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I am solving the problem "Determine the constants $a$ and $b$ such that the curl of the vector $A=(2xy+3yz) \hat i+(x^2+axz-4z^2) \hat j-(3xy+byz) \hat k$ is zero. When I find the curl I get $[-x(3+a)+z(8-b)]\hat i+6y \hat j+ z(-3+a) \hat k$(equation 1) and since it is equal to zero they have taken the solution in the next step as $3+a=0\ and\ 8-b=0\ gives\ a=-3\ and\ b=8$, and $-3+a=0 \ gives\ a=3$.

First question: So does $a$ have two values?\

Second question:When the vector in equation(1) is equated to zero each component gets equal to zero as $-x(3+a)+z(8-b)=0 (equation(2)),6y=0,z(-3+a)=0$, isn't it? Then from the equation (2) how did we take that $x,z$ are not going to be zero but only $3+a=0\ and\ 8-b=0$? (and continued in the steps similarly)

Buddhini Angelika
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  • The problem is, no values of $a$ and $b$ make $\nabla \times A = 0$ across all points $(x,y,z)$. The problem is either miscopied or unsolvable. The fact that $a$ requires two different values to make the curl vanish is one of the indications that there is no solution, as $a$ can only have one value in an expression. But a clearer indication is that the $\hat j$ coefficient is not $0$ and doesn't depend on $a$ or $b$, so no values for them will ever make it $0$. – Paul Sinclair Jul 05 '18 at 23:43
  • Ok, thanks a lot! – Buddhini Angelika Jul 06 '18 at 09:01

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