How to solve this line integral

In summary: In this problem you are asked to find a "line integral" of a scalar function and that is what I did.In summary, the conversation was about finding a line integral and using x(t) and y(t) to solve for C. There was confusion about whether the problem involved a scalar or vector and how to determine that. The expert summarizer clarified that the problem was purely scalar and provided a step-by-step solution for finding the line integral.
  • #1
aruwin
208
0
I have no idea how to even start with this problem. I know the basics but this one just gets complicated. Please guide me!


Find the line integral:
∫C {(-x^2 + y^2)dx + xydy}
When 0≤t≤1 for the curved line C, x(t)=t, y(t)=t^2
and when 1≤t≤2, x(t)= 2 - t , y(t) = 2-t.
Use x(t) and y(t) and C={(x(t),y(t))|0≤t≤2}
Help!
 
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  • #2
aruwin said:
I have no idea how to even start with this problem. I know the basics but this one just gets complicated. Please guide me!


Find the line integral:
∫C {(-x^2 + y^2)dx + xydy}
When 0≤t≤1 for the curved line C, x(t)=t, y(t)=t^2
and when 1≤t≤2, x(t)= 2 - t , y(t) = 2-t.
Use x(t) and y(t) and C={(x(t),y(t))|0≤t≤2}
Help!

[tex]x(t)=t\,\,,\,y(t)=t^2\Longrightarrow x'(t)=1\,\,,\,y'(t)=2t\,\,,\,\,t\in [0,1]\Longrightarrow \oint_{C_1}(-x^2+y^2)dx+xydy=\int_0^1\left[(-t^2+t^4)\,dt+t^3(2tdt)\right]=[/tex]
[tex]=\int_0^1(3t^4-t^2)dt=\frac{3}{5}-\frac{1}{3}=\frac{4}{15}[/tex]
and now you do the other part of the curve, with [itex]x(t)=2-t\,\,,\,y(t)=2-t\,\,,x'(t)=y'(t)=-1\,\,,\,t\in [1,2][/itex], following

the argument as above.

DonAntonio
 
  • #3
DonAntonio said:
[tex]x(t)=t\,\,,\,y(t)=t^2\Longrightarrow x'(t)=1\,\,,\,y'(t)=2t\,\,,\,\,t\in [0,1]\Longrightarrow \oint_{C_1}(-x^2+y^2)dx+xydy=\int_0^1\left[(-t^2+t^4)\,dt+t^3(2tdt)\right]=[/tex]
[tex]=\int_0^1(3t^4-t^2)dt=\frac{3}{5}-\frac{1}{3}=\frac{4}{15}[/tex]
and now you do the other part of the curve, with [itex]x(t)=2-t\,\,,\,y(t)=2-t\,\,,x'(t)=y'(t)=-1\,\,,\,t\in [1,2][/itex], following

the argument as above.

DonAntonio

Oh, got it! Well, one important question, how do you know if it's a scalar or vector? Because if it's scalar then we must find the magnitude of the derivatives. In this case, you didn't find the magnitude which means it's a vector but how do you know it is a vector?
 
  • #4
You start by knowing the difference between a "scalar" and a "vector". Since there was no mention of a vector in this problem, do you have a specific problem where that question might arise?

"Because if it's scalar then we must find the magnitude of the derivatives."
I've never heard of such at thing. The problem here was entirely a "scalar" problem and there is no finding the magnitudes of anything.
 

1. What is a line integral?

A line integral is a type of integral that is used to calculate the total value of a function along a curve or a line. It is also known as a path integral and is commonly used in physics and engineering to solve problems related to work, energy, and fluid flow.

2. How do you solve a line integral?

To solve a line integral, you need to first parameterize the curve or line in terms of a single variable. Then, you can plug this parameterization into the given function and integrate it with respect to the variable. Finally, evaluate the integral and the result will be the value of the line integral.

3. What are the different types of line integrals?

There are two main types of line integrals - the first type is a line integral of a scalar function, which gives the total value of a scalar function along a curve. The second type is a line integral of a vector function, which gives the total work done by a vector field along a curve.

4. What is the difference between a line integral and a double integral?

A line integral is a one-dimensional integral that is calculated along a curve, while a double integral is a two-dimensional integral that is calculated over a region in the xy-plane. Line integrals are used to find the total value of a function along a curve, while double integrals are used to find the volume under a surface.

5. Are there any applications of line integrals in real life?

Yes, line integrals have many real-life applications. They are commonly used in physics and engineering to calculate work and energy, as well as to solve problems related to fluid flow and electromagnetic fields. They are also used in economics and finance to model and predict market trends.

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