Gilberto S. answered • 10/07/20
Experienced College Professor
What class is this for? It looks like it would be more appropriate for Calc 2 or Calc 3.
What I would suggest is try to view the problem using polar coordinates.
I would either try to imagine the origin sitting in the middle of the purple region. Divide the purple region into the 3 congruent pieces and figure out the area of one piece and multiply by three
OR
imagine the origin at C. figure out the area of the sector in between angle ACB.. And then use calculus to find the areas of the other two leftover pieces.
Gilberto S.
Ok, calculus isn't necessary. A and B are on the circle centered at C. And A and C are on the circle centered at B and so on. the common radius is r. So A, B, C are all r units apart. So triangle ABC is an equilateral triangle with side = r. So first focus on that part. You can use facts about a 30-60-90 triangle or an argument based on the Pythagorean theorem to show that the area of the triangle is r^2 sqrt(3) / 410/07/20
Gilberto S.
Then work on the area for the three "bits" that bulge out.10/07/20
Gilberto S.
Take angle ACB for instance. Look at the part of the purple area which is between the rays. If you look at it the right way, that is 1/6 of a circle with radius r. So the area of that section is (1/6) pi r^2 We just figured out that the area of the triangle is r^2 sqrt(3) / 4. So the area of the part which bulges out over the triangle is just the difference. 1/6) pi r^2 - r^2 sqrt(3) / 4. = r^2 [ pi/6 -sqrt(3)/4 ] So then putting all this together, the purple area is made up of one triangle and three "bulges" so the total area will be: r^2 sqrt(3) / 4 + 3 ( r^2 [ pi/6 -sqrt(3)/4 ]) which when you simplify will get you the answer.10/07/20
Joseph C.
Thank you so much! Much appreciated! Now I have an understanding of this.10/08/20
Joseph C.
Hello, well I tried but it just doesn't seem to add up. Is there any way you could help me to prove the expression?10/07/20