# B Confusion regarding area of this figure

1. Nov 5, 2016

### Prem1998

Maybe, it's a useless question. The figure which I'm talking about consists of two parallel lines each of length 'b' and are separated by a distance 2r. Their ends on one side is closed by a semicircle which in pointing inwards and decreases the area and the ends on the other side are joined by another semicircle of the same radius 'r' but this time it is bulging outwards and contributes to the area. So, we have a closed figure. Finding its area is simple because the amount of area decreased by the first semicircle gets added again by the second semi-circle. So, the area remains the same as that of a rectangle with length 2r and breadth b. A=2rb.
But I get a different answer by this reasoning:
The given figure can be thought to be made up of an infinite number of semicircular arcs from top to bottom. The figure is filled with semicircles. So, the area of this figure can be thought to be the sum of the lengths of these infinite number of semicircles. The length of each elementary semicircle, i.e. pi*r is constant. And, these semicircles are distributed over a length 'b'. So, the area of the figure = pi*r*b, which is wrong. But, what is wrong with this reasoning?

2. Nov 5, 2016

### Staff: Mentor

The fact it is not three dimensional. Your adding semicircles, but actually you only have flat straights. And you stretched these straights to a semicircle. If this were correct, you could create even larger numbers by choosing longer curves than a semicircle.

In space as a surface of a cylinder you were right, because there is no shorter way than going around. But this one is flat.

3. Nov 5, 2016

### Prem1998

I know it's not correct and that this reasoning would lead to contradictions.. But, what is wrong with the points I've made to prove that the area should be pi*r*b. I mean, consider a rectangle of length 'pi*r' and breadth 'b'. This rectangle also has an infinite number of 'pi*r lengths' distributed over a breadth 'b'. But, this rectangle has an area pr*r*b. Even if the number of semicircular arcs in the figure is infinite, but a breadth 'b' should be able to accommodate the same number of lines both in the this rectangle of length 'pi*r' and the figure that I'm talking about. After all, both the semicircular arcs in the figure and the straight lines in this rectangle have breadth equal to that of a point, i.e zero (according to definition).

4. Nov 5, 2016

### Staff: Mentor

You are calculating in a space with curvature.

You extended the green line which you should calculate with, to become the read line. Why the red one? Why not the blue one?

5. Nov 5, 2016

### Prem1998

How does this lead to the conclusion that I can't add semicircles? Where is it proved that I can't add curve lengths in two dimensions?

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