 Hello students, let's solve the following problem. It says, find the area of the region in the first quadrant enclosed by the x-axis. The line x is equal to root 3 y and the circle x square plus y square is equal to 4. Whereas now move on to the solution. Now here we have to find the area enclosed in the first quadrant the line x is equal to root 3 y. The x-axis and the circle x square plus y square is equal to 4. So we have to find the area of the region in the first quadrant enclosed by this line x-axis and the circle. So this is the area which we have to find. So here we need to divide this area into two parts. Right? One is triangle this and this area and this line is the point of intersection of the circle and the line. So the given circle is x square plus y square is equal to 4 and the line is x is equal to root 3 y. Now to find the point of intersection we will put x is equal to root 3 y in this. So we have 3 y square plus y square is equal to 4 and this implies 4 y square is equal to 4 and this implies y is equal to plus minus 1. But since we have to find the area in the first quadrant we will ignore the negative sign and we have y is equal to 1. Now if y is equal to 1 then x is equal to root 3 into 1 that is x is equal to root 3. So the point of intersection is root 3 1. Now the area of this triangle is given by the integral x upon root 3 where the limit of the integral goes from 0 to root 3 since from here y is equal to x upon root 3 and here we are taking the limit of the integral as 0 to root 3 because here this line can take the maximum value up to root 3 and the area of this region will be given by integral root 3 2 2 under the root 4 minus y square because this is the area covered by this circle where x goes from root 3 to 2 and this point is 2 because the radius of the circle is 2 because equation of the circle is x square plus y square is equal to 2 square and here this 2 is the radius of the circle. So now the required area denoted by a is given by integral 0 to root 3 x upon root 3 dx plus integral root 3 2 2 under the root 4 minus x square from this equation we get and this implies y is equal to under the root 4 minus x square and here we have taken the positive square root because we need to find the area in just the first quadrant. Now again this can be written as 1 by root 3 integral 0 to root 3 x dx plus integral root 3 2 2 under the root 2 square minus x square dx. Now again integral of x dx is x square by 2 where the lower limit of the integral is 0 and the upper limit is 3 plus. Now here we will apply the formula for the integral of under the root a square minus x square which is given by x by 2 into under the root a square minus x square plus a square by 2 sin inverse x upon a. Here a is 2 and the lower limit is root 3 and the upper limit is 2. Now we will apply the second fundamental theorem. So this becomes 1 by root 3 into root 3 square by 2 minus 1 by root 3 0 square by 2 we put root 3 first then we put 0 in place of x plus now here we will put x is equal to 2 first minus now we put x is equal to root 3 so this becomes root 3 by 2 into under the root 2 square that is 4 minus root 3 square is 3 plus 2 sin inverse root 3 by 2 now again this is equal to 1 by root 3 into 3 by 2 0 square by 2 is 0 plus 2 by 2 is 1 2 square minus 2 square is 0 so it is 0 plus 2 into sin inverse 1 is pi by 2 minus root 3 by 2 4 minus 3 is 1 so it is under root 1 is 1 plus 2 into sin inverse root 3 by 2 is pi by 3 now this is equal to root 3 by 2 plus pi minus root 3 by 2 minus 2 pi by 3 which is again equal to pi minus 2 pi by 3 which is equal to pi by 3 hence the required area is pi by 3. So this completes the question and the session by for now take care have a good day.