Calculate the flow integral $$\iint_{Y} \text{curl} (\vec{F}) · \hat{N} dS$$ where $Y$ is part of the sphere $x^2 + y^2 + (z − 2)^2 = 8$ that lies above the $xy$-plane and $\hat{N}$ is the outward unit normal vector on $Y$ and $$F(x, y , z) = (y^2 \cos(xz), x^3e^{yz} , −e^{−xyz}).$$
I got the answer $48\pi$ but the correct answer is $12\pi$. What am I doing wrong?
In your work it seems that you are evaluating the flux of the curl of $\mathbf{F}$ through the disc $S$ (upward oriented) at the bottom of the truncated sphere $Y$: $$\begin{align} S&=\{(x,y,z): x^2+y^2+(z-2)^2\leq 8, z=0\}\\ &=\{(x,y,0): x^2+y^2\leq 8-(0-2)^2=4\}. \end{align}$$ That's fine because $S$ and $Y$ have the same boundary (with the same induced orientation). However the radius of such disc is $2$ (not $\sqrt{8}$), therefore, the final integral should be $$\pi\int_0^23r^3\,dr=\pi\left[\frac{3r^4}{4}\right]_0^2=12\pi.$$
The same result can be obtained by applying Stokes' Theorem in a more direct way: $$\begin{align} \iint_{Y} &\text{curl} (\mathbf{F})\cdot\mathbf{n}\, dS= \int_{\gamma}\mathbf{F}\cdot d\mathbf{s}\\ &=\int_0^{2\pi} ((2\sin(t))^2 \cos(0), (2\cos(t))^3e^{0} , −e^{0})\cdot(-2\sin(t),2\cos(t),0)dt\\ &=\int_0^{2\pi}(-8\sin^3(t)+16\cos^4(t)) dt=0+ 16\cdot \frac{3\pi}{4}=12\pi \end{align}$$ where $\gamma$ is the boundary of $Y$ (and $S$) counterclockwise oriented, that is the circle $$\gamma(t)=(2\cos(t),2\sin(t),0)\quad\text{ with $t\in [0,2\pi]$}.$$