$\text{Calculate using Stokes' theorem } I=\int_C(y-2z)\text{d}x + (x-z) \text{d}y + (2x-y)\text{d}z, \text{ where } C \text{ is the intersection between } x^2+y^2+z^2=a^2, a > 0, \text{ and } x - y + z = 0$
I'm having trouble reaching the value in my textbook. Here's what I did. By Stokes' theorem, we know that $I = \int\int_D\text{rot}(\vec{V})\cdot(T_x\times T_y)dxdy$. First, I have to calculate $\text{rot}(\vec{V})$, where $\vec{V} = (y-2z)\vec{i} + (x-z)\vec{j} + (2x-y)\vec{k}$ and where $$\text{rot}(\vec{V}) = \begin{vmatrix}\vec{i} & \vec{j} & \vec{k} \newline \frac{d}{dx} & \frac{d}{dy} & \frac{d}{dz} \newline (y-2z) & (x-z) & (2x-y) \end{vmatrix}$$
After calculating, I get $\text{rot}(\vec{V})=0\cdot\vec{i}-4\vec{j}+0\cdot\vec{k}$
Let $\begin{cases}x = x \newline y = y \newline z = y - x\end{cases}$. Then: $$T_x\times T_y=\begin{vmatrix}\vec{i} & \vec{j} & \vec{k} \newline 1 & 0 & -1 \newline 0 & 1 & 1\end{vmatrix} = \vec{i} - \vec{j} + \vec{k}$$
Thus, $I = \int\int_D<0, -4, 0>\cdot<1, -1, 1>\text{d}x\text{d}y = 4\int\int_D\text{d}x\text{d}y$. Let $x=a\cdot r\cos t, y=a\cdot r\sin t, r \in[0, 1], t \in[0, 2pi]$, then $$x^2+y^2+z^2=a^2 \iff x^2+y^2+x^2+y^2-2xy = a^2 \iff 2a^2r^2-2a^2r^2\cos t\sin t = a^2 \iff r^2(2-2\cos t\sin t)= 1 \iff r=\frac{1}{\sqrt{2-2\cos t\sin t}}$$
The determinant of the Jacobian will be $\text{det}(J) = a^2\cdot r$
Thus, $I = 4\int_{0}^{2\pi}\int_{0}^{\frac{1}{\sqrt{2-2\cos t\sin t}}}a^2\cdot r \text{d}r \text{d}t$. However, computing this on Desmos gives a different number from the answer of the textbook, which is $\frac{16\pi a^2\sqrt{3}}{9}$, so I must be doing something incorrectly along the way. I've redone my steps a few times already, and I can't find where I am wrong. Any help is much appreciated!
Too long for a comment:
Could it be that the textbook is wrong? With Wolfram Alpha I get $$ I=4\int_0^{2\pi}\int_0^{\frac{1}{\sqrt{2-2\cos t\sin t}}}a^2 r \,dr \,dt=4\frac{\pi a^2}{\sqrt{3}}\,. $$ This is correct because the length of $T_x\times T_y$ is $\sqrt{3}$ and by $D$ we parametrize a disk of radius $a\,.$ Its area can be expressed as $$ A=\sqrt{3}\int_0^{2\pi}\int_0^{\frac{1}{\sqrt{2-2\cos t\sin t}}}a^2 r \,dr \,dt=\pi a^2\,, $$ fully compatible with $I\,.$