The region between the graph f(x)=x^(1/2) and the x-axis for x∈[0,16] is revolved about the line y=8. Which integral will determine the volume of the solid that is generated?
I got π(x-64)dx, x∈[0,16] using the yA=1/2[(f(x))^2 - (g(x))^2]dx x∈[a,b] equation, but the answer is ∫π[64−(8−x^(1/2))^2]dx x∈[0,16]. No idea where I went wrong, but maybe this isn't the right equation to use?

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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

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