Circumference of a Circle is $2 \pi r$ , what about a ring though ? It will have two radiuses : Which one will we take into consideration while measuring the perimeter ? Will it have two different perimeters ?
I am asking this because I was studying calculus and the Example of a circle was taken and it was broken down into several rings etc etc and there the Professor was taking the Circumference of the inner Circle I.e Professor was taking the inner radius into consideration : Why not the outer radius ? I know that if you "kind of" open this circle hypothetically , it will look like a sort of rectangle with the length being $2 \pi r$ so it should not matter which we take. The inner radius and outer radius should be the same ? It is not same , so I am confused.
Best Answer
Here , we see various Circles , Discs , rings , Etc :
A Circle is a thin line , which is at the Same Distance from a Point , which we call Center. That Distance , we call radius.
Mathematically , the interior area is technically not Part of the Circle.
That is the first figure where radius is $X$.
Perimeter is $2 \pi X$.
A Disc is the Area inside that Circle.
The Points in the Disc are not at Same distance from the Canter.
In other words , the Points are at a Distance less than the radius.
(When we want to include the Circumference in the Disc , we call it Closed Disc , otherwise we call it Open Disc : this is some advanced Issue which you might want to ignore at the moment)
That is the Second figure where radius is $X$.
Perimeter is $2 \pi X$.
Disc Area = $\pi X^2$.
A ring is a "Disc" where a Disc has been removed.
In other words , the Points in that Area are at a Distance "less than the outer radius" & "more than the inner radius".
That is the third figure , where outer radius is $X$ , inner radius is $x$.
Outer Perimeter is $2 \pi X$.
Inner Perimeter is $2 \pi x$.
In other words , there is no "Perimeter of ring" , rather we have "outer Perimeter of ring" & "inner Perimeter of ring".
Area of ring = $\pi X^2 - \pi x^2$ , where Subtraction indicates removal.
In Calculus , we sometimes consider very thin rings , where the outer radius is $x+\Delta x$ & inner radius is $x$.
That is the fourth figure , where the thin ring has thickness $(\Delta x)$.
Outer Perimeter of thin ring is $2 \pi (x)$.
Inner Perimeter of thin ring is $2 \pi (x + \Delta x)$.
Area of ring = $\pi (x+\Delta x)^2 - \pi (x)^2$ , where Subtraction indicates removal.
In Calculus , we take $(\Delta x)$ to tend to $0$ , where we get $(\Delta x)^2$ tending to $0$ too.
When you get through this , your confusion will get resolved.