Were given the integral of a function over region, were asked to use polar coordinates to evaluate this integral. The integral is the double winds girl over the region de of co sign square root of exports y square d A where region de is the disc with center at the origin in a radius of two. So let's sketch a graph of this region. So we have the line data equals pi over two and they did equals zero. We see that this is going to be if I draw the region in red when you have a boundary of our equals two and it will contain all points in that boundary. So this red shaded region, this is the region de and in polar coordinates. This region is given by set of all our data such that are is going to lie between zero and two and data will lie between zero and to pipe square. Root function is defined as long as its argument is greater than or equal to zero. We have that X squared plus y square is always greater than or equal to zero so that the square of it is also defined and we have X squared is a continuous function. Why square his continuous function so X squared plus y squared is continuous. Therefore, it's where that is also continuous, and so the coastline of that is also continuous. So it follows that the function that call it F of X y equals co sign of square root of X squared plus y squared an F is continuous, including on the region de since our region de is a simple region and since our function is continuous on this region, it follows that the double integral over G of our function co sign of square root of X squared plus y squared D A is well defined and is given by formula in polar coordinates integral from 0 to 2 pi integral from zero to of co sign of square of and in polar coordinates. X squared plus y squared is R squared says is the square root of R squared which, because our is always positive, this is simply co sign or and then because of the change in variable to polar coordinates or DRG data using food beanies. The're, um because the bounds on the integral, our constant weaken right This is the product of two into girls and to go from 0 to 2 pi of deep data times the integral from 0 to 2 of our co sign of or d R first into girls Easy to evaluate. Second, Integral does not have a simple anti derivative, but we can use integration by parts to find the anti derivative. So we just look at this integral for a second. It's label you as so suppose that you was equal to our and D V is equal to co sign of our Dior. Therefore, you have that he is going to be equal to sign of our and d you is simply going to be d r and by integration by parts we have that the integral from 0 to 2 of our co sign of our d r is equal to you. Times the judges are sign are evaluated from you are too minus the integral from 0 to 2 of this was you DVDs and I were looking for V. D. U, which is simply sign our g r. This is equal to to sign to minus zero or to sign it too, minus taking anti derivatives negative co sign of ours. This is the same as plus co sign of our evaluated from It's Hard to, which is to sign up to plus co sign of to minus co sign of zero, which is one therefore returning to the original Integral we have that the integral from 0 to 2 pi a dif data taking a derivatives and solving is simply going to be two pi and then our second integral is equal to to sign too Waas co signing too minus one This is our answer.