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are we using shells or discs?

shells add the area of rectangles: 2pi r
discs add the area of circles: pi r^2

it looks like disc method to me

|dw:1442770213833:dw|

we have a circle of radius 8, that have a center removed from them of a radius of 8-sqrt(x)

pi (8^2 - (8-sqrt(x))^2) is the area of a disc for a given value of x

washer method is another name for removing a disc out of a disc ...

pi(R^2 - r^2) gives us the portion that we are concerned with.

Thank you very much for you answers~!

youre welcome