# Search results

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...
16. ### Double Integral, tricky

Ok, so how do I determine the new ranges? Couldn't find anything about that in our study material.
17. ### Double Integral, tricky

I tried that but I got something that is just wrong.. am I supposed to get a sum of 2 integrals when changing the order?
18. ### Double Integral, tricky

Homework Statement Calculate the following: integrate [ (e^(x/y)) dy y=x^(1/3)..x ] dx x=1..8 Homework Equations The Attempt at a Solution In this current state the dy part is not integrable, so there must be some trick with changing the variables or something else. Couldn't find a proper...