Understanding Green's Theorem in 2-Dimensional Vector Fields

[member 428835]

Hi PF!

So let's say I have some vector field, call it $\vec{F}$ and let $\vec{F}$ be 2-Dimensional and suppose I wanted to compute $\iint_D \nabla \cdot \vec{F} dD$. Using green's theorem we could write $\iint_D \nabla \cdot \vec{F} dD = - \int_{\partial D} \vec{F} \cdot \hat n dS$ where $\hat n$ is the outward oriented surface normal. Now if $D = [0,1] \times [0,1]$ I may use the formula to find the solution. Can you tell me what I'm doing wrong? $$\int_{\partial D} \vec{F} \cdot \hat n dS = \int_0^1 \vec{F} \cdot (- \hat j)dx + \int_0^1 \vec{F} \cdot ( \hat i)dy+\int_1^0 \vec{F} \cdot (\hat j)dx+\int_1^0 \vec{F} \cdot (\hat i)dx$$

I know there is another way to write Green's Theorem but how to do it this way?

Thanks a ton!

Josh

 

[member 428835]

Please help me out! I'm not looking for an answer, just looking for what I am doing wrong.

 

[member 428835]

For the record what I have on the R.H.S is what is wrong.

 

[slider142]

Hi PF!

So let's say I have some vector field, call it $\vec{F}$ and let $\vec{F}$ be 2-Dimensional and suppose I wanted to compute $\iint_D \nabla \cdot \vec{F} dD$. Using green's theorem we could write $\iint_D \nabla \cdot \vec{F} dD = - \int_{\partial D} \vec{F} \cdot \hat n dS$ where $\hat n$ is the outward oriented surface normal. Now if $D = [0,1] \times [0,1]$ I may use the formula to find the solution. Can you tell me what I'm doing wrong? $$\int_{\partial D} \vec{F} \cdot \hat n dS = \int_0^1 \vec{F} \cdot (- \hat j)dx + \int_0^1 \vec{F} \cdot ( \hat i)dy+\int_1^0 \vec{F} \cdot (\hat j)dx+\int_1^0 \vec{F} \cdot (\hat i)dx$$

I know there is another way to write Green's Theorem but how to do it this way?

Thanks a ton!

Josh

You've almost got it, except for the errant - sign in the red equation, and the latter two integrals, which are not proper line integrals, as the upper limit must be greater than the lower limit (or you must used a signed differential to represent that they are parameterized in the opposite orientation to the prior two integrals. Either way is fine.) Using a standard positive differential, we could use the following explicit parameterizations to traverse the unit square in the counterclockwise orientation:
$$r_1(t) = (t, 0), 0 \leq t \leq 1$$
$$r_2(t) = (1, t), 0 \leq t \leq 1$$
$$r_3(t) = (1 - t, 1), 0 \leq t \leq 1$$
$$r_4(t) = (0, 1 - t), 0\leq t \leq 1$$
And your integrals would then be:
$$\int_0^1 \vec{F}(t, 0) \cdot (- \hat j)\, dt + \int_0^1 \vec{F}(1, t) \cdot ( \hat i)\, dt+\int_0^1 \vec{F}(1 - t, 1) \cdot (\hat j)\, dt+\int_0^1 \vec{F}(0, 1 - t) \cdot (-\hat i)\, dt$$
When it comes to vector calculus, you just have to be very careful about orientations. They are as sneaky as sign errors. :-)