 Hi and welcome to the session. Let's work out the following question. The question says, find the area of the region lying between the parabolas, y square is equal to 4ax and x square is equal to 4a y where a is positive. So let us start with the solution to this question. First of all we need to draw these two graphs that is, y square is equal to 4ax and x square is equal to 4a y. We see that the points lying on x, y square equals to 4ax are when x is 0, y is 0, when x is plus minus a, y is plus minus 2a, when x is plus minus 4a, y is plus minus 4a. Similarly for x square equals to 4a y, we see that when x is 0 or when y is 0, x is 0, when y is plus minus a, x is plus minus 2a and when y is plus minus 4a, x is plus minus 4a. So by this we can tell that their point of intersection is 4a 4a and the point 0, 0. So we draw the graphs, this is x square equals to 4a y, this is y square equals to 4ax. We draw these two graphs by plotting the points. Now this is the point of intersection. So we see that the required area is the region oavco because this is a common portion covered by the two curves. So we see that area oavco is equal to area of odvco or we can say that this area is area by this curve minus by area by this curve that is equal to integral where limit goes from 0 to 4a, a limit is going from 0 to the point 4a into y, here the y is from this curve that is y square equals to 4ax dx minus integral where limit goes from 0 to 4a where y is from the curve x square equals to 4a y dx, this is equal to integral where limit goes from 0 to 4a into of square root 4ax dx minus integral where limit goes from 0 to 4a x square by 4adx that is equal to 2 square root of a. Now integral of square root of x is x raise to power 3 by 2 divided by 3 by 2, limit goes from 0 to 4a minus 1 upon 4a into x cube by 3 where limit goes from 0 to 4a this is equal to 4 into square root of a divided by 3 into 4a raise to power 3 by 2 minus 1 upon 12a into 4a the whole cube that is equal to 16a square by 3 square units. So this is our answer to this question. I hope that you understood the solution and enjoyed the session. Have a good day.