- #1
opus
Gold Member
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My question is, why is the circled integral the chosen integral for this case?
My thoughts are that we don't just use ##\int_0^1e^{-x}## because we need to make this two dimensional area into a three dimensional volume by doing 360 degrees of rotation. This would correspond to ##2πr##. ##2π## is a constant and can be brought out front so I see why the ##2π## is there.
My question is, the radius is just 1, so why is there an ##x## in the integrand?
My thoughts are that we don't just use ##\int_0^1e^{-x}## because we need to make this two dimensional area into a three dimensional volume by doing 360 degrees of rotation. This would correspond to ##2πr##. ##2π## is a constant and can be brought out front so I see why the ##2π## is there.
My question is, the radius is just 1, so why is there an ##x## in the integrand?