- #1
chwala
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- TL;DR Summary
- I am going through the notes on pde. I want to be certain that i am getting it right. From my understanding of determinants of 3 by 3 matrices ...my approach is indicated
My interest is on how they arrived at ##r^2 \sin θ##
My approach using the third line, is as follows
##\cos θ[r^2 \cos θ \sin θ \cos^2 ∅ + r^2 \sin θ \cos θ cos^2∅ ] + r\sin θ [r\sin^2 θ \cos^2∅ + r \sin^2 θ \sin^2 ∅]=##
##\cos θ[r^2 \cos θ \sin θ[\cos^2 ∅+ \sin^2 ∅]] + r\sin θ [r\sin^2 θ [cos^2 ∅+ \sin^2 ∅]]=r^2\cos^2θ\sin^2 θ + r^2\sin^2θ\sinθ=##
##r^2\sinθ(cos^2θ+\sin^2θ)=r^2\sin θ##
If correct then we could also use second row but with negative place value i.e ##-\sin θ\sin ∅ [ a-b] + r\cos θ sin ∅[c-d]...## to realize the same.
Cheers.
My approach using the third line, is as follows
##\cos θ[r^2 \cos θ \sin θ \cos^2 ∅ + r^2 \sin θ \cos θ cos^2∅ ] + r\sin θ [r\sin^2 θ \cos^2∅ + r \sin^2 θ \sin^2 ∅]=##
##\cos θ[r^2 \cos θ \sin θ[\cos^2 ∅+ \sin^2 ∅]] + r\sin θ [r\sin^2 θ [cos^2 ∅+ \sin^2 ∅]]=r^2\cos^2θ\sin^2 θ + r^2\sin^2θ\sinθ=##
##r^2\sinθ(cos^2θ+\sin^2θ)=r^2\sin θ##
If correct then we could also use second row but with negative place value i.e ##-\sin θ\sin ∅ [ a-b] + r\cos θ sin ∅[c-d]...## to realize the same.
Cheers.