Find the volume bound by the curve $\gamma(t) = (e^{\sin(t)}\cos(t),e^{\cos(t)}\sin(t))$ Find the volume bound by the curve $\gamma(t) = (e^{\sin(t)}\cos(t),e^{\cos(t)}\sin(t))$
I've tried computing directly with $\int_0^{2\pi} \int_0^R r \, dr \, d\theta$ where $R = \sqrt{(e^{\sin(t)}\cos(t))^2+(e^{\cos(t)}\sin(t))^2}$ but this seems unfruitful. 
I've also tried considering the path as a line integral over the vector field $F=(-y,x)$, which gives us the area by Green's theorem. Here's why...
\begin{equation}
\int_\gamma F \cdot dr = \int_\gamma -y\,dx+x\,dy = \iint\limits_\Omega dA = \mathrm{Area}(\Omega)
\end{equation}
Where $\Omega$ is the region bound by the curve. But this attempt also seems unfruitful. To be honest I'm not sure if this is even possible without numerical methods. Thanks for the help in advance. 
 A: $$\gamma(t)=(x(t),y(t))=\left(e^{\sin t}\cos t,\,e^{\cos t}\sin t\right)$$
is a closed simple curve around the origin, counter-clockwise oriented. The enclosed area is given by:
$$ A = \frac{1}{2}\int_{0}^{2\pi}\left(xy'-yx'\right)\,dt = \frac{1}{8}\int_{0}^{2\pi}e^{\cos t+\sin t}\left(4-\cos t-\sin t+\cos(3t)-\sin(3t)\right)\,dt $$
or, by setting $t=x+\frac{\pi}{4}$,
$$ A = \frac{1}{8}\int_{0}^{2\pi}e^{\sqrt{2}\cos x}\left(4-\sqrt{2}\cos x-\sqrt{2}\cos(3x)\right)\,dx.\tag{1} $$
Since for any $\alpha\in\mathbb{R}$ and $n\in\mathbb{N}$ we have:
$$ \int_{0}^{2\pi} e^{\alpha\cos x}\cos(nx)\,dx = 2\pi\cdot I_{n}(\alpha)\tag{2}$$
by the integral definition of the modified Bessel functions of the first kind, it follows that:
$$ A = \color{red}{\frac{\pi}{4}\left(4\, I_0(\sqrt{2})-\sqrt{2}\, I_1(\sqrt{2})-\sqrt{2}\, I_3(\sqrt{2})\right)}=\color{blue}{3.84713365\ldots}.\tag{3} $$
The area has so a fast-converging series representation given by:
$$ A = \color{purple}{\frac{\pi}{4}\sum_{m=0}^{+\infty}\frac{1}{2^m\,m!^2}\left(4-\frac{1}{m+1}-\frac{1}{2(m+1)(m+2)(m+3)}\right)}.\tag{4}$$
