My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree5Thanks
  • 1 Post By v8archie
  • 1 Post By romsek
  • 1 Post By v8archie
  • 1 Post By v8archie
  • 1 Post By v8archie
Reply
 
LinkBack Thread Tools Display Modes
September 4th, 2017, 03:17 PM   #1
Senior Member
 
Joined: Jan 2017
From: Toronto

Posts: 152
Thanks: 2

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.
zollen is offline  
 
September 4th, 2017, 03:48 PM   #2
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,031
Thanks: 2342

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
v8archie is offline  
September 4th, 2017, 03:58 PM   #3
Senior Member
 
Joined: Jan 2017
From: Toronto

Posts: 152
Thanks: 2

It was a typo. I have fixed it.

Quote:
Originally Posted by v8archie View Post
It's probably not the only problem, but your dot product is wrong. $2 \times 2 \ne 2$.
zollen is offline  
September 4th, 2017, 05:30 PM   #4
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: Southern California, USA

Posts: 1,602
Thanks: 816

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
romsek is offline  
September 4th, 2017, 05:34 PM   #5
Senior Member
 
Joined: Jan 2017
From: Toronto

Posts: 152
Thanks: 2

It isn't a square base pyramid at all. z = x^2 + y^2 is a inverse circular cone.


Quote:
Originally Posted by romsek View Post
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
File Type: jpg shape.jpg (19.2 KB, 1 views)
zollen is offline  
September 4th, 2017, 06:56 PM   #6
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,031
Thanks: 2342

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.
v8archie is offline  
September 5th, 2017, 09:32 AM   #7
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,031
Thanks: 2342

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by v8archie View Post
It does have cross-section parallel to the $xy$ plane...
That should be a circular cross-section.
Thanks from zollen
v8archie is offline  
September 5th, 2017, 04:54 PM   #8
Senior Member
 
Joined: Jan 2017
From: Toronto

Posts: 152
Thanks: 2

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 View Post
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.
zollen is offline  
September 5th, 2017, 05:14 PM   #9
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,031
Thanks: 2342

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.
v8archie is offline  
September 5th, 2017, 05:17 PM   #10
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,031
Thanks: 2342

Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by zollen View Post
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.
Thanks from zollen

Last edited by v8archie; September 5th, 2017 at 05:23 PM.
v8archie is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
integral, problem, surface



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Generic surface integral problem zollen Calculus 6 September 4th, 2017 02:47 PM
Another Surface Integral problem... zollen Calculus 2 September 4th, 2017 10:05 AM
What is a surface integral? henrymerrild Calculus 2 May 1st, 2014 11:33 AM
surface integral kriko Calculus 1 August 21st, 2010 11:55 AM
Surface Integral sansar Calculus 1 May 10th, 2009 05:21 AM





Copyright © 2017 My Math Forum. All rights reserved.