Verifying Stoke's Theorem using path integration along a parametrized ellipse 
Verify the Stoke's Theorem by showing that the line integral
$$ \int\limits_{C} -y^3 dx + x^3 dy - zdz = \frac{3\pi}{2} \tag{1},$$
through direct computation, where $C$ is the intersection of the cylinder $x^2 + y^2 =1$ and the plane $x+y+z =1$. (Hint : Find a parametrization  of $C$ then compute the line integral.)

My Attempt
The ellipse that results from the intersection of the cylinder and the plane is parameterized as
\begin{align}
C(t) &= (r\cos\theta, r\sin\theta, 1-r\cos\theta - r\sin\theta) \\
\implies C'(t) &= (-r\sin\theta, r\cos\theta, r\sin\theta - r\cos\theta).
\end{align}
Then using the definition of the line integral we get
\begin{align}
\int\limits_{C} -y^3 dx + x^3 dy - zdz &= \int_{0}^{2\pi} \int_{0}^{1} F(C(t)) \cdot C'(t) \ drd\theta \\
&= \int_{0}^{2\pi} \int_{0}^{1}(-r^{3}\sin^{3}\theta , r^{3}\cos^{3}\theta, -1+4\cos\theta + r\sin\theta) \\
&\quad \cdot (-r\sin\theta, r\cos\theta , r\sin\theta - r\cos\theta) \ drd\theta \\
&= \int_{0}^{2\pi} \int_{0}^{1} r^{4}(1-2\sin^{2}\theta \cos^{2}\theta)-r\sin\theta+r\cos\theta - r^2 \cos(2\theta) \ drd\theta
\end{align}
Plugging the above integral expression in Wolfram alpha gives $\dfrac{3\pi}{10}$ , but the correct answer is $\dfrac{3\pi}{2}$. I don't understand where I have gone wrong ,does anyone have an idea of what's going on ?

EDIT
THe correct parametrization is
\begin{align}
C(\theta) &= (\cos\theta, \sin\theta, 1-\cos\theta - \sin\theta) \\
\implies C'(\theta) &= (-\sin\theta, \cos\theta, \sin\theta - \cos\theta).
\end{align}
Then using the definition of line integral we get
\begin{align}
\int\limits_{C} -y^3 dx + x^3 dy - zdz &= \int_{0}^{2\pi} F(C(\theta)) \cdot C'(\theta) \ d\theta \\
&= \int_{0}^{2\pi}(\sin^{3}\theta , \cos^{3}\theta, -1+\cos\theta + \sin\theta) \\
&\quad \cdot (-\sin\theta, \cos\theta , \sin\theta - \cos\theta) \ d\theta \\
&= \int_{0}^{2\pi} (1-2\sin^{2}\theta \cos^{2}\theta)-\sin\theta+\cos\theta -  \cos(2\theta) \ d\theta = \frac{3\pi}{2}.
\end{align}

Assuming that the $ z$ in $ \vec F$ is a typo, found in an exercise of integral transformation, I consider $ \vec F=-y^3\hat i + x^3\hat j - z^3\hat k$
 Stoke's theorem tells us that $ \int_C \vec F.d\vec{r}=\iint_S (\vec\nabla\times\vec F) d\vec S$ where $ C$ is the curve bounded by the region $ S$.  
So $ \vec\nabla\times\vec F= 
    \begin{vmatrix}
    \hat i & \hat j & \hat k \\
    \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z}\\
    -y^3 & x^3 & -z \\
    \end{vmatrix}
 =3(x^2+y^2)\hat k$ 
We take S to be the region in
the plane $ h(x, y, z) = x + y + z = 1$ with boundary $ C$.
A unit normal to $ S$ is
given by $ \hat n= \dfrac{\vec\nabla.f}{\vert \vec\nabla.f \vert}=\dfrac{\hat i+\hat j+\hat k}{\sqrt 3}$
Therefore \begin{align} \iint_S (\vec\nabla\times\vec F) d\vec S &=\iint_S3(x^2+y^2)\hat k.\dfrac{(\hat i+\hat j+\hat k)}{\sqrt 3}\;\dfrac{dxdy}{\vert\hat n.\hat k\vert}\\
&=\iint_S3(x^2+y^2)\;dxdy\\
&= 3\int_{\theta=0}^{2\pi}\int_{r=0}^1r^2\cdot r\cdot dr dθ\;[\text{putting} \quad x=r\cos\theta, y=r\sin\theta]\\  
&=6\pi\cdot\dfrac{r^4}{4}\Big |_0^1\\
&=\dfrac{3\pi}{2}
\end{align}

Parameterizing the curve $ C$ we can
write $ x = \cos t, y = \sin t$ and $ z = 1 − x − y = 1 − \cos t − \sin t, 0 \le t \le 2\pi$ and write
\begin{align}
 \int_C −y^3dx + x^3dy − z^3dz &= \int_0^{2\pi} \left(−y^3\dfrac{dx}{dt} + x^3\dfrac{dy}{dt} − z^3\dfrac{dz}{dt}\right) dt\\
&= \int_0^{2\pi}\sin^4 t+\cos^4t+(1 − \cos t − \sin t)^3(\sin t − \cos t) dt\\
\end{align}
But this is too much work to calculate.
