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# Calculate $\iint_{\lambda S} \mathbf{F} \cdot \mathbf{n} d S$ for each of the following. Looked at the right way, all are quite easy and some are even trivial.(a) $\mathbf{F}=(2 x+y z) \mathbf{i}+3 y \mathbf{j}+z^2 \mathbf{k}$; $S$ is the solid sphere $x^2+y^2+z^2 \leq 1$.(b) $\mathbf{F}=\left(x^2+y^2+z^2\right)^{5 / 3}(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) ; S$ as in part (a).(c) $\mathbf{F}=x^2 \mathbf{i}+y^2 \mathbf{j}+z^2 \mathbf{k}$; S is the solid sphere $(x-2)^2+y^2+z^2 \leq 1$.(d) $\mathbf{F}=x^2 \mathbf{i} ; S$ is the cube $0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1$.(e) $\mathbf{F}=(x+z) \mathbf{i}+(y+x) \mathbf{j}+(z+y) \mathbf{k} ; S$ is the tetrahedron cut from the first octant by the plane $3 x+4 y+2 z=12$.(f) $\mathbf{F}=x^3 \mathbf{i}+y^3 \mathbf{j}+z^3 \mathbf{k} ; S$ as in part (a).(g) $\mathbf{F}=(x \mathbf{i}+y \mathbf{j}) \ln \left(x^2+y^2\right)$; S is the solid cylinder $x^2+y^2 \leq 4,0 \leq z \leq 2$.

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### (a)

Let's find the divergence of the given vector field. Since ${\bf F}(x,y,z)=(2x+yz){\bf i}+3y{\bf j}+z^2{\bf k}$ then $P(x,y,z)=2x+yz, Q(x,y,z)=3y$ and $R(x,y,z)=z^2$ and therefore:

\begin{aligned} \nabla\cdot{\bf F}&=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}\\ &=\dfrac{\partial (2x+yz)}{\partial x}+\dfrac{\partial (3y)}{\partial y}+\dfrac{\partial (z^2)}{\partial z}\\ &=2+3+2z\\ &=5+2z \end{aligned}

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