My Math Forum Another Surface Integral problem... #2

 Calculus Calculus Math Forum

 September 4th, 2017, 02:17 PM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 179 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 02:57 PM.
 September 4th, 2017, 02:48 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2457 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, 02:58 PM   #3
Senior Member

Joined: Jan 2017
From: Toronto

Posts: 179
Thanks: 2

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, 04:30 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,009 Thanks: 1042 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, 04:34 PM   #5
Senior Member

Joined: Jan 2017
From: Toronto

Posts: 179
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 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, 05:56 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2457 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 06:51 PM.
September 5th, 2017, 08:32 AM   #7
Math Team

Joined: Dec 2013
From: Colombia

Posts: 7,330
Thanks: 2457

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, 03:54 PM   #8
Senior Member

Joined: Jan 2017
From: Toronto

Posts: 179
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 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 03:57 PM.

 September 5th, 2017, 04:14 PM #9 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2457 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: 7,330
Thanks: 2457

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 04:23 PM.

 Tags integral, problem, surface

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post zollen Calculus 6 September 4th, 2017 01:47 PM zollen Calculus 2 September 4th, 2017 09:05 AM henrymerrild Calculus 2 May 1st, 2014 10:33 AM kriko Calculus 1 August 21st, 2010 10:55 AM sansar Calculus 1 May 10th, 2009 04:21 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top