Question

F(r) = \textit{xy}\hat{x} + \textit{yz}\hat{y} + \textit{zx}\hat{z}. A closed path is defined as the circle on the \textit{xy} plane, $P: x^2 + y^2 = a^2$. The path $P$ is counterclockwise. $S_1$ is the hemispherical surface, $x^2 + y^2 + z^2 = a^2$ with $z \ge 0$. $S_2$ is the area on the \textit{xy} plane enclosed by $P$. $S$ is the closed surface including $S_1$ and $S_2$ and $V$ is the volume enclosed by $S$. (c) Verify the divergence theorem using the volume $V$ and the closed surface $S$. (f) Can the vector F be an electrostatic field? Explain.

          F(r) = \textit{xy}\hat{x} + \textit{yz}\hat{y} + \textit{zx}\hat{z}.
A closed path is defined as the circle on the \textit{xy} plane, $P: x^2 + y^2 = a^2$. The path $P$ is counterclockwise.
$S_1$ is the hemispherical surface, $x^2 + y^2 + z^2 = a^2$ with $z \ge 0$. $S_2$ is the area on the \textit{xy} plane enclosed
by $P$. $S$ is the closed surface including $S_1$ and $S_2$ and $V$ is the volume enclosed by $S$.
(c) Verify the divergence theorem using the volume $V$ and the closed surface $S$.
(f) Can the vector F be an electrostatic field? Explain.

F(r) = xyx̂ + yzŷ + zxẑ.
A closed path is defined as the circle on the xy plane, P: x^2 + y^2 = a^2. The path P is counterclockwise.
S1 is the hemispherical surface, x^2 + y^2 + z^2 = a^2 with z ≥ 0. S2 is the area on the xy plane enclosed
by P. S is the closed surface including S1 and S2 and V is the volume enclosed by S.
(c) Verify the divergence theorem using the volume V and the closed surface S.
(f) Can the vector F be an electrostatic field? Explain.

Added by Sonia Y.

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