 Hello everyone, myself, Mrs. Mayuri Kangray, Assistant Professor of Mathematics from the Department of Humanities and Sciences, Walchen Institute of Technology, Solapur. Today, we are going to see multiple integrals, part 3. 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 previous two videos, we have seen how to evaluate double integral over the region R, where the region R is expressed in Cartesian system or polar system. Now in this video, we will see how to evaluate double integral over the region R, where the region R is expressed in Cartesian system but reducible to polar system. Before going to see the procedure, please pause the video for a minute and give the answer of this question. What is the substitution to convert Cartesian system to polar system? I hope you all have written the solution. Let us check the solution. We put x equals to r cos theta, y equals to r sin theta and dx dy equals to r dr d theta to convert the given Cartesian system to polar system. Now let us see the steps to solve the double integral over the region R. Step one, draw the region R and trace out the region of integration. Step two, put x equals to r cos theta, y equals to r sin theta and dx dy equals to r dr d theta. By this substitution, we convert the given Cartesian system to polar system. Then step three, find the limits of region R using radial strip. As we have converted the given Cartesian region to the polar region, to find the limits, we will draw a radial strip by using which we will find out the limits of region R. Step four, solve the integral. Now let us go for the examples. Select the integral, double integral over R, x plus y dx dy, where R is an annular region between x square plus y square equals to 1 and x square plus y square equals to 4 on left half of the circles. Here the region R is left half of annular region between the circles x square plus y square equals to 1 and x square plus y square equals to 4. As here region of integration is a circular part, we will convert the given Cartesian example to the polar one. Before that we will draw the region of integration. Here x square plus y square equals to 1 is a circle with center at origin and radius 1, which is the inner circle of this diagram. Similarly, x square plus y square equals to 4, which can be written as x square plus y square equal to 2 square, which is a circle with center at origin and radius equals to 2. Now the region of integration is left half of this annular region. So this left part is the region of integration. Now we will put x equals to R cos theta, y equals to R sin theta and dx dy equal to R dr d theta to convert the given Cartesian example to the polar one. Now by this substitution, we know that x square plus y square equals to R square. Therefore, the inner circle equation x square plus y square equals to 1 becomes R square equals to 1 as this x square plus y square is equals to R square, which gives us R equals to 1 as an equation of inner circle. Similarly, the outer circle equation is x square plus y square equals to 4, which becomes R square equals to 4. Therefore, the equation of outer circle is R equals to 2. The same region is now written in polar form with the inner circle equation R equals to 1 and outer circle equation R equals to 2. The region of integration is left half of this annular region. As the example is now expressed in polar form, we have to draw a strip which is a radial. So let us draw the radial strip. It moves from theta equal to pi by 2 to theta equal to 3 pi by 2 within the region of integration. Therefore, the outer integral limits are theta equals to pi by 2 to theta equals to 3 pi by 2. Now, to find out the inner integral limits, look at the ends of the strip. Its lower end is on the circle R equals to 1 and upper end is on the circle R equals to 2. Therefore, the inner integral limits are R equals to 1 to R equals to 2. So we will rewrite the given integral in polar form as double integral over R x plus y dx dy can be written as integration from pi by 2 to 3 pi by 2, integration from 1 to 2. Now x is replaced by R cos theta, y is replaced by R sin theta and dx dy is replaced by R dr d theta. Therefore, the integral becomes integration from pi by 2 to 3 pi by 2, integration from 1 to 2 R cos theta plus R sin theta into R dr d theta. Now here, the order of integration is first with respect to R and then with respect to theta. Therefore, to integrate first with respect to R, we will take R common from the bracket. So, we can write the given integral as double integral over R x plus y dx dy can be written as integration from pi by 2 to 3 pi by 2. The integration is with respect to R, so theta is treated as constant with respect to R. It can be taken outside the inner integral, so it can be written as integration from pi by 2 to 3 pi by 2 cos theta plus sin theta. This R and one more R gives us integration from 1 to 2 R square dr into d theta. Therefore, the integral becomes double integral over R x plus y dx dy equals to integration from pi by 2 to 3 pi by 2 cos theta plus sin theta into integration from 1 to 2 R square dr into d theta. Now let us evaluate this. The integration of R square is R cube upon 3, so the integral becomes integration from pi by 2 to 3 pi by 2 cos theta plus sin theta into R cube upon 3 with the limits 1 to 2 d theta. Now we will put the limits. The R is replaced by the upper limit 2, so we get 2 cube minus the lower limit 1 cube is 1 upon 3. Therefore, the integral can be written as integration from pi by 2 to 3 pi by 2 cos theta plus sin theta into bracket 2 cube minus 1 upon 3 into d theta. 2 cube is 8 minus 1 is 7 upon 3. Now this 7 upon 3 is a constant which can be taken outside the integral sign. Now we will integrate this cos theta and sin theta with respect to theta. So we can write down the integral as double integral over R x plus y dx dy equals to 7 upon 3. The integration of cos theta is sin theta and the integration of sin theta is minus cos theta. So we can write it equals to 7 upon 3 sin theta minus cos theta with the limits pi by 2 to 3 pi by 2. Now let us evaluate this. For this first of all we will substitute the upper limit. So we can write it as 7 upon 3 is a constant, so kept as it is into the bracket. Sine 3 pi by 2 minus cos 3 pi by 2 minus now we will put the lower limit. So into the bracket sin pi by 2 minus cos pi by 2. We know that sin 3 pi by 2 is minus 1, cos 3 pi by 2 is 0, sin pi by 2 is 1 and cos pi by 2 is 0. So we can write it as 7 upon 3 into the bracket minus 1 minus 0 minus of 1 minus 0. Now minus 1, minus 1 becomes minus 2. So we can write it as 7 upon 3 into minus 2 that is equals to minus 14 upon 3. Therefore the value of the integral double integral over r x plus y dx dy is equal to minus 14 upon 3. Thank you.