Line Integrals

[nasi112]

Line integrals are known for finding the area under a function along a path or curve.

If I have a function [MATH]f(x,y) = 2x[/MATH] and I have a line segment C from [MATH](1,2)[/math] to [math](3,5)[/MATH], the evaluation of the line integral is

[MATH]\int_C f(x,y) \ ds = \frac{\sqrt{13}}{2} \int_{1}^{3} 2x \ dx = 4\sqrt{13}[/MATH]
The function [MATH]f(x,y) = 2x[/MATH] is in the xz-plane while the line segment C is in the xy-plane. The path will never be under the function! How can this integral be visualized? Or What area I am calculating here?

 

[skeeter]

don’t know if this will help, but [MATH]f(x,y) =2x[/MATH] is a plane .

..

 

[Dr.Peterson]

Line integrals are known for finding the area under a function along a path or curve.

If I have a function [MATH]f(x,y) = 2x[/MATH] and I have a line segment C from [MATH](1,2)[/math] to [math](3,5)[/MATH], the evaluation of the line integral is

[MATH]\int_C f(x,y) \ ds = \frac{\sqrt{13}}{2} \int_{1}^{3} 2x \ dx = 4\sqrt{13}[/MATH]
The function [MATH]f(x,y) = 2x[/MATH] is in the xz-plane while the line segment C is in the xy-plane. The path will never be under the function! How can this integral be visualized? Or What area I am calculating here?

Wikipedia has a nice animation that shows what it means by the area under a curve (your line segment) along a surface (your plane z = 2x). I'd word it differently, as the given curve is in the plane, not on the surface; I'd say it's the area above the curve, under the surface.

Line integral - Wikipedia

en.wikipedia.org

 

[skeeter]

[MATH]\frac{\sqrt{13}}{2}(2+6) = 4\sqrt{13}[/MATH]
area of a trapezoid

 

[nasi112]

don’t know if this will help, but [MATH]f(x,y) =2x[/MATH] is a plane .

..View attachment 26096

Nice visualizing. I forgot that it will be an infinite plane. My bad, I was treating the graph as a Line?

Wikipedia has a nice animation that shows what it means by the area under a curve (your line segment) along a surface (your plane z = 2x). I'd word it differently, as the given curve is in the plane, not on the surface; I'd say it's the area above the curve, under the surface.

Line integral - Wikipedia

en.wikipedia.org

Yeah, Dr. Your wording makes it more clear to understand the concept.

[MATH]\frac{\sqrt{13}}{2}(2+6) = 4\sqrt{13}[/MATH]area of a trapezoid

That was beautiful. I have also tested the area without Calculus and got the same result. That was a magic, but honestly, I tested it after seeing the graph you sent.


Thanks a lot skeeter and Dr Peterson.

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