If "\\overrightarrow{v}" denotes the flow velocity, then for a velocity potential function "\\Phi", "\\overrightarrow{v}" can be represented as the gradient of a scalar function "\\Phi" :
"{\\displaystyle \\overrightarrow{v} =\\nabla \\Phi \\ ={\\frac {\\partial \\Phi }{\\partial x}}\\mathbf {i} +{\\frac {\\partial \\Phi }{\\partial y}}\\mathbf {j} +{\\frac {\\partial \\Phi }{\\partial z}}\\mathbf {k} \\,.}"
In our case, "\\frac {\\partial \\Phi }{\\partial x}=u=ay\\sin xy,\\ \\ \\frac {\\partial \\Phi }{\\partial y}=v=ax\\sin xy."
Then "\\Phi(x,y)=\\int ay\\sin xy dx=-a\\cos xy+C(y)", and therefore,
"\\frac {\\partial \\Phi }{\\partial y}=ax\\sin xy+C'(y)".
It follows that "ax\\sin xy+C'(y)=ax\\sin xy", and consequently, "C'(y)=0". Then "C(y)=C", and we conclude that
"\\Phi(x,y)=-a\\cos xy+C."
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