Evaluate the line integral by two methods

[tony873004]

Homework Statement

Evaluate the line integral by two methods: (a) directly and (b) using Green's Theorem. \oint_C {xy^2 \,dx\, + \,x^3 \,dy}. C is the rectangle with vertices (0,0), (2,0), (2,3) and (0,3).

The Attempt at a Solution

I get the correct answer using Green's theorem. But I don't know how to do the "directly" method. I came up with 4 parametric equations, one for each line of the rectangle. Let's look at the first one for the line (0,0) - (2,0): x=t, y=0, t goes between 0 and 2.

\oint_C {xy^2 \,dx\, + \,x^3 \,dy} = \int\limits_0^2 {t\left( 0 \right)^2 \,\, + \,t^3 \,dt} = \int\limits_0^2 {t^3 \,dt} = \left[ {\frac{{t^4 }}{4}} \right]_0^2 = \left( {\frac{{2^4 }}{4}} \right) - \left( {\frac{{0^4 }}{4}} \right) = \frac{{16}}{4} = 4.

I'm pretty sure I did this step wrong. How do I go from the C integral to the 2-0 integral?

Thanks

Homework Equations

 

[dx]

Think of {xy^2 \,dx\, + \,x^3 \,dy} as the dot product of the vectors (dx,dy) and (xy^2, x^3). On the x axis, the x component is zero, so the integral from (0,0) to (0,2) should be zero.

 

[tony873004]

sorry, I still don't understand. How do I get from dx+dy to dt?

 

[nrqed]

Homework Statement

Evaluate the line integral by two methods: (a) directly and (b) using Green's Theorem. \oint_C {xy^2 \,dx\, + \,x^3 \,dy}. C is the rectangle with vertices (0,0), (2,0), (2,3) and (0,3).

The Attempt at a Solution

I get the correct answer using Green's theorem. But I don't know how to do the "directly" method. I came up with 4 parametric equations, one for each line of the rectangle. Let's look at the first one for the line (0,0) - (2,0): x=t, y=0, t goes between 0 and 2.

\oint_C {xy^2 \,dx\, + \,x^3 \,dy} = \int\limits_0^2 {t\left( 0 \right)^2 \,\, + \,t^3 \,dt} = \int\limits_0^2 {t^3 \,dt} = \left[ {\frac{{t^4 }}{4}} \right]_0^2 = \left( {\frac{{2^4 }}{4}} \right) - \left( {\frac{{0^4 }}{4}} \right) = \frac{{16}}{4} = 4.

I'm pretty sure I did this step wrong. How do I go from the C integral to the 2-0 integral?

Thanks

Homework Equations

Your line is along the x-axis so dy is zero for that section of the rectangle!

 

[Hootenanny]

sorry, I still don't understand. How do I get from dx+dy to dt?

So you have paremerised the line,

x = t \Rightarrow \frac{dx}{dt} = 1 \Rightarrow dx = dx

y = 0 \Rightarrow dy = 0

Do you follow?

 

[tony873004]

Thanks, I think I'm getting it now. So in the next line segment (the right vertical one), dx is zero, so

\oint_C {xy^2 \,dx\, + \,x^3 \,dy} = \int\limits_0^3 {2t^2 \left( 0 \right)\, + \,2^3 \,dt} = \int\limits_0^3 {8\,dt} = \left[ {8t} \right]_0^3 = 24

 

[Hootenanny]

Thanks, I think I'm getting it now. So in the next line segment (the right vertical one), dx is zero, so

\oint_C {xy^2 \,dx\, + \,x^3 \,dy} = \int\limits_0^3 {2t^2 \left( 0 \right)\, + \,2^3 \,dt} = \int\limits_0^3 {8\,dt} = \left[ {8t} \right]_0^3 = 24

Looks good to me .

Although from a formalism point of view, I will point out that it isn't technically correct to say that your path integral is equal to the integral over this segment, but I know what you mean.