# Evaluate the line integral by two methods

Gold Member

## 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

## The Attempt at a Solution

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dx
Homework Helper
Gold Member
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.

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

nrqed
Homework Helper
Gold Member

## 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

## The Attempt at a Solution

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

Hootenanny
Staff Emeritus
Gold Member
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?

Gold Member
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
Staff Emeritus
$$\oint_C {xy^2 \,dx\, + \,x^3 \,dy} = \int\limits_0^3 {2t^2 \left( 0 \right)\, + \,2^3 \,dt} = \int\limits_0^3 {8\,dt} = \left[ {8t} \right]_0^3 = 24$$
Looks good to me .