# 求解大學數vector

In cylindrical polar coordinates, vector fields F and G are defined as

F = r^2 sin\$ er and G= r z^2 e\$

Verify, using cylindrical polar coordinates that

div(F x G) = G . (curl F) - F.(curl G)

### 1 個解答

• c
Lv 5
7 年前
最愛解答

Actually the equation required to be verified isa property relating div and curl.

Now use the same convention of notation stated in yourquestion.

firstwe know FxG =(r^3 * sin \$ * z^2) ez With the formula list in wolfram alpha,[reference link: http://mathworld.wolfram.com/CylindricalCoordinate... (no. 116)Remark: one notation is different in the abovewebsite, theta as \$]for any F can be written as F = F_r er + F_z er +F_\$ e\$,[kindly note that F_r is the scalar component ofF in the direction on r, and so on]Div F= 1/r * d/dr (r*F_r) + 1/r* d/d\$ (F_\$) + d/dz(F_z) R.H.S.=div (F x G)=0+0+d/dz (r^3 * sin \$ * z^2)=2 * r^3 * z * sin \$ L.H.S.=G . (curl F) – F . (curl G)=G . { (0 – 0) er + [ d/dz (r^2 * sin\$)- 0]e\$ + 1/r*[ 0 - d/d\$ (r^2 * sin\$)] ez }-F. { [0 – d/dz (r*z^2)] er + (0- 0) e\$ + 1/r*[d/dr (r^2 * z^2) - 0] ez }[formula of curl is also in the formulalist (no. 117)]=[r* z^2 e\$] . [-r*cos\$ ez] – [r^2 *sin\$ er]. [-2*r*z er + 2*z^2 ez]=2*r^3 *z*sin\$ done.

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