How to calculate the line integral and substitute $dx\,\ dy$ in a question on Green's theorem

Question

The question states that: Verify Green's Theorem on the plane for $\oint_C (2x-y^3)dx-xydy$ where C is the boundary of the region enclosed by the circles $x^2+y^2=1$ and $x^2+y^2=9$.

My attempt:

1. While verifying the L.H.S. I divided the line integral into 2 parts for the 2 circles and substituted $x=r\cos \theta$ and $y=r\sin \theta$ where the limits for $r$ will be different for the 2 circles. I then put $dx=dr\cos \theta -r\sin \theta d\theta$ and $dy=dr\sin \theta +r\cos \theta d\theta$. The integral became very complex and after computing both of them and adding them, I got $60\pi +\frac{273}{12}$. But the correct answer should be just $60\pi$.
2. For the R.H.S. I don't know what to substitute $dx\,\ dy$ with and hence cannot proceed.

Answer 1

On a circle the radius $r$ is a constant, so we have: $$ x=r\cos \theta \quad \Rightarrow \quad dx=-r\sin \theta d\theta $$ and $$ y=r\sin \theta \quad \Rightarrow \quad dy=r \cos \theta d \theta $$ (this is the mistake in your solution, I suppose).

So, for a circle of radius $r$ we have: $$ \oint_C (2x-y^3)dx-xydy= \oint_C\left[(2r\cos \theta-r^3\sin^3 \theta)(-r\sin \theta)-r^3\cos^2 \theta \sin \theta \right]d\theta= $$ $$ =\oint_C\left[-2r^2\cos \theta\sin \theta+r^4\sin^4 \theta-r^3\cos^2 \theta \sin \theta \right]d\theta= $$ $$ =\int_0^{2\pi}-r^2\sin 2\theta d\theta + \int_0^{2\pi}r^4\sin^4 \theta d \theta -\int_0^{2\pi} r^3\cos^2 \theta \sin \theta d\theta $$ I suppose that you can integrate these functions and find the result $\frac{3}{4}r^4 \pi$, that, for $r=1$ is $\frac{3}{4} \pi$ and for $r=3$ is $\frac{3}{4}(81) \pi$, so the difference (the boundary of the region enclosed) is $60 \pi$.

For the question 2. In the OP I don't see an RHS. (?)

If the RHS is from Green's theorem, than it is: $$ \int \int _D\left(\dfrac{\partial}{\partial y}(2x-y^3)-\dfrac{\partial}{\partial x}xy \right) da $$

where $da$ is the area element. So, using the same substitution and $da= rdrd\theta$ in polar coordinates, we have:

$$ \int \int _D\left(-y+3y^2 \right)dx dy= $$ $$ \int_0^{2\pi} \int_1^3 \left(-r\sin \theta+3r^2 \sin^2 \theta \right)r dr d\theta= \int_0^{2\pi}\left(\frac{-26}{3}\sin \theta+60\sin^2 \theta \right)d \theta = 60 \pi $$

Answer 2

Verify Green's theorem for the integral

$\displaystyle \oint_{c}(2x-y^{3})dx-xydy$; where $c$ is the boundary between the two circles $x^{2}+y^{2}=1$ and $x^{2}+y^{2}=9.$

Ans.

$\bullet$ Evaluating the double integral: The Double Integral

$M=2x-y^{3}, \displaystyle \frac{\partial M}{\partial y}=-3y^{2},$ $N=-xy, \displaystyle \frac{\partial N}{\partial x}=-y$

$I_{d}=\displaystyle \int_{R}\int(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dA$

$=\displaystyle \int_{R}\int(-y+3y^{2})dA$

$=\displaystyle \int_{0}^{2\pi}\int_{1}^{3}(-r\sin\theta+3r^{2}\sin^{2}\theta)rdrd\theta=60\pi$

$\bullet$ Evaluating the line integral: The Line Integral

Consider $c=c_{1}\cup c_{2}\cup c_{3}\cup(-c_{2})\cup c_{4}$

$I_{l}=I_{c_{1}}+I_{c_{2}}+I_{c_{3}}+I_{-c_{2}}+I_{c_{4}}$

$I_{l}=I_{c_{1}}+I_{c_{3}}+I_{c_{4}}$

Evaluating $I_{c_{1}}$ :

$c_{1}:x=3\cos t, y=3\sin t;t:0\rightarrow\pi$

$I_{c_{1}}=\displaystyle \int_{0}^{\pi}[(6\cos t-27\sin^{3}t)(-3\sin t)-(3\cos t)(3\sin t)(3\cos t)]dt$

$=\displaystyle \int_{0}^{\pi}[-18\cos t\sin t+81\sin^{4}t-27\cos^{2}t\sin t]dt$ $$ I_{c_{1}}=\frac{243}{8}\pi-18 $$ Evaluating $I_{c_{3}}$ :

$c_{3}:x=\cos t, y=\sin t;t:\pi\rightarrow-\pi$

$I_{c_{3}}=\displaystyle \int_{0}^{\pi}[-\cos t\sin t+\sin^{4}t-\cos^{2}t\sin t]dt$ $$ I_{c_{3}}=-\frac{3}{4}\pi $$ Evaluating $I_{c_{4}}$ :

$c_{4}:x=3\cos t, y=3\sin t; t:\pi\rightarrow 2\pi$

$I_{c_{4}}=\displaystyle \int_{\pi}^{2\pi}[-18\cos t\sin t+81\sin^{4}t-27\cos^{2}t\sin t]dt$ $$ I_{c_{4}}=\frac{243}{8}\pi+18 $$ Therefore;

$ I_{l}=(\displaystyle \frac{243}{8}\pi-18)+\left(\begin{array}{l} -\frac{3}{4}\pi \end{array}\right)+(\frac{243}{8}\pi+18)=60\pi$. Verified