- #1

Master1022

- 611

- 117

- Homework Statement
- Evaluate the line integral for the vector field ## v = (2x - y) \hat i + (-yz^2) \hat j + -(y^2 z) \hat k ## from (1,1) to (0,1)

- Relevant Equations
- Integration

Hi,

I apologise as I know I have made similar posts to this in the past and I thought I finally understood it. However, this solution seems to disagree on a technicality. I know the answer ends up as 0, but I still want to understand this from a conceptual point.

Given that we are in the xy-plane where ## z = 0 ## and at ## y = 1 ##, the vector field becomes:

$$ v = (2x - 1) \hat i $$

Then when I set up the integral, ## d \vec r = -dx \hat i ## and thus:

$$ \int_0^1 \vec v \cdot d \vec r = \int_0^1 (2x - 1) \hat i \cdot -dx \hat i = - \int_0^1 (2x - 1) dx = 0 $$

However, the answer writes the integral as:

$$ - \int_1^0 (2x - 1) dx = 0 $$

I know that the answers are the same, but if the integral wasn't 0, then the answers would be different. I thought the convention was to define ## d \vec r ## in the direction of the path and the limits in terms of the increasing parameter.

I am sure this has been answered in similar posts but am unable to find them. Any guidance would be greatly appreciated.

Thanks.

I apologise as I know I have made similar posts to this in the past and I thought I finally understood it. However, this solution seems to disagree on a technicality. I know the answer ends up as 0, but I still want to understand this from a conceptual point.

**Question:**Evaluate the line integral for the vector field ## v = (2x - y) \hat i + (-yz^2) \hat j + -(y^2 z) \hat k ## from (1,1) to (0,1) when we are in the xy-plane (i.e. z = 0).**Approach:**Given that we are in the xy-plane where ## z = 0 ## and at ## y = 1 ##, the vector field becomes:

$$ v = (2x - 1) \hat i $$

Then when I set up the integral, ## d \vec r = -dx \hat i ## and thus:

$$ \int_0^1 \vec v \cdot d \vec r = \int_0^1 (2x - 1) \hat i \cdot -dx \hat i = - \int_0^1 (2x - 1) dx = 0 $$

However, the answer writes the integral as:

$$ - \int_1^0 (2x - 1) dx = 0 $$

I know that the answers are the same, but if the integral wasn't 0, then the answers would be different. I thought the convention was to define ## d \vec r ## in the direction of the path and the limits in terms of the increasing parameter.

I am sure this has been answered in similar posts but am unable to find them. Any guidance would be greatly appreciated.

Thanks.