Line Integrals 2: Evaluate Triangle on Vertices (0,0), (3,3), (0,3)

[bugatti79]

Homework Statement

Evaluate this integral directly

Homework Equations

$\int cos x sin y dx +sin x cos y dy$ on vertices (0,0), (3,3) and (0,3) for a triangle

The Attempt at a Solution

Does this have to evaluated parametrically using $r(t)=(1-t)r_0+tr_1$ for $0 \le t\le 1$

or can I just proceed evaluating each line segment in a counter clockwise manner starting with the vertices:

from (0,0) t (3,3) which gives the line y=x hence dy=dx, substituting in the above

$\displaystyle I_{ab} = \int_0^3 (cos x sin x + sin x cos x ) dx = \int_0^3 sin 2x dx$...?

 

[A. Bahat]

How directly is directly? Because using Green's theorem makes this trivial.

 

[bugatti79]

Yes, i can use green's but i need to verify it directly...

 

[bugatti79]

Homework Statement

Evaluate this integral directly

Homework Equations

$\int cos x sin y dx +sin x cos y dy$ on vertices (0,0), (3,3) and (0,3) for a triangle

The Attempt at a Solution

Does this have to evaluated parametrically using $r(t)=(1-t)r_0+tr_1$ for $0 \le t\le 1$

or can I just proceed evaluating each line segment in a counter clockwise manner starting with the vertices:

from (0,0) t (3,3) which gives the line y=x hence dy=dx, substituting in the above

$\displaystyle I_{ab} = \int_0^3 (cos x sin x + sin x cos x ) dx = \int_0^3 sin 2x dx$...?

Any thoughts on how I calculate this directly?

 

[A. Bahat]

I guess the best way would be to just split it up over each side and evaluate it parametrically.

 

[HallsofIvy]

Homework Statement

Evaluate this integral directly

Homework Equations

$\int cos x sin y dx +sin x cos y dy$ on vertices (0,0), (3,3) and (0,3) for a triangle

The Attempt at a Solution

Does this have to evaluated parametrically using $r(t)=(1-t)r_0+tr_1$ for $0 \le t\le 1$

or can I just proceed evaluating each line segment in a counter clockwise manner starting with the vertices:

from (0,0) t (3,3) which gives the line y=x hence dy=dx, substituting in the above

$\displaystyle I_{ab} = \int_0^3 (cos x sin x + sin x cos x ) dx = \int_0^3 sin 2x dx$...?

There is no "or" here- the methods you mention are identical. Of course, because the path itself has "corners", you can't have a single polynomial for the entire path- you need a "piecewise" polynomial which is what you have started with- the line from (0,0) to (3,3). Now, the lline from (3, 3) to (0, 3) is given by x= 3- t, y= 0 so that dx= -dt, dy= 0 with t from 0 to 3 Or, equivalently, x= t, y= 0 so that dx= dt, dy= 0 with t from 3 to 0. Note that integrating from 3 to 0 reverses the sign of the integral, giving the same result as the "-dx" when integrating from 0 to 3.

Now, how would you handle the path from (0,3) to (0,0)?

 

[bugatti79]

There is no "or" here- the methods you mention are identical. Of course, because the path itself has "corners", you can't have a single polynomial for the entire path- you need a "piecewise" polynomial which is what you have started with- the line from (0,0) to (3,3). Now, the lline from (3, 3) to (0, 3) is given by x= 3- t, y= 0 so that dx= -dt, dy= 0 with t from 0 to 3 Or, equivalently, x= t, y= 0 so that dx= dt, dy= 0 with t from 3 to 0. Note that integrating from 3 to 0 reverses the sign of the integral, giving the same result as the "-dx" when integrating from 0 to 3.

Now, how would you handle the path from (0,3) to (0,0)?

My attempt for (3,3) back to (0,3) is as follows

THis is a horizontal line at y=3 implies dy=0 therefore the original integral of ∫cos x sin y dx+ sin x cos y dy (and subbing y=3 )reduces to

$\int_3^0 cos x sin 3 dx$...? Why isn't this right?

 

[bugatti79]

My attempt for (3,3) back to (0,3) is as follows

THis is a horizontal line at y=3 implies dy=0 therefore the original integral of ∫cos x sin y dx+ sin x cos y dy (and subbing y=3 )reduces to

$\int_3^0 cos x sin 3 dx$...? Why isn't this right?

I have it thank you.