13.7 #7

Dear Professor Taylor, This problem has me really stumped I don't know what it is asking for the second to last portion because I have entered r dr dtheta as the usual substitution of dA. But it wants it in vector form. Can you please clarify what WebWork wants exactly.

Thank you,

























ok, you have a couple of different confusions going on here:

1) this is a good example of the diversity of notation surrounding this integral: what this problem wants to call a vector dA is what we have been calling n dS. (In support of my rant earlier today about "Flux Integrals",  you might want to note that wikipedia distinguishes two kinds of "surface integrals" as I do, surface integrals of functions and surface integrals of vector fields, that the textbook calls flux integrals)

2) just to confuse the issue a little bit more, this is a parametric integral, not an integral in polar coordinates.  What this means is that the element of area for polar coordinates r dr dθ is NOT the appropriate element of area for this problem.  What you need is just (r_r x r_θ)dr dθ, (note the additional confusion of r being the parametric equation while r is just a parameter) and the correction term r is not used (it sort of gets absorbed into the  r_r x r_θ). Since

  r_r x r_θ=<cos(θ),sin(θ),1>x<-r sin(θ), r cos(θ),0>=<-r cos(θ), -r sin(θ), r>, which is pointing upward and not downward, so your correct answer to part (b) would be
- r_r x r_θ drdθ.  
This is *almost* what you have, you just forgot to multiply the drdθ times the first two components of your vector.