# Integral 2,-2 sqrt (4-x^2)dx ,circle?

### 3 Answers

- Ian HLv 71 month ago
y = 2 - 2√[4 – x^2] is the lower half of an ellipse

https://www.wolframalpha.com/input/?i=y+%3D+2+-+2%...

2√[4 – x^2] = 2 - y

4(4 – x^2) = (2 – y)^2

16 – 4x^2 = (y – 2)^2

4x^2 + (y – 2)^2 = 16

x^2/2^2 + (y – 2)^2/4^2 = 1

The related full ellipse is centred at (0, -2)

semi-axes 2 along x and 4 along y

https://www.wolframalpha.com/input/?i=x%5E2%2F2%5E...

For the integral, use x = 2sin(u) = 2s, dx = 2cos(u)du and

2[cos(u)]^2 du = [1 + cos(2u)] du and

sin(2u) = 2sin u * cos u and

2sin u = x and 2 cos u = √(4 – x^2)

Applying limits x from -2 to 2, I = 8 – 4𝛑

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- 1 month ago
As david said it is a semicircle, so you could find what the area would be for a circle of radius 2 then cut it in half. To actually carry out the calculation, are you familiar with trigonometric substitution? if not I could explain it.

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