Understanding the given transformation - PDE

[chwala]

TL;DR

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.

 

[Mark44]

Is your question about how to evaluate a 3 x 3 determinant? If so, you can choose to go across any row or down any column. To go across the 2nd row, the determinant is $(-a_{2, 1})$ times its cofactor + $(+a_{2, 2})$ times its cofactor + $(-a_{2, 3}$) times its cofactor. For each of these terms the cofactor is the 2 x 2 determinant that consists of the entries not in that row and not in that column.

 

[pbuk]

$\cos θ[r^2 \cos θ \sin θ \cos^2 ∅ + r^2 \sin θ \cos θ cos^2∅ ] + r\sin θ [r\sin^2 θ \cos^2∅ + r \sin^2 θ \sin^2 ∅]=$

The symbol $\phi$ is called "phi" and is entered in $\LaTeX$ as \phi ($\phi$). The Unicode character ∅ represents the empty set and should not be used as a substitute for $\phi$.

 

[chwala]

My intention was to distinguish phi and theta. Noted though.

 

[pbuk]

My intention was to distinguish phi and theta. Noted though.

Maybe, but the way to do this is to use a symbol that represents phi to represent phi, not a symbol that represents the empty set. The fact that sometimes this symbol looks a bit like some representations of phi is irrelevant.

It's like writing the equation for the circumference of a circle as 2r.

 

[pbuk]

Also, due to a quirk of the $\LaTeX$ implementation, the Unicode character ∅ is translated to the symbol \emptyset which renders as $\emptyset$: if you want a character that looks like ∅ you need to use \varnothing: $\varnothing$.