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?

Lightning
Jul 22, 2018

#1**+1 **

\(f(x)=1-\sqrt{1-x^2} \)

\(-1\le x\le 1 \)

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

CPhill
Jul 23, 2018