 Hello students, let's solve the following question. It says find the area of the smaller region bounded by the ellipse x square upon 9 plus y square upon 4 is equal to 1 and the straight line x upon 3 plus y upon 2 is equal to 1. Let us now proceed on with the solution. Here we have to find the area bounded by the ellipse x square upon 9 plus y square upon 4 is equal to 1 and the line x upon 3 plus y upon 2 is equal to 1. This line is in the intercept form where 3 is the x-intercept and 2 is the y-intercept. So this line cuts the x-axis at the point 3 0 and it cuts the y-axis at the point 0 2. So we have to find the area bounded by this ellipse and the line. So we have to find the smaller area. So this is the area which we need to find. And we see that it lies in the first quadrant only. So from here we say that y is equal to 2 y 3 under the root 9 minus x square. We are just taking the positive square root since the area lies only in the first quadrant. And from here we see that y is equal to 2 y 3 into 3 minus x. Now to find this area we first need to find the point where this ellipse cuts the x-axis that is where y is equal to 0. So put y is equal to 0. So we have under the root 9 minus x square is equal to 0 and this implies x square is equal to 9 and this implies x is equal to plus minus 3. So the ellipse cuts the x-axis at the point 3 0 also this line passes through the point 3 0 and it also passes through the point 0 2. Now to find the area of the shaded region we need to subtract area of this triangular region from the area of the ellipse in the first quadrant. So the required area t drill 0 to 3 2 by 3 under the root 9 minus x square. This is the area of the ellipse in the first quadrant since we have just taken the positive square root minus integral 0 to 3 2 by 3 into 3 minus x dx. And this is the area of the triangular region covered by this line x upon 3 plus y upon 2 is equal to 1 where x can go maximum up to the point 3 0. So we have taken the limit of the integral as 0 to 3 and here also we have taken the limit as 0 to 3 because in the first quadrant x is going from 0 to 3. Now again this is equal to 2 by 3 integral 0 to 3 under the root 3 square minus x square dx minus 2 by 3 integral 0 to 3 3 minus x dx. Now we will calculate this integral and to calculate this integral 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. Now a is 3 plus a square that is 3 square upon 2 into sin inverse x upon a that is x upon 3 and here the lower limit is 0 and the upper limit is 3 minus 2 by 3. Now the integral of 3 with respect to x is 3x minus the integral of x with respect to x is x square by 2 and here lower limit is 0 and the upper limit is 3. Now we will apply the second fundamental theorem so we will put x is equal to 3 first so it becomes 3 by 2 into under the root 3 square minus 3 square plus 9 by 2 sin inverse 3 by 3 minus. Now we will put x is equal to 0 in this expression so this becomes 0 by 2 into under the root 3 square minus 0 square plus 9 by 2 sin inverse 0 by 3 minus 2 by 3. Now we will put x is equal to 3 in this expression so it becomes 3 into 3 minus 3 square by 2 minus now we will put x is equal to 0 in this expression so this whole expression becomes 0. Now this is equal to 2 by 3 into this term is 0 plus 9 by 2 into sin inverse 1 is 9 by 2 into pi by 2 with the sin inverse 1 is pi by 2 and thus whole expression is 0 minus 2 by 3 into 9, 3 into 3 is 9 minus 9 by 2 so again this is equal to 2 by 3 into 9 pi by 4 minus 2 by 3 into 9 minus 9 by 2 is 9 by 2 so this is equal to 3 pi by 2 minus 3. So this is equal to 3 by 2 into pi minus 2 which is the required area. This completes the question and the session. Hope you will be able to solve more of such problems. Bye and take care.