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Given that $$\vec{F}=-x^2y \hat{i}+x y^2\hat{j}$$ $$C:r=a \cos{t}\hat{\imath}+a \sin{t} \hat{\jmath}, 0 \leq t \leq 2 \pi \text{ and } R: x^2+y^2 \leq a^2$$ I have to calculate $\iint_R{ \nabla \times \vec{F} \cdot \hat{n}}\,dA$. $$$$ $$\nabla \times \vec{F}=(x^2+y^2)\hat{k}$$ $$\hat{n}=\hat{k}$$ So $$\iint_R{ \nabla \times \vec{F} \cdot \hat{n}}\,dA=\iint_R{ \nabla \times \vec{F} \cdot \hat{k}}\,dA=\iint_R{x^2+y^2}dA$$ But how can I continue?? Do I have to do something like the following?? $$\iint_R{x^2+y^2}\,dA=\int_{-1}^1 \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}{(x^2+y^2)}\,dy\,dx$$

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That looks fine to me. A transformation into polar coordnates looks like the thing to do next. –  Joshua Biderman Mar 7 at 0:39
    
@StellaBiderman Ok!! Thanks a lot!! :) –  Mary Star Mar 7 at 9:53
    
@StellaBiderman Could check also what I've post as an answer below? –  Mary Star Mar 16 at 10:28

2 Answers 2

up vote 3 down vote accepted

Make the change of variable $(x,y)=(r\cos t, r\sin t)$ into the integral, using $dA = dxdy = rdrdt$:

$$ \int\int_R (x^2 + y^2) dA= {2\pi}\int_0^{a} r^2 rdr = {2\pi}\frac{a^4}4 $$

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Nice!! Thank you very much!! :) –  Mary Star Mar 7 at 9:53
    
To verify the Green's Theorem, it is as followed: $$\oint_C{\overrightarrow{F}}dr=\int_0^{2 \pi}{\overrightarrow{F}\frac{dr}{dt}}dt=\int_0^{2 \pi}{(-a^2 \cos^2{t} a \sin{t} \hat{\imath}+a \cos{t} a^2 \sin^2{t} \hat{\jmath})\cdot (-a \sin{t} \hat{\imath}+a \cos{t} \hat{\jmath})}dt=...=\frac{a^4 \pi}{2}$$. Isn't it?? –  Mary Star Mar 16 at 2:14

To verify also the other formula of the theorem $\oint_C{\overrightarrow{F} \cdot \hat{n}}ds=\iint_R{\nabla \cdot \overrightarrow{F}}dA$, could you tell me if the following is correct?

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

$\oint_C{\overrightarrow{F} \cdot \hat{n}}ds=\int_0^{2 \pi}{(-a^3 \cos^2{t} \sin{t} \hat{\imath}+a^3 \cos{t} \sin^2{t} \hat{\jmath})(a \cos{t} \hat{\imath}+a \sin{t} \hat{\jmath})}dt=\int_0^{2 \pi}{(-a^4 \cos^3{t} \sin{t}+a^4 \cos{t} \sin^3{t})}dt=...=0$

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

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