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September 4th, 2017, 02:17 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3  Another Surface Integral problem... #2
Evaluate ∫∫ ( 2, 3, 4 ) * N dS, where D is given by z = x^2 + y^2, 1 <= x <= 1, 1 <= y <= 1, oriented up. Official Answer: 60 My solution: $\displaystyle x = r cos \theta $ $\displaystyle y = r sin \theta $ $\displaystyle z = r^2 $ $\displaystyle G(r, \theta) = ( r cos \theta , r sin \theta , r^2 ) $ $\displaystyle dG/dr = ( cos \theta , sin \theta , 2r ) $ $\displaystyle dG/d \theta = ( r sin \theta , r cos \theta , 0 ) $ $\displaystyle dG/dr ~cross~product~ dG/d \theta = ( 2r^2 cos \theta , 2r^2 sin \theta , r ) $ Dot Product: $\displaystyle \int_{0}^{2\pi} \int_{0}^{1} ( 2, 3, 4) * ( 2r^2 cos \theta , 2r^2 sin \theta , r ) ~r~dr~d \theta $ $\displaystyle \int_{0}^{2\pi} \int_{0}^{1} ( 4r^2 cos \theta + 6r^2 sin \theta + 4r ) ~r~dr~d \theta $ My Answer: $\displaystyle {8 \pi} / 3 $ Any idea what I did wrong? Last edited by zollen; September 4th, 2017 at 02:57 PM. 
September 4th, 2017, 02:48 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
It's probably not the only problem, but your dot product is wrong. $2 \times 2 \ne 2$.

September 4th, 2017, 02:58 PM  #3 
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3  
September 4th, 2017, 04:30 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,500 Thanks: 1372 
Your entire strategy is flawed I believe. There is no reason to go to cylindrical coordinates here. The surface is a square based pyramid. There are 5 normal vectors in total, 4 sides and the top. Find these, dot them with the field vector $(2, 3, 4)$ and integrate over each appropriate surface. 
September 4th, 2017, 04:34 PM  #5  
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 
It isn't a square base pyramid at all. z = x^2 + y^2 is a inverse circular cone. Quote:
 
September 4th, 2017, 05:56 PM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
It does have crosssection parallel to the $xy$ plane, but the top is not flat. For example, the points $(1,1,2),(1,1,2),(1,1,2),(1,1,2)$ are on the surface, but the rest of the circle formed by the intersection of the cylinder $x^2 + y^2 = 2$ and the plane $z=2$ is not on the surface. The projection onto the $xy$ plane is a square. Here's a plot of the surface: Academo Surface Plotter Following your working with the revised surface, you then get $$G(s,t) = (s,t,s^2+t^2) \quad \text{where} \begin{cases}1 \le s \le 1 \\ 1 \le t \le 1\end{cases}$$ Thus \begin{align*}\frac{\partial G}{\partial s} &= \langle 1,0,2s \rangle \\ \frac{\partial G}{\partial t} &= \langle 0,1,2t \rangle \\ \text{and} \quad \frac{\partial G}{\partial s} \times \frac{\partial G}{\partial t} &= \langle 2s, 2t, 1 \rangle \end{align*} You then have \begin{align*}\int_{1}^1 \int_{1}^{1} \langle 2,3,4 \rangle \cdot \langle 2s, 2t, 1 \rangle \, \mathrm ds \,\mathrm dt &= \int_{1}^1 \int_{1}^{1} 4s+ 6t + 4 \, \mathrm ds \,\mathrm dt \\ &= \int_{1}^1 \big[2s^2 + 6st + 4s\big]_{s=1}^1 \,\mathrm dt \\ &= \int_{1}^1 12t + 8 \,\mathrm dt \\ &= \big[ 6t^2 + 8t\big]_{t=1}^1 \\ &= 16 \end{align*} Now, I'm rusty on surface integrals, so I may well have made an error. But equally, you (or somebody) might have misheard 16 as 60. What I would say is that I'm expecting the normal vector to be a unit normal vector and I don't think this one is. Last edited by v8archie; September 4th, 2017 at 06:51 PM. 
September 5th, 2017, 08:32 AM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra  
September 5th, 2017, 03:54 PM  #8  
Senior Member Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 
Correct me if I am wrong. But.. between 1 <= z <= 2 is a square, but below z <= 1 is still a circular cone. Should I break down the surface integration into two parts and compute them separately? Quote:
Last edited by zollen; September 5th, 2017 at 03:57 PM.  
September 5th, 2017, 04:14 PM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 
No, because $z=x^2+y^2$ and $1 \le x \le 1$, $1 \le x \le 1$ specifies the whole surface. That is, by specifying $x$ and $y$, we fix $z$. An alternative way to see this is that the parameterisation $\langle s,t,s^2+t^2\rangle$ specifies the surface with two degrees of freedom. The two limits on the integrals (w.r.t. $s$ and $t$ respectively) therefore cover the whole surface. 
September 5th, 2017, 04:17 PM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra  No. Above $z=1$ the surface is comprised of circular arcs cutting the corners of a square. The sides of the square are not part of the surface. The site I linked to for the surface plot also has a contour plotter which shows this clearly. Unfortunately I can't get a URL to the exact plot. Last edited by v8archie; September 5th, 2017 at 04:23 PM. 

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