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find the volume of the solid generated by revolving the region bounded by the curve y = sqrt(16 - x^2) and the line y = 0 about the x axis.

this is how I solved it

[tex] \int_{0}^{4} \Pi (16 - y^{2}) \; dy [/tex]

[tex] \Pi \int_{0}^{4} (16 - y^{2}) \; dy [/tex]

pi(16*y - y^3/3) from 0 to 4

pi(16*4 - 64/3 - 0)

so the answer I got is (128 pi)/3

is this correct.