Calculus Calculus Math Forum

 September 4th, 2017, 03:17 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 03:57 PM. September 4th, 2017, 03:48 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra It's probably not the only problem, but your dot product is wrong. $2 \times 2 \ne 2$. Thanks from zollen September 4th, 2017, 03:58 PM   #3
Senior Member

Joined: Jan 2017
From: Toronto

Posts: 209
Thanks: 3

It was a typo. I have fixed it.

Quote:
 Originally Posted by v8archie It's probably not the only problem, but your dot product is wrong. $2 \times 2 \ne 2$. September 4th, 2017, 05:30 PM #4 Senior Member   Joined: Sep 2015 From: USA Posts: 2,641 Thanks: 1475 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. Thanks from zollen September 4th, 2017, 05:34 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:
 Originally Posted by romsek 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.
Attached Images shape.jpg (19.2 KB, 1 views) September 4th, 2017, 06:56 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra It does have cross-section 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 mis-heard 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. Thanks from zollen Last edited by v8archie; September 4th, 2017 at 07:51 PM. September 5th, 2017, 09:32 AM   #7
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,697
Thanks: 2681

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by v8archie It does have cross-section parallel to the $xy$ plane...
That should be a circular cross-section. September 5th, 2017, 04:54 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:
 Originally Posted by v8archie It does have cross-section 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.

Last edited by zollen; September 5th, 2017 at 04:57 PM. September 5th, 2017, 05:14 PM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 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, 05:17 PM   #10
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,697
Thanks: 2681

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by zollen But.. between 1 <= z <= 2 is a square,
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 05:23 PM. Tags integral, problem, surface Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post zollen Calculus 6 September 4th, 2017 02:47 PM zollen Calculus 2 September 4th, 2017 10:05 AM henrymerrild Calculus 2 May 1st, 2014 11:33 AM kriko Calculus 1 August 21st, 2010 11:55 AM sansar Calculus 1 May 10th, 2009 05:21 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      