I got 2.705 for the semi-minor and 4.2890 for the semi-major. As far as I was aware these were just lengths, I solved them by equating those two q/p equations and rearranging to give ##λ^2-(u+w)λ+uw-v^2=0## and solving.
I basically followed the steps on these notes...
Homework Statement
Given the ellipse
##0.084x^2 − 0.079xy + 0.107y^2 = 1 ##
Find the semi-major and semi-minor axes of this ellipse, and a unit vector in the
direction of each axis.
I have calculated the semi-major and minor axes, I am just stuck on the final part.
Homework Equations
this...
I think so. No flux of V is passing through the sides so it all must be through the ABEF surface (as I type that I'm beginning to think I don't understand)
Yes, apologies. I have a habit of explaining how my brain remembers things rather than the mathematical reasoning. It occasionally causes me to lose marks. OAF and BED are along the y plane, thus dS will contain the unit vector j which would cause the integral to be zero.
Okay, so I have shown...
Okay, ADEF is in the x plane, which makes the dS=dydzi. This would mean ∫C(∇xV)⋅dS=∫S(3x2+3y2)k⋅dydzi. I know that dotting two different unit vectors = 0, so does that mean ∫C(∇xV)⋅dS is equal to 0 for ADEF and by extension, BED and OAF are 0 too?
Sorry it was in an edit that I didn't save.
The vector field is V(r)=-y3i+x3j+z3k
The C from the first question was the square path OABD.
Also w.r.t. what you highlighted in red, in my notes ∫S(∇xV)⋅dS is called the area integral. That's what I was referring to.
Homework Statement
. In the diagram below, the line AB is at x = 1 and the line BD is at y = 1.
Use Stokes’ theorem to find the value of the integral ∫s2(∇xV)⋅dS where S2 where S2 is
the curved surface ABEF, given that AF and BE are straight lines, and the curve EF is in the y-z plane (i.e...