Vtyjkyuk

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 $frac12int_0^2pi(-1-sin t+tcos t)dt=-pi$

Is it correct?

The form of Green's theorem that you used:
$$
A(D) = iint_D dx,dy = frac12oint_partial D (x,dy - y,dx)
$$
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 $x = rcostheta$, $y = rsintheta$. So use
$$
gamma(t) = ((1+sin t)cos t,(1+sin t)sin t)
$$

As for the former, you can compute
beginalign*
dx &= costheta ,dr - r sintheta ,dtheta \
dy &= sintheta ,dr + r costheta ,dtheta
endalign*
Then through a little exterior algebra, we get
$$
dx ,dy = r,dr,dtheta
$$
and
$$
x,dy - y,dx = r^2,dtheta
$$
So Green's theorem tells us
$$
A(D) = int_D r,dr,dtheta = frac12oint_partial D r^2,dtheta
$$
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 $r = 1 + sintheta$, then
$$
A(D) = frac12int_0^2pi(1+sintheta)^2,dtheta
$$

While I think Matthew Leingang work is more elegant and more general. Here is a slightly different approach.

Green's Theorem:
$iint (frac partial Qpartial x - frac partial Ppartial y) dx dy = oint P dx + Q dy$

Choose $P, Q$ such that $(frac partial Qpartial x - frac partial Ppartial y) = 1.$ We have a wide range of options here. The point is to choose the easiest to work with. e.g. Let $P,Q = 0,x$

$r = 1 + sin theta\
x = rcos theta = (1+sintheta)costheta\
dx = -sintheta + cos^2theta - sin^2theta dtheta\
y = rsin theta = (1+sintheta)sintheta\
dy = costheta + 2sinthetacostheta dtheta$

$oint x dy = int_0^2pi (costheta + frac 12 sin 2theta)(costheta + sin 2theta) dtheta = frac 3pi2$