Evaluating surface integral over vectorfield out of a sphere with spherical cordinates

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This is a question thats been bothering my mind. I've tried to solve it but its not working as I was thinking.

I know the values.

I know also div F = 0+0+2x = 2x

R highest and lowest value is 0 < $\theta$ < $1/\pi$

I know theta $\theta$ is betwen 0 < $\theta$ < $\pi/2$

I know $\phi$ is between $\pi/4$ < $\theta$ < $\pi/2$

But out of all this information I cant find out to solve the tripple integral? Im quite unsure on how I shouldput the equation up when using spherical coordinates and I would love some advice. Thank you.

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This is a line integral, you can relate it to a surface integral of $\nabla \times \mathbf F$, not to a volume integral of $\nabla \cdot \mathbf F$. Since $\nabla \times \mathbf F = 0$, its surface integral is zero. You can take a different path between the same endpoints, the simplest being $\mathbf r(t) = (t, t, 0)$, and evaluate $$\int_0^{1/\sqrt 2} (y + z^2, z + x, y + 2 x z) \cdot \mathbf r'(t) \bigg\rvert_{(x, y, z) = \mathbf r(t)} dt.$$ Alternatively, you can find a potential of $\mathbf F$ and apply the gradient theorem: $$\mathbf F = \nabla (x y + y z + x z^2), \\ \int_\gamma \mathbf F \cdot d\mathbf r = (x y + y z + x z^2) \bigg\rvert_{(x, y, z) = (1/\sqrt 2, 1/\sqrt 2, 0)}.$$