 Hello and welcome to the session let us discuss the following question it says find the area of the smaller part of the circle x square plus y square is equal to a square cut off by the line x is equal to a by root 2. Let us now proceed on with the solution here we are given a circle x square plus y square is equal to a square the circle has center 0 0 and radius a we have to find the area of the smaller part of the circle x square plus y square is equal to a square cut off by the line x is equal to a by root 2 so we have to find this area. Now we know that the area of any region bounded by the curve y is equal to fx and the ordinate x is equal to ax is equal to b is given by the integral a to b fx dx and the integral of this function is again a function of x where the lower limit of the integral is a and the upper limit of the integral is b which is again equal to fb minus fa. So now y square is equal to a square minus x square and this implies y is equal to under the root a square minus x square. Now the region for which we have to find the area is the part of the circle x square plus y square is equal to a square and in this region x goes from a by root 2 to a and also since the curve is in the form y square is equal to a square minus x square this is symmetric about x axis so the area of the region above the x axis is same as the area of the region below x axis so the required area denoted by a is given by the integral a by root 2 to a fx dx here fx is under the root a square minus x square dx and since we have taken just the positive square root so this gives us the area of the region in just above the x axis so to find the complete area we need to multiply the area of the region above x axis by 2 because the area of the region above x axis is same as the area of the region below x axis now here will apply the formula for the integral of under the root a square minus x square so this is 2 into and the formula is x by 2 into under the root a square minus x square plus a square by 2 into sin inverse x upon a and here the lower limit of the x is a by root 2 and the upper limit is a now we'll apply the second fundamental theorem so we'll put x is equal to a first so it becomes a by 2 into under the root a square minus a square plus a square by 2 sin inverse a upon a minus now we put x is equal to a by root 2 so this becomes a by root 2 by 2 into under the root a square minus a square by 2 the square of a by root 2 is a square by 2 2 plus a square by 2 sin inverse x is a by root 2 upon a now again this is equal to 2 into a by 2 into 0 a square minus a square is 0 plus a square by 2 sin inverse 1 is pi by 2 minus a by root 2 into 2 under the root a square minus a square by 2 is under the root a square by 2 plus a square by 2 sin inverse 1 by root 2 again this is equal to 2 into a square pi by 4 minus a by 2 to the power 3 by 2 into a upon a to the power 1 by 2 plus a square by 2 sin inverse 1 by root 2 is pi by 4 and this is again equal to 2 into a square pi by 4 minus a square upon is 2 square 2 to the power 3 by 2 into 2 to the power 1 by 2 is 2 square minus a square pi by 8 again this is equal to 2 into a square pi by 4 minus a square pi by 8 is a square pi by 8 minus a square by 4 again this is equal to a square pi by 4 minus a square by 2 now this is equal to a square by 2 into pi by 2 minus 1 hence the area of the region is a square by 2 into pi by 2 minus 1 so this completes the question and the session why for now take care have a good day