/sci/ - Science & Math

Would someone mind checking my work? I'm supposed to calculate the flux out of this hemisphere (r=4, quadrants 6 and 7) using the divergence theorem for the vector [math]\vec{F}= yx^2\hat{i} + (xy^2-3z^4)\hat{j} + (x^3+y^2)\hat{k}[/math]
however, when working the triple integral in spherical coordinates, I get a problematic term that cancels the flux to 0

[math]\nabla\cdot\vec{F} = 2xy+2xy+0[/math]
[math]\iint_S \vec{F}\cdot dS = \iiint_V \nabla\cdot\vec{F} dV \Rightarrow 4\iiint_V xydV[/math]
converting to spherical coordinates, where [math]\phi[/math] is the angle from the z axis,
[math]4\iiint_V xydV = 4\int_{\frac{\pi}{2}}^{\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{4}(\rho sin(\phi)cos(\theta)\cdot \rho sin(\phi)sin(\theta))\cdot \rho^2sin(\phi)d\rho d\theta d\phi[/math]
[math] = 4\int_{\frac{\pi}{2}}^{\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{4} \rho^4 sin^3(\phi)cos(\theta)sin(\theta)d\rho d\theta d\phi[/math]
evaluating,
[math] = \frac{4}{5}\int_{\frac{\pi}{2}}^{\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} (\rho^5 sin^3(\phi)cos(\theta)sin(\theta))\bigg\rvert_{\rho=0}^{\rho=4}d\theta d\phi[/math]
[math]= \frac{4096}{5}\int_{\frac{\pi}{2}}^{\pi} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}sin^3(\phi)cos(\theta)sin(\theta)d\theta d\phi[/math]

evaluating theta is where I get an issue: [math]\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} sin(\theta)cos(\theta)d\theta = 0[/math]
which cancels the entire expression out. Where am I going wrong?