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**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. [tex]**

S= \int_{y_1} ^{y_2} 2 \pi x ds \ where \ ds=\sqrt{1+ \left( \frac{dx}{dy} \right) ^2} dy

[/tex]

S= \int_{y_1} ^{y_2} 2 \pi x ds \ where \ ds=\sqrt{1+ \left( \frac{dx}{dy} \right) ^2} dy

[/tex]

**3. [tex]**

S=2\pi \int _{0} ^{1} x\sqrt{1+\left(\frac{dx}{dy}\right)^2} dy

[/tex]

Let

[itex]

u=6e^{2y} \Rightarrow \frac{du}{dy}=12e^2y \Rightarrow \frac{du}{4}=3e^{2y}dy

[/itex]

[tex]

S=2\pi \int \frac{1}{4} \sqrt{1+u^2} du

[/tex]

I then got 2(pi) integral sec(theta)^3 d(theta) by substituting "u" for tan(theta)

S=2\pi \int _{0} ^{1} x\sqrt{1+\left(\frac{dx}{dy}\right)^2} dy

[/tex]

Let

[itex]

u=6e^{2y} \Rightarrow \frac{du}{dy}=12e^2y \Rightarrow \frac{du}{4}=3e^{2y}dy

[/itex]

[tex]

S=2\pi \int \frac{1}{4} \sqrt{1+u^2} du

[/tex]

I then got 2(pi) integral sec(theta)^3 d(theta) by substituting "u" for tan(theta)

I have been working on this problem for a while, and I am getting frustrated. What is the final result?

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