Calculating Surface Area of Revolution for a Rotated Curve

[Gauss177]

Homework Statement

Find the area of the surface obtained by rotating the curve about the x-axis:
y=cos 2x, 0<=x<=pi/6

Homework Equations

Surface area about the x-axis = Integral of 2pi * f(x) * sqrt(1+[f'(x)]^2) dx

The Attempt at a Solution

I think I set up my integral correctly, so my problem is figuring out how to solve the integral. I have:

2pi * Integral of cos 2x * sqrt(1+4sin^2 2x) dx on [0, pi/6]

I can't figure out how to solve that. And can someone teach me how to use Latex for integrals so it looks better.

 

[mjsd]

you integral in LaTeX is (hope I got it right):
\displaystyle{2\pi \int_{0}^{\pi/6} \cos 2x \sqrt{1+4\sin^2 2x} \;dx}

can be done in two stages (instructive way and require only substitution):
first use substitution to make the sq root looks like \sqrt{1+u^2}, in doing so you will find that the integrand has been simplified some what
then
do the u integral with trig substitution, and invoke trig identities to eliminate sq root, then...you should be fine... remember to fix up the limits accordingly.

 

[Gauss177]

Thanks mjsd.

I ran into another problem I'm having trouble with, this time with:
x=e^2y, 0<=y<=1/2 rotated about y-axis.

So, the integral is:

\displaystyle{2\pi \int_{0}^{1/2} y \sqrt{1 + 4e^{4y}} \;dx}

Here I don't think I can use trig substitutions because I can't get rid of the e. Thanks for your help

 

[mjsd]

given your data, I don't think you get such integral. the way it is given, it is much easier to interchange the dummy variable (x <-> y) and do the integral as usual. Draw a diagram if unsure. The resultant integral should be similar to previous one.

 

[HallsofIvy]

Homework Statement

Find the area of the surface obtained by rotating the curve about the x-axis:
y=cos 2x, 0<=x<=pi/6

Homework Equations

Surface area about the x-axis = Integral of 2pi * f(x) * sqrt(1+[f'(x)]^2) dx

The Attempt at a Solution

I think I set up my integral correctly, so my problem is figuring out how to solve the integral. I have:

2pi * Integral of cos 2x * sqrt(1+4sin^2 2x) dx on [0, pi/6]

I can't figure out how to solve that. And can someone teach me how to use Latex for integrals so it looks better.

This is cute. Let u= sin(2x). That will reduce to a square root function and then you use a trig substitution!

Thanks mjsd.

I ran into another problem I'm having trouble with, this time with:
x=e^2y, 0<=y<=1/2 rotated about y-axis.

So, the integral is:

\displaystyle{2\pi \int_{0}^{1/2} y \sqrt{1 + 4e^{4y}} \;dx}

Here I don't think I can use trig substitutions because I can't get rid of the e. Thanks for your help

Don't confuse the variables. Since you are rotating around the y-axis, f(y)= e^2y so your 2\pi f is 2\pi e^{2y}, not y. And, of course, you are integrating with respect to y, not x.

 

[geomajor]

Hi, I have the exact same problem--finding the surface area of y=cos(2x) from 0,\pi/6 rotated about the x-axis.

I already did the substitution and trig substitution to get rid of the square root. But now I'm left with \pi \intfrom 0, pi/6 of sec^3 of \theta d\theta. I want to go back at the very end and replace everything so that I don't have to change the limits...but I'm stuck. I don't know if I did the work correctly so far, and if I did, how do I go on from here?

 

[Gib Z]

\pi \int^{\pi/6}_0 2 \cos (2x) \sqrt{1+4\sin^2 (2x)} dx. Let u= 2 sin (2x), then du = 4 cos (2x) dx. Let's ignore the limits of the integral and just switch back to the original variable at the end, because changing limits accordingly for this is not very fun. Anyway, so it becomes
\frac{\pi}{2} \int \sqrt{1+u^2} du.

Now geomajor, I'm guessing you let u= tan theta, in which case you would get the sec cubed integral. That integral is one of the more harder elementary integrals out there. Usually you repeat integration by parts on it twice, or If you get that integral as a separate question to start with you let u= tan theta to get back to the integral just above, then continue as such:

\int \sqrt{1+u^2} du
Let u= sinh t, then du = cosh t dt. Using the identity \cosh^2 t - \sinh^2 = 1 the new integral follows: \int \cosh^2 t dt.

now use the identity \cosh^2 t = \frac{1+ \cosh (2t)}{2}.

This last integral is very very easy, then just switch back to the original variable x, and plug in your limits as normal.

 

[geomajor]

Thanks! In the end my work finally came together and made sense-yay! :)