Area of Cardioid $r=1+\sin(\theta)$ Using Green's Theorem 
Find the area enclosed by $r=1+\sin \theta$ using Green's theorem.

What I have is
$\gamma(t)=(t,1+\sin t)$
$\gamma'(t)=(1,\cos t)$
Then $\frac{1}{2}\int_{0}^{2\pi}(-1-\sin t+t\cos t)dt=-\pi$
Is it correct?
 A: The form of Green's theorem that you used:
$$
    A(D) = \iint_D dx\,dy = \frac{1}{2}\oint_{\partial D} (x\,dy - y\,dx)
$$
applies when $x$ and $y$ are cartesian coordinates.  Your expression $\gamma(t) = (t,1+\sin t)$ expresses the curve in polar coordinates.  So you need to either state Green's theorem in polar coordinates, or express $\gamma$ in cartesian coordinates.
Let's do the latter.  The change of variables is $x = r\cos\theta$, $y = r\sin\theta$.  So use
$$
    \gamma(t) = ((1+\sin t)\cos t,(1+\sin t)\sin t)
$$
As for the former, you can compute
\begin{align*}
    dx &= \cos\theta \,dr - r \sin\theta \,d\theta \\
    dy &= \sin\theta \,dr + r \cos\theta \,d\theta
\end{align*}
Then through a little exterior algebra, we get
$$
    dx \,dy = r\,dr\,d\theta
$$
and
$$
    x\,dy - y\,dx = r^2\,d\theta
$$
So Green's theorem tells us
$$
    A(D) = \int_D r\,dr\,d\theta = \frac{1}{2}\oint_{\partial D} r^2\,d\theta
$$
I think in many undergraduate multivariable calculus courses this identity isn't derived from Green's theorem, but it can be.  In any case, if $D$ is the region enclosed by the cardioid $r = 1 + \sin\theta$, then
$$
    A(D) = \frac{1}{2}\int_{0}^{2\pi}(1+\sin\theta)^2\,d\theta
$$
A: While I think Matthew Leingang work is more elegant and more general. Here is a slightly different approach.
Green's Theorem: 
$\iint (\frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y})\ dx\ dy = \oint P dx + Q dy$
Choose $P, Q$ such that $(\frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y}) = 1.$  We have a wide range of options here.  The point is to choose the easiest to work with.   e.g. Let $P,Q = 0,x$
$r = 1 + \sin \theta\\
x = r\cos \theta = (1+\sin\theta)\cos\theta\\
dx = -\sin\theta + \cos^2\theta - \sin^2\theta\ d\theta\\
y = r\sin \theta = (1+\sin\theta)\sin\theta\\
dy = \cos\theta + 2\sin\theta\cos\theta\ d\theta$
$\oint  x\ dy = \int_0^{2\pi} (\cos\theta + \frac 12 \sin 2\theta)(\cos\theta + \sin 2\theta)\ d\theta = \frac {3\pi}{2}$
