`y=sqrt(4-x^2) , -1<=x<=1` Set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the x-axis.

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To calculate the surface area generated by curve `y=f(x)` revolving about `x`-axis between `a` and `b`, we use the following formula

`S_x=2pi int_a^b y sqrt(1+y'^2)dx`

Let us therefore first find the derivative `y'.`

`y'=1/(2sqrt(4-x^2))cdot(-2x)=-x/sqrt(4-x^2)`

`y'^2=x^2/(4-x^2)`

We can now calculate the surface.

`S_x=2pi int_-1^1sqrt(4-x^2)sqrt(1+x^2/(4-x^2))dx=`

`2pi int_-1^1sqrt(4-x^2)sqrt(4-x^2+x^2)/sqrt(4-x^2)dx=`

`2pi int_-1^1 2dx=4pi x|_-1^1=4pi(1+1)=8pi`

The area of surface generated by revolving the given curve about `x`-axis between `-1` and `1` is `8pi`.

Graphs of the curve and the surface generated by curve's revolution can be seen in the images below.

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