# Surface Area obtained by rotating a curve?

1. Apr 27, 2008

### seichan

Find the area of the surface obtained by rotating the curve
y=2e^(2y)
from y=0 to y=4 about the y-axis.

Any help on this would be greatly appreciated. This has my whole hall stumped. We know that you have to use the equation 2pi*int(g(y)sqrt(1+(derivative of function)^2), but cannot figure out how to integrate this correctly.

What I have gotten so far:
y=2e^(2y)
[when u=2y, du/2=dx]
y=e^u
New bounds: 1 to e^4

2pi*int(e^u*sqrt(1+(e^u)^2)

How do you go from there? Any help would be greatly appreciated.

2. Apr 27, 2008

### seichan

Sorry, the problem is x=2e^(2y)

3. Apr 28, 2008

### rock.freak667

$x=2e^{2y} \Rightarrow \frac{dx}{dy}=4e^{2y}$

$$S=2\pi \int _{0} ^{4} x\sqrt{1+\left(\frac{dx}{dy}\right)^2} dy$$

$$S=2\pi \int _{0} ^{4} 2e^{2y}\sqrt{1+(4e^{2y})^2} dy$$

Let $u=4e^{2y} \Rightarrow \frac{du}{dy}=8e^2y \Rightarrow \frac{du}{4}=2e^{2y}dy$

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

I think a hyperbolic trig. substitution will work here e.g.t=sinhx (OR if you want t=secx)

4. Apr 28, 2008

### seichan

Thank you very much. I got this far, but tried to use normal trig substitution. IT goes without saying that it didn't really work for me.

5. Apr 28, 2008

### rock.freak667

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

Let $u=sect \Rightarrow \frac{du}{dt}=secttant \Rightarrow du=secttant dt$

$$\frac{\pi}{2}\int \sqrt{1+sec^2t}secttant dt \equiv \frac{\pi}{2}\int \sqrt{tan^2t}secttant dt$$

$$\frac{\pi}{2}\int secttan^2t dt \equiv \frac{\pi}{2}\int sect(sec^2t-1) dt$$

Long and tedious but it should work.

6. Apr 28, 2008

### tron_2.0

you could just use trig substitution where something in the form sqrt(A^2+X^2) should be approached making the substitution u=tangent(theta)

so you could plus u into sqrt(1+u^2) and end up with sec(theta)
then have du=sec^2(theta)

then for 2pi*integtal(0.25sqrt(1+u^2))du you would get

pi/2*integral(sec(theta)*sec^2(theta))d(theta)

then just use u substitution and get your answer

7. Apr 28, 2008

### rock.freak667

Ahhh yes my mistake..sect would be the wrong trig fn....tant is much better...My mistake....Though I prefer the hyperbolic ones to the trig ones

8. Sep 22, 2009

### rock.freak667

it is not good to thread-jack, so start a new thread showing your work

use this formula

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