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 Calculus Calculus Math Forum

 September 6th, 2017, 10:58 PM #1 Math Team   Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित parameterization in Surface integration I want to surface integrate over a surface of the plane $\displaystyle S : 2x+3y+6z =12$ which lie in the 1st octant. Should I use parameterization, if I should, how? September 7th, 2017, 03:52 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 A surface, being two dimensional, requires two parameters. You are given 2x+3y+6z=12 which you can write as x= 6- (3/2)y- 3z. Take y and z as parameters- or, if you prefer, s and t, writing x= 6- (3/2)s- 3t, y= s, z= t. Obviously, you can do that with any of the variables: we can just as well write y= 4- (2/3)x- 2z and take x= s, y= 4- (2/3)s- 2t, z= t or we can write z= 2- (1/3)x- (1/2)y and take x= s, y= t, z= 2-(1/3)s- (1/2)t. Or, if you don't like fractions, take x= 3s, y= 2t, z= 2- s- t. September 7th, 2017, 05:40 AM   #3
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 Originally Posted by Country Boy Or, if you don't like fractions, take x= 3s, y= 2t, z= 2- s- t.
Poor fractions September 7th, 2017, 08:11 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Yeah, I'm so mean to fractions- but they started it! September 7th, 2017, 08:31 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,689 Thanks: 2669 Math Focus: Mainly analysis and algebra I think you're being irrational. September 7th, 2017, 04:52 PM #6 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,838 Thanks: 653 Math Focus: Yet to find out. Get real! September 11th, 2017, 05:10 AM #7 Math Team   Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित Suppose i need to integrate a function F = 36z - 36 + 18y over the surface, how should i place the limits? September 12th, 2017, 04:56 AM #8 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I thought I had already responded to this! In my first post I suggested the parameterization x= 3s, y= 2t, z= 2- s- t. Since we can write s and t in terms of x and y separately, we want to project the surface down to the xy-plane. The plane 2x+3y+6z=12 crosses the z= 0 plane where 2x+ 3y= 12 and that cuts the x-axis where 2x= 12 or x= 6. To integrate over that, we can take x going from 0 to 6 and, for each x, y going from 0 to y= (12- 2x)/3= 4- (2/3)x. Since x= 3s and y= 2t, s= x/3 goes from 0 to 2 and, for each s, t goes from 0 to (4- (2/3)x)/2= 2- (1/3)x= 2- (1/3)(3s)= 2- s. That is, integrate $\displaystyle \int_0^6 \int_0^{2- s} .... dtds$. Tags integration, parameterization, 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 3 September 6th, 2017 05:51 PM KyVanchhay Calculus 4 July 27th, 2013 01:06 AM zell^ Calculus 13 April 13th, 2012 08:41 AM Zilee Calculus 4 November 25th, 2011 09:35 AM FataLIdea Calculus 1 March 28th, 2008 04:46 PM

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