Answer:
-6re−r [sin(6θ) - cos(6θ)]
Step-by-step explanation:
the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ
x = e−r sin(6θ), y = er cos(6θ)
δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)
δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)
∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ
= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]
= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))
= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]
= -6re−r [sin(6θ) - cos(6θ)] since [cos²(6θ) + sin²(6θ)] = 1
