 Hello students. Let's solve the following problem. It says find the area of the region bounded by the ellipse X square upon 16 plus Y square upon 9 is equal to 1. Let us first understand how do we find the area of the region? bounded by the curve Y is equal to Fx and the ordinate X is equal to A, X is equal to B is given by integral A to B Fx dx which is Fx A to B which is equal to Fb minus Fa. This knowledge will work as the idea. Let us now move on to the solution. Here we are given an ellipse X square upon 16 plus Y square upon 9 is equal to 1. This can be written as Y square upon 9 is equal to 1 minus X square upon 16. So this implies Y square is equal to 9 upon 16 into 16 minus X square. Now we have to find the area of the ellipse and here we are not given the limits of X that is we are not given where the ellipse cut the X axis. So to find the limits we put Y is equal to 0 because when the ellipse that is where the ellipse cut the X axis that is where Y is 0. So we put Y is equal to 0. So if Y is 0, this implies 9 by 16 into 16 minus X square is equal to 0. So this implies X square is equal to 16 and this implies X is equal to 4. That's plus minus 4. So the ellipse cut the X axis at a point 4 0 and minus 4 0. Now Y square is equal to 9 by 16 into 16 minus X square. So this implies Y is equal to 3 by 4 under the root 16 minus X square. Now we have to find the area of the region bounded by the curve Y equal to 3 by 4 into under the root 16 minus X square. X is equal to 0. X is equal to 4. So the area denoted by A of the region bounded by the curve this and X is equal to 0 and X is equal to 4 is given by 3 by 4 into 16 minus X square. But when we take the limit of X varying from 0 to 4 it gives us the area of the region in just the first quadrant. So to find the area of the complete ellipse will multiply this area by 4 because we have four symmetric regions, right? So to find the area of the complete ellipse we multiply this with 4. So this is equal to 3 into integral 0 to 4 under the root 16 minus X square dx. Now we'll use substitution method to solve this integral and we put X is equal to 4 sine theta. So dx is equal to 4 cos theta d theta. So the integral becomes 3 into integral 16 minus 16 sine square theta and dx is 4 cos theta d theta. Now as we put X is equal to 4 sine theta, limits will also be changed and we have to find limits in terms of theta. So if X is 0 then 0 is equal to 4 sine theta and this implies theta is equal to 0 and if X is equal to 4 then 4 is equal to 4 sine theta and this implies sine theta is equal to 1 and this implies theta is equal to pi by 2. So here the limits will vary from 0 to pi by 2. Now this is equal to 3 into integral 0 to pi by 2 under the root taking 16 common it is 1 minus sine square theta into 4 cos theta d theta. Again, this is equal to 3 into integral 0 to pi by 2. Square root of 16 is 4 and 1 minus sine square theta is cos square theta and square root of cos square theta is cos theta into 4 cos theta d theta. So this is equal to 12 into integral 0 to pi by 2 cos square theta d theta. 3 into 4 is 12 into 4 which can be again written as 24 into integral 0 to pi by 2 2 cos square theta 4 can be written as 2 into 2. So 12 into 2 is 24 into 2 cos square theta d theta. Now we'll apply the formula of 2 cos square theta which is 24 into integral 0 to pi by 2. 2 cos square theta is equal to 1 plus cos 2 theta d theta and integral of this is 24 integral of 1 with respect to theta is theta and the integral of cos 2 theta is sine 2 theta upon 2 and limits are 0 to pi by 2. This is again equal to 24 into applying the limits pi by 2 plus first put theta equal to pi by 2 sine 2 pi by 2 upon 2 minus put theta equal to 0 0 plus sine 2 into 0 upon 2 which is equal to 24 into pi by 2 plus 2 gets cancelled with 2 and it becomes sine pi by 2 minus 0 0 plus 0 sine 0 is 0 Now again, this is equal to 24 into pi by 2. Since sine pi is 0 So this becomes 0 and we are left with pi by 2 into 24 which is equal to 12 pi hence the area of the region or the given ellipse is 12 pi and this completes the question and the session. Bye for now. Take care. Have a good day