Multivariable calculus

Curl of the Vector Field (sin y, sin z, sin x)

The curl of the field with components sin y, sin z and sin x is the vector with components -cos z, -cos x and -cos y. In each difference the first partial is zero and the second is a cosine, so every component of the answer is a single negative cosine. At the origin the curl is (-1, -1, -1).

P(x,y,z)=sin⁡yP(x,y,z) = \sin y
Q(x,y,z)=sin⁡zQ(x,y,z) = \sin z
R(x,y,z)=sin⁡xR(x,y,z) = \sin x
∇×F=⟨−cos⁡z, −cos⁡x, −cos⁡y⟩\nabla \times \mathbf{F} = \langle -\cos z,\ -\cos x,\ -\cos y \rangle

Each component depends on one variable, and it is the wrong one

PP contains only yy, QQ only zz, RR only xx. Lining that up against the curl formula, three of the six partials are derivatives with respect to an absent variable and die, and three survive.

∂R∂y−∂Q∂z=0−cos⁡z=−cos⁡z\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = 0 - \cos z = -\cos z
∂P∂z−∂R∂x=0−cos⁡x=−cos⁡x\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 0 - \cos x = -\cos x
∂Q∂x−∂P∂y=0−cos⁡y=−cos⁡y\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 - \cos y = -\cos y

Put this beside ⟨x2,y2,z2⟩\langle x^2, y^2, z^2 \rangle, whose curl is zero. There each component held its own variable and every crossed partial vanished. Here each component holds the next variable instead, and the curl is as large as it can be.

Where the spin stops

The curl vanishes exactly where all three cosines vanish, which is where xx, yy and zz are each an odd multiple of π2\frac{\pi}{2}. Those are isolated points, not a region, so the field is nowhere conservative.

∇×F=0whencos⁡x=cos⁡y=cos⁡z=0\nabla \times \mathbf{F} = \mathbf{0} \quad \text{when} \quad \cos x = \cos y = \cos z = 0

At (π2,π2,π2)\left( \frac{\pi}{2}, \frac{\pi}{2}, \frac{\pi}{2} \right) the field is ⟨1,1,1⟩\langle 1,1,1 \rangle and the curl is 0\mathbf{0}: the flow is moving briskly and shearing not at all. At the origin the field is 0\mathbf{0} and the curl is ⟨−1,−1,−1⟩\langle -1,-1,-1 \rangle: nothing is moving and everything is turning.

Those two points together are the cleanest argument that the size of a field and the size of its curl are unrelated quantities.

The mistake: differentiating sine without checking the variable

The reflex is that the derivative of sin⁡\sin is cos⁡\cos, and it fires before the variable is read. So ∂∂zsin⁡y\frac{\partial}{\partial z} \sin y gets written as cos⁡y\cos y when it is 00, and the answer comes out with the wrong three cosines in it.

Say the variable out loud before differentiating: with respect to zz, the expression sin⁡y\sin y is a constant. The reflex is right only when the argument contains the variable you are differentiating in.

A quick check at the origin catches the error. There cos⁡x=cos⁡y=cos⁡z=1\cos x = \cos y = \cos z = 1, so the correct curl is ⟨−1,−1,−1⟩\langle -1,-1,-1 \rangle. Any answer that gives something else at the origin has a partial in the wrong slot.

Each cosine above is confirmed on every build by numerically differentiating the three sine components at seeded sample points.

Frequently asked questions

Why is there no chain rule factor anywhere in the answer?

Because each argument is a bare variable, so the inner derivative is 11 when the variable matches and 00 when it does not. Change PP to sin⁡(yz)\sin(yz) and the picture changes: ∂P∂z=ycos⁡(yz)\frac{\partial P}{\partial z} = y \cos(yz), and that extra factor of yy shows up in the second component of the curl.

How large can this curl get?

Its length never exceeds 3\sqrt{3}, since each component is a cosine and so lies between −1-1 and 11. The maximum is reached where cos⁡x\cos x, cos⁡y\cos y and cos⁡z\cos z all equal 11 or all equal −1-1, for instance at the origin, where the curl is ⟨−1,−1,−1⟩\langle -1,-1,-1 \rangle with length 3\sqrt{3}.