Calculating a Line Integral with Triangle C

kliker
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Homework Statement


hello again, sorry for asking so many questions, i just want to make sure if I am correct or not

calculate the line integral y^2dx+x^2dy where line C is the triangle with sides x=1, y=0 and y=x

The Attempt at a Solution



first of all i tried to find a customization of the line

we know that x = 1 hence it will be like this

r(t) = (1,t) but I am not sure if it's correct

then i said that the integral would be this

[URL]http://www2.wolframalpha.com/Calculate/MSP/MSP470019b34815c34hai2900000d7egii1beg78b2i?MSPStoreType=image/gif&s=9&w=88&h=37[/URL]

could I just use Green's theorem?

I mean using Greens theorem I get the same result

[URL]http://www2.wolframalpha.com/Calculate/MSP/MSP167119b34a9g72bg50ia000021ai53gc37dac65a?MSPStoreType=image/gif&s=3&w=186&h=37[/URL]

im 99% sure that greens theorem is correct, i mean the way i implemented it, but is the first way i showed also correct?

thanks in advance
 
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kliker said:

Homework Statement


hello again, sorry for asking so many questions, i just want to make sure if I am correct or not

calculate the line integral y^2dx+x^2dy where line C is the triangle with sides x=1, y=0 and y=x



The Attempt at a Solution



first of all i tried to find a customization of the line

we know that x = 1 hence it will be like this

r(t) = (1,t) but I am not sure if it's correct

then i said that the integral would be this

[URL]http://www2.wolframalpha.com/Calculate/MSP/MSP470019b34815c34hai2900000d7egii1beg78b2i?MSPStoreType=image/gif&s=9&w=88&h=37
[/URL]
"we know that x=1"? Did you leave out much of the problem? We know that x= 1, y= any number from 0 to 1 is one side of triangle and so one part of the path over which we want to integrate. Taking x= 1, y= t as parametric equations, dx= 0, dy= dt so the integral becomes
\int_0^1 1 dx= 1[/itex]<br /> <br /> But you still have to do the other two sides of the triangle.<br /> <br /> On the line y= 0, we can use parametric equations x= t, y= 0 with t from 0 to 1. Then dx= dt, dy= 0 but y^2 dx= 0dt so the integral is <br /> \int_0^1 0dt= 0[/itex].&lt;br /&gt; &lt;br /&gt; On the line y= x, where we are integrating from (1, 1) to (0, 0) (we got counterclockwise around the closed path), we can take x= t, y= t so that dx= dt, dy= dt and the integral is&lt;br /&gt; \int_1^0 2t^2 dt= -\int_0^1 2t^2 dt= -2/3[/itex] and the entire integral is 1- 2/3= 1/3.&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; &amp;lt;blockquote data-attributes=&amp;quot;&amp;quot; data-quote=&amp;quot;&amp;quot; data-source=&amp;quot;&amp;quot; class=&amp;quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&amp;quot;&amp;gt; &amp;lt;div class=&amp;quot;bbCodeBlock-content&amp;quot;&amp;gt; &amp;lt;div class=&amp;quot;bbCodeBlock-expandContent js-expandContent &amp;quot;&amp;gt; could I just use Green&amp;amp;#039;s theorem? &amp;lt;blockquote data-attributes=&amp;quot;&amp;quot; data-quote=&amp;quot;&amp;quot; data-source=&amp;quot;&amp;quot; class=&amp;quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&amp;quot;&amp;gt; &amp;lt;div class=&amp;quot;bbCodeBlock-content&amp;quot;&amp;gt; &amp;lt;div class=&amp;quot;bbCodeBlock-expandContent js-expandContent &amp;quot;&amp;gt; Yes, you certainly could!&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; &amp;lt;blockquote data-attributes=&amp;quot;&amp;quot; data-quote=&amp;quot;&amp;quot; data-source=&amp;quot;&amp;quot; class=&amp;quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&amp;quot;&amp;gt; &amp;lt;div class=&amp;quot;bbCodeBlock-content&amp;quot;&amp;gt; &amp;lt;div class=&amp;quot;bbCodeBlock-expandContent js-expandContent &amp;quot;&amp;gt; I mean using Greens theorem I get the same result&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; [URL]http://www2.wolframalpha.com/Calculate/MSP/MSP167119b34a9g72bg50ia000021ai53gc37dac65a?MSPStoreType=image/gif&amp;amp;amp;s=3&amp;amp;amp;w=186&amp;amp;amp;h=37[/URL]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; im 99% sure that greens theorem is correct, i mean the way i implemented it, but is the first way i showed also correct?&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; thanks in advance &amp;lt;/div&amp;gt; &amp;lt;/div&amp;gt; &amp;lt;/blockquote&amp;gt; &amp;lt;/div&amp;gt; &amp;lt;/div&amp;gt; &amp;lt;/blockquote&amp;gt; &amp;lt;/div&amp;gt; &amp;lt;/div&amp;gt; &amp;lt;/blockquote&amp;gt;
 
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thanks a lot for your help

that cleared up everything in my mind
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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