I have the vector field $F = (3xy,-x)$ along the circle $c$ (counter clockwise) which has a radius $a$ and centre $(a,0)$.
I want to try and apply Green's Theorem to this, where I obtain $\int\int(-1 – 3x )dxdy$.
Now I try and use polar coordinates, where I replace $x = a + r\cos\theta$ and $dxdy =rdrd\theta$.
The limits of $r$ are from $0$ to $a$, and the limits of $\theta$ from $0$ to $2\pi$.
When I solve the integral I get $-a^{2}\pi – 3a^{3}\pi$ . Is this correct? I don't know the answer and im not entirely sure of my working. Many thanks!
Best Answer
If you think about the integral as an average value, then $$\int\int_Rf(x,y)d^2A=\langle f\rangle_R\times A(R)$$ that is the average value of the function $f$ over the region $R$ times the area of region $R$. The average value of $x$ is just the $x$-coordinate of the center of mass, $a$, and the average value of $1$ is $1$, so $$\int\int_R(-1-3x)d^2A=(-1-3a)\cdot\pi a^2=-\pi a^2-3\pi a^3$$ You translated the origin of coordinates to $(a,0)$ but you could have chosen the parameterization $x=r\cos\theta$, $y=r\cos\theta$, $0\le r\le2a\cos\theta$ so that you might get $$\begin{align}\int_{-\frac{\pi}2}^{\frac{\pi}2}\int_0^{2a\cos\theta}(-1-3r\cos\theta)r\,dr\,d\theta&=\int_{-\frac{\pi}2}^{\frac{\pi}2}(-2a^2\cos^2\theta-8a^3\cos^4\theta)d\theta\\ &=-2a^2\left(\frac{\pi}2\right)-8a^3\left(\frac{3\pi}8\right)=-\pi a^2-3\pi a^3\end{align}$$ Where we have again used the average value of $\frac12$ for $\cos^2\theta$ and of $\frac38$ for $\cos^4\theta$.