Vector Line Integral Direction of Limits

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The discussion revolves around evaluating a line integral for a vector field in the xy-plane, specifically from the point (1,1) to (0,1). The vector field simplifies to v = (2x - 1) i when z = 0. The integral is calculated as -∫(2x - 1) dx, which results in 0, regardless of the limits used. There is confusion regarding the direction of the integral and the convention for defining dr, leading to a belief that the book's answer may be incorrect. Clarifications from other users confirm the understanding of the integral's evaluation and the conventions used.
Master1022
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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.

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.
 
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Seems to me the book answer is wrong ...
(@haruspex ?)
 
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BvU said:
Seems to me the book answer is wrong ...
(@haruspex ?)
I agree.
(Odd to write the range with endpoints expressed as only 2D.)
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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