MHB Green's Theorem Verification for Vector Field F and Region R

[mathmari]

Hey!

I have to verify the Green's Theorem $\oint_C{ \overrightarrow{F} \cdot \hat{n}}ds=\iint_R{\nabla \cdot \overrightarrow{F}}dA$.
The following are given:
$$\overrightarrow{F}=-y \hat{\imath}+x \hat{\jmath}$$
$$C: r=a \cos{t} \hat{ \imath}+a \sin{t} \hat{\jmath}, 0 \leq t \leq 2 \pi$$
$$R: x^2+y^2 \leq a^2$$

I have done the following:
$$\hat{n}=\frac{dy}{dt} \hat{ \imath}-\frac{dx}{dt} \hat{\jmath}=a \cos{t} \hat{\imath}+a \sin{t} \hat{\jmath}$$
$$\oint_C{\overrightarrow{F} \cdot \hat{n}}ds=\int_0^{2 \pi} {(-a \sin{t} \hat{\imath}+ a \cos{t} \hat{\jmath})(a \cos{t} \hat{\imath}+a \sin{t} \hat{\jmath}) } a dt=\int_0^{2 \pi}{(-a^2 \sin{t} \cos{t}+a^2 \cos{t} \sin{t})a}dt=\int_0^{2 \pi}{0}dt=0$$

$$ \nabla \cdot \overrightarrow{F}=0$$
$$\iint_R{\nabla \cdot \overrightarrow{F}}dA=\iint_R{0}dA=0$$

Have I calculated correct these two integrals? Is the change of variables at the first integral right?

 

[I like Serena]

Hi!

$$\hat{n}=\frac{dy}{dt} \hat{ \imath}-\frac{dx}{dt} \hat{\jmath}=a \cos{t} \hat{\imath}+a \sin{t} \hat{\jmath}$$

Isn't $\hat n$ supposed to be a unit vector?

 

[mathmari]

Hi!

Isn't $\hat n$ supposed to be a unit vector?

So can I not use this $\hat{n}$? which $\hat{n}$ can I use instead?

 

[I like Serena]

So can I not use this $\hat{n}$? which $\hat{n}$ can I use instead?

To make it a unit vector, you should divide by its length, which happens to be $a$:
$$\hat{n}= \cos{t} \hat{\imath}+ \sin{t} \hat{\jmath}$$

 

[mathmari]

To make it a unit vector, you should divide by its length, which happens to be $a$:
$$\hat{n}= \cos{t} \hat{\imath}+ \sin{t} \hat{\jmath}$$

Aha, ok! And besides from that is the exercise correct?

 

[I like Serena]

Aha, ok! And besides from that is the exercise correct?

Yep!

 

[mathmari]

Yep!

Ok! Thanks a lot! (Smirk)