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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
avatar+89953 
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\(f(x)=1-\sqrt{1-x^2} \)

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

 

 

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

 

 

cool cool cool

CPhill  Jul 23, 2018

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