 Hello friends, let's solve the following question, it says find the area of the region bounded by the ellipse x square upon 4 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. So this will be the key idea. Let us now move on to the solution here we are given an ellipse x square upon 4 plus y square upon 9 is equal to 1 which can be written as y square upon 9 is equal to 1 minus x square upon 4 and this implies y is equal to 3 by 2 n to under the root 4 minus x square and we have to find the area of this ellipse but here we are not given the limits of x. So to find the limits of x we will put y is equal to 0 because where the ellipse cut the x axis then y is equal to 0 so put y equal to 0 so this implies 0 is equal to 3 by 2 n to under the root 4 minus x square and this implies under the root 4 minus x square is equal to 0 and this implies x square is equal to 4 and this implies x is equal to plus minus 2. So the ellipse cut the x axis at the point 2 0 and minus 2 0 and we have to find the area of the region bounded by the curve 3 by 2 into 4 minus x square x is equal to 0 and x is equal to 2 and we see that by taking the limits of x as 0 to 2 we find the area of just the first quadrant and to find the area of this complete ellipse we need to multiply the area of this region with 4 because we have 4 symmetric regions so the area denoted by a is equal to 4 into integral 0 to 2 3 by 2 into under the root 4 minus x square dx. I hope you must have understood why we have multiplied this with 4 this is because when we take limit 0 to 2 it gives us the area of the first quadrant and since all the four quadrants have the same area we multiply the area of the one quadrant with 4 so this is again equal to 6 into the integral 0 to 2 under the root 4 minus x square dx. Now we will use the substitution method to find this integral and we will put x is equal to 2 sin theta so this implies dx is equal to 2 cos theta d theta and when we do the substitution limits will also change so if x is equal to 0 then 2 sin theta is equal to 0 and this implies sin theta is equal to 0 and this implies theta is equal to 0 and when x is equal to 2 then 2 is equal to 2 sin theta and this implies sin theta is equal to 1 and this implies theta is equal to pi by 2 so the integral becomes six into the integral zero to pi by two into under the root four minus four sine square theta into two cos theta d theta because dx is two cos theta d theta so again this is equal to six into the integral zero to pi by two taking four common and taking it out of the root it becomes two and one minus sine square theta is cos square theta and square root of cos square theta is cos theta into two cos theta d theta and this is equal to twelve into the integral zero to pi by two two cos square theta d theta now we apply the formula of two cos square theta two cos square theta is equal to one plus cos two theta d theta now we integrate this 12 into integral of one with respect to theta is theta and the integral of cos two theta is sine two theta upon two over zero to pi by two now we'll apply the limits 12 into first put theta equal to pi by two that is pi by two plus sine two into pi by two upon two minus now put theta is equal to zero zero plus sine two into zero upon two and this is equal to 12 into pi by two plus sine pi upon two minus zero the sine zero is zero again this is equal to 12 into pi by two this is because sine pi is zero so zero upon two is zero which is equal to six pi hence area of the ellipse is six pi so this completes the question and the session bye for now take care have a good day