Find the volume of the solid that results when the region enclosed by the given curves is revolved about the x-axis (using washer method)

`y=(sqrt(100-x^2))^2` , `y=8`

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The volume of the solid generated by revolving the area enclosed by `y = (sqrt(100 - x^2))^2` and y = 8 about the x-axis has to be determined.

`y = (sqrt(100 - x^2))^2`

=> `y = 100 - x^2`

The region enclosed by y = 100-x^2 and y = 8 is:

The intersection of the two curves is at the solution of 8 = 100 - x^2

=> x = +-sqrt 92

The volume of the solid is:

`int_-sqrt92^(sqrt 92) pi*y^2 dx`

=> `int_-sqrt92^(sqrt 92) pi*(100 - x^2)^2 dx`

=> `pi*int_-sqrt92^(sqrt 92) 10000 + x^4 - 200x^2 dx`

=>` ``pi*[10000x + x^5/5 - (200x^3)/3]_-sqrt92^(sqrt 92)`

=> `pi*(10000*2*sqrt 92 + (1/5)*2*(sqrt 92)^5 - (200/3)*2*(sqrt 92)^3)`

`~~ 335047.91`

**The required volume is approximately 335047.91 cube units.**

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