Inactive Tutor answered 10/17/20
Tutor
New to Wyzant
S = ∫072πg(x)√[1 + (g'(x))2]dx = 2π∫07e-9x√(1 + 81e-18x)dx = 19.55
To evaluate integral I use TI-84
Frank O.
asked 10/17/20Inactive Tutor answered 10/17/20
S = ∫072πg(x)√[1 + (g'(x))2]dx = 2π∫07e-9x√(1 + 81e-18x)dx = 19.55
To evaluate integral I use TI-84
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