- #1
Flexer5
- 2
- 0
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]
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)
I have been working on this problem for a while, and I am getting frustrated. What is the final result?
2. [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)
I have been working on this problem for a while, and I am getting frustrated. What is the final result?
Last edited: