Calculating Work and Flux Using Green's Theorem: A Square Example

[mathmari]

Hey!

"Use the Green's Theorem to calculate the work and the flux for the closed anti-clockwise direction that consists of the square which is determined by the lines $x=0$, $x=1$, $y=0$ and $y=1$ if $\overrightarrow{F}=2xy\hat{\imath}+3x^2y\hat{ \jmath }$."I have done the following:

$$\text{Work }: \oint_C{\overrightarrow{F}}dR=\int \int_R {\nabla \times \overrightarrow{F} \cdot \hat{n}}dA=\iint_R {\nabla \times \overrightarrow{F} \cdot \hat{k}}dA=\iint_R {(6xy-2x)}dA=\int_0^1 \int_0^1(6xy-2x)dxdy=\int_0^1(3y-1)dy=\frac{1}{2}$$

$$\text{Flux }: \oint_C{\overrightarrow{F} \cdot \hat{n}}ds=\iint_R{\nabla \cdot \overrightarrow{F}}dA=\iint_R{(2y+3x^2)}dA=\int_0^1 \int_0^1{(2y+3x^2)}dxdy=\int_0^1{(2y+1)}dy=2$$

Is this correct?

 

[I like Serena]

Hey!

"Use the Green's Theorem to calculate the work and the flux for the closed anti-clockwise direction that consists of the square which is determined by the lines $x=0$, $x=1$, $y=0$ and $y=1$ if $\overrightarrow{F}=2xy\hat{\imath}+3x^2y\hat{ \jmath }$."I have done the following:

$$\text{Work }: \oint_C{\overrightarrow{F}}dR=\int \int_R {\nabla \times \overrightarrow{F} \cdot \hat{n}}dA=\iint_R {\nabla \times \overrightarrow{F} \cdot \hat{k}}dA=\iint_R {(6xy-2x)}dA=\int_0^1 \int_0^1(6xy-2x)dxdy=\int_0^1(3y-1)dy=\frac{1}{2}$$

$$\text{Flux }: \oint_C{\overrightarrow{F} \cdot \hat{n}}ds=\iint_R{\nabla \cdot \overrightarrow{F}}dA=\iint_R{(2y+3x^2)}dA=\int_0^1 \int_0^1{(2y+3x^2)}dxdy=\int_0^1{(2y+1)}dy=2$$

Is this correct?

Yep. All correct! (Happy)Btw, I would write $$\oint_C{\overrightarrow{F}} \cdot d\vec R$$ or $$\oint_C{\overrightarrow{F}} \cdot \hat c dR$$ to be notationally correct.

 

[mathmari]

Yep. All correct! (Happy)

Great! (Yes)

Btw, I would write $$\oint_C{\overrightarrow{F}} \cdot d\vec R$$ or $$\oint_C{\overrightarrow{F}} \cdot \hat c dR$$ to be notationally correct.

Ok! Thank you! (Smirk)

 

[mathmari]

At the calculation of work, we took $\hat{n}=\hat{k}$, because the square is at the plane $xy$, right?

 

[I like Serena]

At the calculation of work, we took $\hat{n}=\hat{k}$, because the square is at the plane $xy$, right?

Yes.

 

[mathmari]

Yes.

Ok!