Thank you guys for all your help.
Char. Limit made a great point, one that I overlooked. Here is what I did:
u == 6*E^(2y) + 1
du == 12*E^(2y)
integral_0^1 6e^(2y) pi sqrt(u) du
integral (pi sqrt(u))/2 du
After taking the integral I got: pi/(4 sqrt(u)).
I then plugged in for u...
1. Let the area of the surface obtained by rotating the curve x = 3e^(2y) from y=0 to y=1 about the y-axis
2.
S= \int_{y_1} ^{y_2} 2 \pi x ds \ where \ ds=\sqrt{1+ \left( \frac{dx}{dy} \right) ^2} dy
3.
S=2\pi \int _{0} ^{1} x\sqrt{1+\left(\frac{dx}{dy}\right)^2} dy
Let
u=6e^{2y}...