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Find the area of the surface of revolution generated by revolving the curve

e^(8θ), 0≤θ≥½π around the y-axis.

θ= Theta

½π= 1/2*pi

sqrt= sqare root

² = squared

Here's my work:

ds = sqrt( (e^(8θ))² + (e^(8θ))² )dθ

dθ = (e^(8θ)sqrt(2)

x = rcos(θ)= e^(8θ)cos(θ)

A = 2*½π*sqrt(2)*integral of (e^(16θ)cos(θ)) between 0 and ½π

I solved the integral and I got

2*½π*(2)/257*(e^(8θ)-16))

When I plugged the answer online, I got it wrong. Can someone please figure out what's wrong?

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