>>11598616Alright, as we know from multivariable calculus the area of a thing D is given by the double integral:
In our case we have been given the D, which I shall not describe as it is too hard but we all know what I'm talking about. We now split the D in two parts () to help us with our calculations. If we call As projection to OB C then D1 shall be defined from the lines OA, AC, OC and D2 from AC, CB, BA. Then (1)
At this point it becomes obvious that we need to calculate the x coordinate of C. It is the same as that of A as OB lies on the x axis. To find the x coordinate of A we simply need to solve the equation
We observe that the numbers -8 and 2 are both solutions of the equation. Since the equation is of second degree it only has two (2) solutions. We can also observe that since A lies in the first quadrant (would be obvious if I had done the difficult task of describing the D earlier) it must have a positive x coordinate. Therefore 2 is the x coordinate of A and C. We also need the x coordinate of B. We can obtain that by solving the equation
Similarly we observe that the (only) two soloutions are -4 and 4 and that which matches B is 4. Continuing now from where we left off at (1)
Thank you OP! It was really cool!!
