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Verify the divergence theorem for a cylinder
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[QUOTE="HallsofIvy, post: 4739464, member: 637751"] No, S can have any value of z between 0 and 2: S: <cos(t), sin(t), z>. A [b]two dimensional[/b] surface always requires two parameters! The NORMAL VECTOR has no k component: writing [itex]\vec{r}= cos(t)\vec{i}+ sin(t)\vec{j}+ z\vec{k}[/itex], we have [itex]\vec{r}_t= -sin(t)\vec{i}+ sin(t)\vec{j}[/itex] and [itex]\vec{r}_z= \vec{k}[/itex] and their cross product is [itex]cos(t)\vec{i}+ sin(t)\vec{j}[/itex] The top is NOT "<cos(t), sin(t), z>". That has r fixed at 1 and both t and z varying so is the cylindrical side. The top has both r and t varying and z fixed at 2: <r cos(t), r sin(t), 2> (the bottom is <r cos(t), r(sin t), 0>). The derivative with respect to r, for the top, is <cos(t), sin(t), 0> and the derivative with respect to t <-r sin(t), cos(t), 0>. Their cross product gives [itex]\vec{n}dS= r \vec{k} drdt[/itex]. Similarly, [itex]\vec{n}dS[/itex] is [itex]-r\vec{k}drdt[/itex] for the bottom. [/QUOTE]
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Verify the divergence theorem for a cylinder
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