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Flux/parameterizing a coneby Gameowner
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#1
Apr910, 09:08 PM

P: 45

1. The problem statement, all variables and given/known data
Question B) 2. Relevant equations 3. The attempt at a solution So I know the flux of S1 U S2 is the same as the flux for S1 + flux of S2, that is Double Int of S(surface) = Double Int of S1 U S2 = Double Int S1 + Double Int S2 The problem I'm having is parameterizing S1. I know for a cone(x^2+y^2=z^2) has the parametrization of x=rcos(theta), y=rsin(theta), z=z, and substituting it into the equation of a cone gives z=r, which means when you map it from one coordinate system into another it maps exactly onetoone. But If I were to use the same parametrization for my particular problem, that is, rearranging my equation of S1 so that I get z=f(x,y), and substituting in the parameterization(x=rcos(theta), y=rsin(theta)) then evaluating it, I get z=3/4*r. So I'm just wondering what I do from here? any hints and tips would be much appreciated. 


#2
Apr910, 11:14 PM

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PF Gold
P: 7,708

How do you get z = (3/4)r?? Your equation is
[tex]x^2 + y^2 = (1z)^4[/tex] which is the same as r^{2} = (1z)^{4}. Solve it for z. 


#3
Apr1010, 10:32 AM

P: 45

But I don't see how substituting in x=rcos[tex]\theta[/tex], y=rcos[tex]\theta[/tex] will help. I tried rearranging my original equation [tex]x^2 + y^2 = (1z)^4[/tex] for z, which gave me [tex]z = 1  \left(x^{2}+y^{2}\right)^{1/4}[/tex], this is in the form of z = g(x,y). Let [tex]\Phi[/tex](x,y)=(x , y , g(x,y)) which = ( x , y , [tex]1  \left(x^{2}+y^{2}\right)^{1/4}[/tex] ) so to find the normal vector, [tex]\widehat{n}[/tex] = [tex]\nabla(zg(x,y))[/tex] = (gx ,  gy , 1) gx = [tex]\frac{1}{2}[/tex][tex]\frac{x}{\left(x^2+y^2\right)^3/4}[/tex] gy = [tex]\frac{1}{2}[/tex][tex]\frac{y}{\left(x^2+y^2\right)^3/4}[/tex] so F:=([tex]\Phi(x,y)[/tex]) = ( xy , y , x^2) thus F([tex]\Phi(x,y)[/tex]) doted with [tex]\widehat{n}[/tex](x,y) this is where the problem begins, if i resolve this dot product, I get a long equation which does not cancel, usually, if there are components of z in F:=([tex]\Phi(x,y)[/tex]) = ( xy , y , x^2), I can substitute in my original function [tex]z = 1  \left(x^{2}+y^{2}\right)^{1/4}[/tex] into all the z's, and when I resolve the dot product, some of it will cancel, but this does not happen this case since there are no z's. am I approaching this question the correct way and method? 


#4
Apr1010, 12:23 PM

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PF Gold
P: 7,708

Flux/parameterizing a cone
[tex]\vec R(r,\theta) = \langle r\cos\theta,r\sin\theta, 1r^{\frac 1 2}\rangle[/tex] Do you know how to use the normal vector [tex]\vec N = \pm \vec R_r \times \vec R_\theta[/tex] with the correct sign for the orientation to calculate the flux integral? 


#5
Apr1110, 10:46 AM

P: 45

I totally forgot about the formula [tex]\vec N = \pm \vec R_r \times \vec R_\theta[/tex]. I went away and did the question again, and is again having troubles. so I firstly used [tex]\vec R(r,\theta) = \langle r\cos\theta,r\sin\theta, 1r^{\frac 1 2}\rangle[/tex] to find [tex]_{R}r[/tex] and [tex]_{R}\theta[/tex] which I got [tex]_R{r}[/tex] [tex]\left[ \begin {array}{c} \cos \left( \theta \right)\\ \noalign{\medskip}\sin \left( \theta \right) \\ \noalign{\medskip}0.5\,{r}^{ 0.5}\end {array} \right] [/tex] [tex]_{R}\theta[/tex] [tex]\left[ \begin {array}{c} r\sin \left( \theta \right)\\ \noalign{\medskip}r\cos \left( \theta \right) \\ \noalign{\medskip}0\end {array} \right] [/tex] Taking the cross product of the two gives me. [tex]\left[ \begin {array}{c} 0.5\,{r}^{ 0.5}\cos \left( \theta \right)\\ \noalign{\medskip} 0.5\,{r}^{ 0.5}\sin \left( \theta \right)\\ \noalign{\medskip} \left( \cos \left( \theta \right) \right) ^{2}r+ \left( \sin \left( \theta \right) \right) ^{2}r\end {array}\right][/tex] substituting my [tex]F = \langle xy,y, x^{2}\rangle[/tex] with [tex]\vec R(r,\theta) = \langle r\cos\theta,r\sin\theta, 1r^{\frac 1 2}\rangle[/tex] to get [tex] \left[ \begin {array}{c} {r}^{2}\cos \left( \theta \right) \sin\left( \theta \right) \\ \noalign{\medskip}{\it rsin} \left( \theta\right) \\ \noalign{\medskip}1{r}^{ 0.5}\end {array} \right] [/tex] and thus yielding the integral [tex]\int _{0}^{2\,\pi}\!\int _{0}^{1}\! 0.5\,{r}^{ 2.5} \left( \cos\left( \theta \right) \right) ^{2}\sin \left( \theta \right) + 0.5\,{r}^{ 1.5} \left( \sin \left( \theta \right) \right) ^{2}+2\,{r}^{3}\left( \cos \left( \theta \right)\right) ^{2}{dr}\,{d\theta}[/tex] but again, having problems....any pointers would be much appreciated. 


#6
Apr1110, 11:06 AM

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#7
Apr1110, 11:13 AM

P: 45

Besides that, the original question ask to find the flux out of the bounded region, and here I'm finding the flux of S1 alone, would I then need to find the flux of S2 also? 


#8
Apr1110, 11:31 AM

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PF Gold
P: 7,708

Of course you need to calculate the flux through S_{1}. And be sure to take into account that the outward normal is downward. Did you check your normal for the upper surface was correctly oriented or did you just get lucky? 


#9
Apr1110, 09:59 PM

P: 45

In regards to the orientation, I think I just got lucky heh, to get the right orientation, you just have to change the sign convention of the normal vector? 


#10
Apr1110, 10:11 PM

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PF Gold
P: 7,708

[tex]\vec N = \vec R_r \times \vec R_\theta\hbox{ or } \vec R_\theta \times \vec R_r[/tex] They are both perpendicular to the surface but in opposite directions so one is in the correct direction for your problem and one is opposite. So you calculate one of them and examine it to see whether it is in the correct direction or whether you need to multiply it by 1. In your problem, outward on the top surface is generally upward so look at the z component of N to see if you have it correct. Lucky guess, eh? Don't forget to get it right on the bottom surface. 


#11
Apr1410, 11:09 AM

P: 45

If you're still reading this thread, what double angle formula would I need to use in this case? is it Cos2Theta=cos^2thetasin^2theta? I took back what I said earlier, I'm actually having difficulties integrating this. Also, I've just came back to question c), and was wondering where the question saids that the answer should be the same as a), which I don't understand? as a) is just a graph? lastly, this is what I've done with c) to find the volume integral(triple) and was wondering if you can have a look and let me know if i'm going in the right direction. So I started off rearranging my original equation of the semicone [tex]x^2 + y^2 = (1z)^4[/tex] into [tex]x^2 + y^2 (1z)^4 = 0[/tex] using the formula given, I calculated [tex]\nabla[/tex] which is [tex]\left[\frac{d}{dx}\frac{d}{dy}\frac{d}{dz}\right][/tex] [tex]\nabla=[/tex] [tex] \left[ \begin {array}{c} 2\,x\\ \noalign{\medskip}2\,y \\ \noalign{\medskip}4\, \left( 1z \right) ^{3}\end {array} \right][/tex] [tex]F=[/tex] [tex]\left[ \begin {array}{c} xy\\ \noalign{\medskip}y \\ \noalign{\medskip}{x}^{2}\end {array} \right][/tex] Substituting into the formula given then evaluating the dot product, I get this as a final integral [tex]\int _{1}^{1}\!\int _{\sqrt {1{x}^{2}}}^{\sqrt {1{x}^{2}}}\!\int _ {0}^{1}\!2\,{x}^{2}y+2\,{y}^{2}+4\, \left( 1z \right) ^{3}{x}^{2}{dz} \,{dy}\,{dx}[/tex] 


#12
Apr1410, 12:15 PM

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PF Gold
P: 7,708

Edit  Added later: You should get 9π/20 for the upper surface, π/4 for the lower, and π/5 for the triple integral. 


#13
Apr1410, 11:46 PM

P: 45

Lastly, I went back to check the double integral I did with S1, I realized I made a mistake with the vector F.. I originally had the vector F which is [tex] F = \langle xy,y, x^{2}\rangle [/tex] substituting in [tex] \vec R(r,\theta) = \langle r\cos\theta,r\sin\theta, 1r^{\frac 1 2}\rangle [/tex] to give me [tex] \left[ \begin {array}{c} {r}^{2}\cos \left( \theta \right) \sin\left( \theta \right) \\ \noalign{\medskip}{\it rsin} \left( \theta\right) \\ \noalign{\medskip}1{r}^{ 0.5}\end {array} \right] [/tex] Then I realized the last component isn't z, therefore it should have been substituted with x^2 instead of z. So I'm just wondering if the answer you given in your last post of 9π/20 was adjusted with this error? Thanks for the 3 answers you given me, it helps a lot as now I have something to work towards. 


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