Question

Use intergral to find the Area of a circle of redius1centered at(0,2)?

Answer 1

The area is `pi` square units.

Explanation:

This will have equation

`x^2 + (y- 2)^2 = 1`

I would solve for `y` and integrate with respect to `y`.

`x^2 = 1 - (y - 2)^2`

`x = sqrt(1 -(y- 2)^2)`

Now integrate from `[1, 3]`.

`A = int_1^3 sqrt(1 - (y- 2)^2) dy`

Let `u = y - 2`. Then `du = dy`.

`A = int_(-1)^1 sqrt(1 - u^2) du`

This integral is derived here :https://socratic.org/questions/how-do-you-integrate-int-sqrt-1-x-2-by-trigonometric-substitution-1

`A = [1/2(arcsinx + xsqrt(1 - x^2))]_(-1)^1`

`A = 1/2arcsin(1) - 1/2arcsin(-1) = 1/2(pi/2 - (-pi/2))`

`A = 1/2pi = pi/2`

But this is only the part of the circle when `x` is positive, since we didn't use `+-` when we took the `√`.

Thus

`A = 2 * pi/2 = pi` square units

We would get the same result if we found the area using geometry.

Hopefully this helps!