I could do (b) but not (c)...
Im not sure how they changed to the plane coordinate system of r(radius) and theta.
For part (c): i did up to,
-∬(x^2 + y^2)^(1/2) dxdy
then they changed it to the plane coordinate system which gives,
-∫(r=0 to 3)∫(theta=0 to 2pi) r^2 drd(theta)
How??
Let S be the funnel shaped surface defined by x^2 + y^2 = z^2 for 0<z<9
(a) Sketch S
(b) Determine the outward pointing normal to S
(c) Evaluate ∬F.ds where F = -yi + xj + zk
This is a super long question, but if you could help... it will be great..
Answer for (b): x(x^2 + y^2)^(-0.5) i + y(x^2 + y^2)^(-0.5) j - k
Answer for (c): -(2pi)/3
2 answers
recall polar coordinates:
x^2+y^2 = r^2
dx dy = r dr dθ
x^2+y^2 = r^2
dx dy = r dr dθ