This is an easy question.
http://www.ma.utexas.edu/users/goddardb/M427Lfsum11.pdf
Do problem #3 and #6 make sense?
The problem I have with #3 is that it is that the surface integral is over an unbounded region.
x, y, z ∈ [0, Infinity)
The problem I have with #6 is that... well, I don't even know what it is. I think I need more information. It looks sorta like a triple line integral.
It's not a basic triple integral of the form: ∫∫∫ f(x,y,z) dx dy dz
It's not a line integral since those look like: ∫ F1dx F2dy F3dz where f(x,y,z) =
I imagine he wants us to use Gauss' Theorem.. somehow.
∫∫∫ div F dV = ∫∫ F (dot) dS...
I don't know!! Help. :(
http://www.ma.utexas.edu/users/goddardb/M427Lfsum11.pdf
Do problem #3 and #6 make sense?
The problem I have with #3 is that it is that the surface integral is over an unbounded region.
x, y, z ∈ [0, Infinity)
The problem I have with #6 is that... well, I don't even know what it is. I think I need more information. It looks sorta like a triple line integral.
It's not a basic triple integral of the form: ∫∫∫ f(x,y,z) dx dy dz
It's not a line integral since those look like: ∫ F1dx F2dy F3dz where f(x,y,z) =
I imagine he wants us to use Gauss' Theorem.. somehow.
∫∫∫ div F dV = ∫∫ F (dot) dS...
I don't know!! Help. :(
-
For #6, this integral is independent of path, since the gradient of 2x cos y + ln y + ln z equals
<2 cos y, 1/y - 2x sin y, 1/z>.
So, the integral equals by FTC for line integrals
(2x cos y + ln y + ln z) {for (x,y,z) = (1, π/2, 2) to (0, 2, 1)}
= ln(π/2).
I hope this helps!
<2 cos y, 1/y - 2x sin y, 1/z>.
So, the integral equals by FTC for line integrals
(2x cos y + ln y + ln z) {for (x,y,z) = (1, π/2, 2) to (0, 2, 1)}
= ln(π/2).
I hope this helps!