Recent content by november1992

This problem was given to me by my instructor.

Homework Statement Evaluate ∫∫∫\sqrt{x^{2} + y^{2}} dA where R is the region bounded by the paraboloid y=x^2+z^2 and the plane y=4 Homework Equations I believe this is a problem where cylindrical coordinates would be useful 0 ≤ z ≤ \sqrt{4-x^2} 0 ≤ r ≤ 2 ( I think this is wrong). 0 ≤ θ ≤...

I can't believe I overlooked that. I also removed the parentheses. Now I'm getting 26.89. Thanks for the help

I tried it 3 different ways and I still get 20.39 http://i.imgur.com/mQW6TaO.png http://i.imgur.com/DmTseJc.png http://i.imgur.com/QKfTfoW.png

I'm not getting a number. http://i.imgur.com/mQ4Pfac.png I'm guessing I have incorrect syntax. Edit: I realized i didn't capitalize the 'p' in Pi. now I"m getting 13k though I got 20.39

From my textbook. I scanned the page I found it from. http://i.imgur.com/AivojRL.jpg

Okay, thanks. I have one last question. How can I use the inte\frac{}{}rval <0,4pi> in this equation? I'm just a little confused about why this formula, k = |\frac{r(t)' X r(t)''}{r(t)'^3}| gives a different answer to the formula you posted

Haha, I actually managed to plot it, but I thought it was wrong because I wasn't expecting a huge wave. I was also told to calculate the length of the curvature. Do you think this would be an acceptable answer? http://i.imgur.com/HHYzFy7.png Is there a way to simplify this expression?

It doesn't seem to make a difference. http://i.imgur.com/woWlabm.png http://i.imgur.com/ptUZcVG.png

Homework Statement r(t)={(4+cos20t) cost,+(4+cos20t) sint,+0.4sin20t} Calculate the curvature of r[t] for 0≤t≤4pi Homework Equations k = | r' x r'' | / | r' |^3 The Attempt at a Solution r[t_]:={4+Cos[20t]*Cos[t],4+Cos[20t]*Sin[t],0.4Sin[20t]} k[t_]:=Norm[Cross[r',r'']]/Norm[r']^3...

I was wondering if there is an easier way to solve circuits using matrix inversion if I have complex numbers. So far I've been doing them by hand with 2x2 matrices. It isn't hard, but it takes a while. I have a test coming up in about a week which will have 10 questions that have to be solved in...

Your explanation confused me, but I got the answer right answer. I'm not sure why though. I doubled the distance from the object to the first mirror and added that to the distance of the second image from the first mirror.

What is dn?

Homework Statement Two parallel mirrors are separated by a distance of d = 4 meter. A point object is placed at a position a = 0.4 meter from the mirror 1. How deep is the second image of the object in the mirror 1? How deep is the third image of the object in the mirror 1...

I don't see that formula in my textbook. When I plug in the numbers I get: \frac{2}{s^3} + \frac{e^-s}{s} This is what I did to get my first answer. L[t^2 - t^2 δ(t-1))] \frac{2}{s^3} + \frac{d}{ds} (\frac{d}{ds} (\frac{e^(-s)}{s}) \frac{2}{s^3} +\frac{d}{ds} (\frac{e^-s...