 Hello and welcome to the session. Let's work out the following problem. It says find the area bounded by the circle x square plus y square is equal to 16 and the line y is equal to x in the first quadrant. So let's go move on to the solution. We have to find the area bounded by the circle x square plus y square is equal to 16 and the line y is equal to x. The circle is x square plus y square 16 is 4 square. So the radius of the circle is 4 centimeter. And we have to find the area bounded by the circle and the line in the first quadrant. So we have to find this area and to find this area we need to divide this region into two regions. One is the triangular region and one is this region. Now to find this complete area we first need to find the point of intersection of the line and the circle. So now y is equal to x is the line which cuts the circle. So we'll put y is equal to x here. So we have x square plus x square is equal to 16. That is 2x square is equal to 16 that is x square is equal to 8 that is x is equal to 2 root 2. Now since x is equal to 2 root 2 and y is equal to x so this implies y is also equal to 2 root 2. So now the point of intersection of circle and the line y is equal to x is 2 root 2, 2 root 2. Now the required area is given by integral. Now as I said earlier the required area is sum of this triangular region and this region. Now area of this triangular region is given by integral 0 to 2 root 2. Since this point of intersection is 2 root 2, 2 root 2. So this line is x is equal to 2 root 2. There is integral 0 to 2 root 2 x dx that is the line y is equal to x plus integral 2 root 2 to 4 of the circle which is given by y is equal to 4 square minus x square. This is integral 2 root 2 to 4 of under the root 4 square minus x square dx. Since in this region x is going from 2 root 2 to 4 as we know that the area of radius of this circle is 4 centimeters so this point is 40. Now the integral of x with respect to x is x square by 2 and here the lower limit is 0 and the upper limit is 2 root 2 plus. Now here we will use the formula of 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 by a. Now here a is 4 so we have x by 2 into under the root 4 square minus x square plus a square by 2 that is 4 square by 2 sin inverse x upon a and here the lower limit is 2 root 2 and the upper limit is 4. Now here we will first put x is equal to 2 root 2 so it is 2 root 2 square by 2 minus now we will put x is equal to 0 so it is 0 square by 2 plus here we will put x is equal to 4 first. So we have 4 by 2 into under the root 4 square minus 4 square plus 4 square by 2 is 8 sin inverse 4 by 4 minus now we will put x is equal to 2 root 2 so we have 2 root 2 by 2 under the root 4 square minus 2 root 2 square that is 8. 2 root 2 square is 8 plus again is 4 square by 2 is 8 sin inverse 2 root 2 by 4. Now 2 root 2 square is 8 8 by 2 is 4 plus 4 plus 2 into 0 plus 8 into sin inverse 1 sin inverse 1 is pi by 2 minus root 2 into under the root 4 square minus 8 is 8 4 square is 16 16 minus 8 is 8. Plus 8 into sin inverse root 2 by 2 plus 4 plus 4 pi minus root 2 into under the root 8 is 2 root 2 plus 8 sin inverse root 2 by 2 is 1 by root 2 4 plus 4 pi. minus root 2 into root 2 is 2 2 into 2 is 4 plus 8 sin inverse 1 by root 2 is pi by 4. So this is equal to 4 plus 4 pi minus 4 minus 2 pi plus 4 gets cancelled with minus 4 and this is 4 pi minus 2 pi is 2 pi. Hence the required area is equal to 2 pi square units. So this completes the question and the session by for now take care have a good day.