- #1
mathmari
Gold Member
MHB
- 5,049
- 7
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?
"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?