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'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
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?
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.
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...
Homework Statement
L[t^{2} - t^{2}δ(t-1)]
Homework Equations
L[ t^{n}f(t)] = (-1^{n}) \frac{d^{n}}{ds^{n}} L[f(t)]
L[δ-t] = e^-ts
The Attempt at a Solution
My teacher wrote \frac{2}{s^{3}} -e^{s} as the answer.
I got \frac{2}{s^{3}} + \frac{e^-s}{s} + 2 \frac{e^-s}{s^2} + \frac{2e^-s}{s^3}
Homework Statement
y" + y = 4δ(t-2π); y(0)=1, y'(0)=0
Homework Equations
L[f(t-a) U(t-a)] = e^{-as} L[f(t)]
L[δ(t-c)] = e^{-cs}
The Attempt at a Solution
My answer is: cos(t) + 4U(t-2π)sin(t-2π).
When I used Wolframalpha it gave me 4sin(t)U(t-2π) + cos(t)
I have another question. Can the undetermined coefficients method be used to solve cauchy-euler equations, or does variation of parameters have to be used?
That's what I assumed, but my teacher believes it's still faster to use variation of parameters. I don't like using that method because I tend to make more mistakes when I integrate e and a trig function.
Would this method still be plausible if the right hand side was like this...
I was wondering what a guess would be for the particular solution of the right hand side of an equation if it looked like this:
x^{2}y" - 4xy' + 6y = ln(x)
My textbook has some specific examples of the right side function along with the corresponding form of the particular solution...
Homework Statement
http://i.imgur.com/fSRqw.png
Homework Equations
V_{th} = V_{oc}
The Attempt at a Solution
I got the thevenin voltage by using node voltage
\frac{V2-40}{5} - \frac{80+2V2}{5} -8 + V2 = 0
(\frac{1}{5} + \frac{2}{5} + 1)V2 = 32
\frac{8}{5}V2 = 32
V2 =...
Homework Statement
http://i.imgur.com/YMuGM.png
Homework Equations
\tau = F * r
The Attempt at a Solution
I don't know how to do these types of problems. My professor didn't even go over this. He just said to use that formula to find the answer.
I know the electric field...
I got 12.025. I used mesh current analysis. I misread the section in my textbook about supermeshes. I thought that supermeshes were only used when an independent current source was shared by two meshes. I didn't know that dependent current sources were acceptable. Boy do I feel stupid; I spent...
Yes, I meant short-circuit current. I mixed up the different steps in my textbook. For another method it says when deactivating a current source, it should be replaced with an open circuit.
I'll give that a try.
That equation you wrote looks similar to the one in my textbook, but it deals with maximum power transfer. The way my professor wanted me to solve for the thevenin resistance was to find the thevenin voltage and divide by the open-circuit current.
http://i.imgur.com/OkJNF.png