Recent content by soofjan

1. Rouche's theorem

If you draw the circle, you can see that it is possible for the distance from (0,0) to a z on the circle to be 1 (when the argument is 60 degrees), and then | 6z^2 - 6 | = 0. In that case, g(z) won't be bigger than f(z) - g(z) for every z, as Rouche requires. This is what buffles me about this...
2. Rouche's theorem

I have been trying to solve the following question for a while now. I need to find the number of roots of: f(z) = 6z^2 - 6 + Log(1+z) in the area D = { z is complex : |z-1| < 1 } I assume this is solved by Rouche's theorem that requires me to find 2 analytic functions, h(z) and g(z) in D...
3. Triple Integral, Spherical Coordinates

Well, I tried integrating by dp before dr.. it seems to work. I get: sqrt{r^2+a^2-2ra cos(p)} / {a*r} and when I put the limits 0..pi, I get that: r^2 * {|a+r|-|a-r|} / {a*r} since a>R>r, I can remove the absolute value and my final answer is: {4*pi*R^3} / {3*a} It never occoured to me...
4. Triple Integral, Spherical Coordinates

Homework Statement Calculate: integral B [ 1/sqrt(x^2+y^2+(z-a)^2) ] dx dy dz , when B is the sphere of radius R around (0,0,0), a>R. Homework Equations The Attempt at a Solution I tried spherical coordinates for the integrand: x=rsin(p)cos(t) y=rsin(p)sin(t) z=rcos(p)+a The problem is...

Ok, thanks.
6. Length of a Curve

x' = -sin(t) y' = cos(t) z = cos(t)*sin(t) = sin(2t)/2 --> z' = cos(2t) (x')^2+(y')^2+(z')^2 = [-sin(t)]^2+cos(t)^2+cos(2t)^2 = 1+cos(2t)^2 hence, sqrt[1+cos(2t)^2]
7. Length of a Curve

Homework Statement Find the length of the curve on the sufrace z=xy, whose projection on the xy plane is a 1 radius circle around (0,0). Homework Equations The Attempt at a Solution Length of a curve - usually linear integration of the first kind, when f(x,y,z)=1. So I tried...
8. Area of a given Curve

I see your point. Thanks again.
9. Area of a given Curve

I tried to polar plot the following on Wolfram: sqrt(|r*cos(t)^4|) + sqrt(|r*sin(t)^4|) = 1 r=0..1, t=0..2*pi
10. Area of a given Curve

Homework Statement Find the area, whose edge is given by the following curve: sqrt(|x|) + sqrt(|y|) = 1 Also, draw the area. Guidance: try x = r*cos(t)^k, y = r*sin(t)^k for a fitting k. Homework Equations The Attempt at a Solution I tried: x = r*cos(t)^4, y = r*sin(t)^4 From...
11. Double Integral, tricky

The problem with Exponential Intergal is that we don't learn it in Calculus 2, thus we can't use something we never learned in class. I will ask a teaching assistant about this, since I'm obviously missing something here.
12. Double Integral, tricky

This homework was written by the Vice Dean of Mathematics, so I'm almost sure it is supposed to be messy like that... however, I'm not sure I even remember what to do in case of a series.. but I'll give it some more thought. Thanks everyone =)
13. Double Integral, tricky

Well, I guess that's not the end of it. After integrating: e^(x/y) dx dy, x=y..8, y=2..8 I get: ( y*e^(8/y) - y*e ) dy y=2..8 What do I do about the bolded part? It's not integrable.
14. Double Integral, tricky

edit: didnt see post above. Thank you very much. It works =)
15. Double Integral, tricky

First of all, thanks for your informative post. Now, I drew the functions and it looks like y should be from 1 to 8. As for dx, how exactly to you know the range? I tried this: x^(1/3) = y ==> x = y^3 x=y ==> x=y so the new range for dx is y..y^3 That's far from what wolfram gives, so I have...