Computing areas using Green's theorem

Question

I want to compute the area of the surface $B$ with boundary parametrised by $$ \gamma(t)=\left(\begin{array}{c} \sin t \\ 4 \cos ^{2} t+\cos t \end{array}\right), \quad t \in[0,2 \pi] $$ By greens theorem we have $$ \int_{B}^{} \mathrm{~d}\mu = \int_{B}^{} \!\left( \frac{\partial f _{2}}{\partial x} - \frac{\partial f _{1}}{\partial y} \right) \mathrm{~d}\mu ( x, y) = \int_{\partial B}^{} f \mathrm{~d}\mathbf{s} $$ for $f(x, y) = (0, x)$. However, for the last path integral I find \begin{align*} \int_{\partial B}^{} f \mathrm{~d}\mathbf{s} = \int_{0}^{2\pi } \left\langle f( \gamma ( t)), \gamma '( t)\right\rangle \mathrm{~d}t &= \int_{0}^{2\pi }\!\left( - 8\sin\!\left( t\right)^{2} \cos\!\left( t\right) -\sin\!\left( t\right) ^{2} \right)\mathrm{~d}t \\ &=- \int_{0}^{2\pi } \sin\!\left( t\right) ^{2} = -\pi \end{align*} which makes no sense since the result has to be positive. The boundary is parametrised in counterclockwise direction, so normally I should arrive at a correct result. What did I do wrong?

Answer 1

As pointed out by Robert Lee in the comments, the problem is that the parameterization is clockwise and not counterclockwise. Thus reversing the direction will get rid of the minus sign and get you the correct result.