Let $f(x)$ be the function defined on $-1\le x\le 1$ by the formula $$f(x)=1-\sqrt{1-x^2}.$$This is a graph of $y=f(x)$: If a graph of $x=f(y)$ is overlaid on the graph above, then one fully enclosed region is formed by the two graphs. What is the area of that region, rounded to the nearest hundredth?
\(f(x)=1-\sqrt{1-x^2} \)
\(-1\le x\le 1 \)
We need to find the inverse of the given function....for f(x), we can write, y
y = 1 - √ [1 - x^2] rearrange as
√ [1 - x^2] = 1 - y square both sides
1 - x^2 = (1 - y)^2
1 - x^2 = y^2 - 2y + 1 subtract 1 from both sides
-x^2 = y^2 - 2y mutiply through by -1
x^2 = 2y - y^2 take both roots
x = ±√ [ 2y - y^2 ] "swap" x and y
y = ±√ [2x - x^2 ]
We only need the top half of this graph ⇒ y = √ [2x - x^2 ]
Here is the graph of both functions along with the graphs of y = x and y = 1
https://www.desmos.com/calculator/ozvzaxlwxe
The area enclosed by both functions will be twice the area of the area between y = x and y = 1 - √ [1 - x^2]
Note that the area bounded by the y axis from 0 to 1 , the line y = 1 from 0 to 1 and the line y = x will from a triangle with a base and height of 1....so...it's area = (1/2) base * height = (1/2)(1)(1) = 1/2 units^2
And note that the area formed y axis from 0 to 1, the line y =1 from 0 to 1 and the function y = 1 - √ [1 - x^2] will be the area of a quarter circle with a radius of 1 =
(1/4) pi * (1)^2 = pi /4 units^2
So....this area less the area of the triangle will = 1/2 of the area we are seeking
So....the total area = 2 [ pi/4 -1/2] = [ pi/2 - 1] units^2 ≈ 0.571 units^2