3. Thank you.