# A unit disk centered at the origin is sliced so that the right portion has width h. Derive a formula A(h) for the area of slice. find antiderivatives use WolframAlpha. Show that A(0)=0 and...

A unit disk centered at the origin is sliced so that the right portion has width h. Derive a formula A(h) for the area of slice. find antiderivatives use WolframAlpha.

Show that A(0)=0 and A(2)=pi

The equation given is F(x)=sqrt(1-x^2) The point the circle intersects the x and y axis are (1,0) (-1,0) and (0,1) (0,-1)

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The unit circle is given by the equation `x^2+y^2=1` which means that there are two functions that describe the circle: `y=sqrt{1-x^2}` for the top half and `y=-sqrt{1-x^2}` for the bottom half.

Since you have described the area as having values `A(0)=0` and `A(2)=pi` , the area function is starting from the left of the circle and moving to the right of the circle. By adding up the vertical slices of the circle, noting that we only need to consider twice the top half of the circle, we find the area is given by the integral

`A(x)=int_{-1}^{x-1}2sqrt{1-t^2}dt`

Now from Wolfram Alpha (as part of the question) or from the webpage listed below, we see that

`int 2sqrt{1-t^2}dt=t sqrt{1-t^2}+sin^{-1}t` which means that

`A(x)=(x-1)sqrt{1-(x-1)^2}+sin^{-1}(x-1)+pi/2`

Notice that from this formule, we have `A(0)=0` and `A(2)=pi` .

**The formula for the area function is `A(x)=(x-1)sqrt{1-(x-1)^2}+sin^{-1}(x-1)+pi/2` .**

**Sources:**