WEBVTT

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We want the circulation
of this vector field around one circle.

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Begin by reading every geometric fact
in the question against that circle.

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Its radius is three,

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it lies in the plane y equals four, and
its centre is the point zero, four, zero.

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That plane is perpendicular to the y axis,

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so the circle stands like a coin
in a translated copy of the x z plane.

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Viewed from the positive y axis
while looking toward the origin,

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the red arrows run clockwise.

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This direction will determine
the sign of the answer.

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Stokes' theorem replaces circulation
around the boundary with flux of the curl

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through a spanning surface.

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The boundary direction
and the surface normal must be compatible.

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The surface may be any
surface with this rim.

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A bowl can dip behind it,
and a dome can rise in front of it.

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Both have exactly the same boundary.

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The flat disc is the useful choice.

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It lies in y equals four,

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its unit normal is constant, and its area
is pi times three squared, or nine pi.

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Write the field as three components:
minus y z, four y plus one, and x y.

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These are the three entries
the curl differentiates.

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Each curl slot uses the other
two components in cyclic order.

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The determinant mnemonic
records the same pattern with i, j,

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k on top, the partial derivatives in the
middle, and the field along the bottom.

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For the first slot, differentiate x y

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with respect to y, then subtract
the z derivative of four y plus one.

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Those terms are x and zero,
so the first slot is x.

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For the second slot, the z
derivative of minus y z is minus y.

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Subtract the x derivative
of x y, which is y.

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Minus y minus y is minus two y.

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For the third slot, the x
derivative of four y plus one is zero.

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Subtract the y derivative of minus y z,
which is minus z, and the result is z.

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Together the three slots
give x, minus two y, z.

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Notice that four y plus
one contributed nothing

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because both derivatives taken from it
were in directions it does not depend on.

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Keep that term marked for the final check.

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Now the right-hand rule fixes the normal.

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From the side the normal points toward,

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the positive walk around
the boundary must look counterclockwise.

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Move to the positive y viewpoint.

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The given red walk looks
clockwise from there,

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so the compatible normal points away
from that viewer, back toward the origin.

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Therefore n hat is minus j hat:

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zero, minus one, zero.

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A free vector keeps that direction

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when its base is moved
away from the y-axis line.

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The orientation pair moves together.

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Reverse the boundary walk
and the compatible normal flips to plus j.

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Restore the given clockwise walk
and the normal returns to minus j.

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Because the disc is flat,
that same normal belongs at every point.

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Slide it across the surface
and only its base point changes;

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its direction remains minus j everywhere.

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Dot the curl x, minus two y,
z with the normal zero, minus one, zero.

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The x term is multiplied by zero,
and the z term is multiplied by zero.

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They vanish.

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The middle product is minus two y
times minus one, leaving plus two y.

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Every point of this disc
has y equal to four.

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As a point crosses the disc,
its reading stays at y equals four:

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two y is eight at every point,
even while x and z change.

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So the surface integral is eight
times the area of a radius-three disc.

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The area is nine pi, and eight
times nine pi is seventy-two pi.

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Therefore the requested
circulation is seventy-two pi.

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The unused middle component
has a structural explanation.

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Four y plus one in the j direction
is the gradient of two y squared plus y,

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and every gradient has zero curl.

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A direct parametrization gives
r of t equal to three cosine t,

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four, three sine t.

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Its dot product with the field
simplifies to thirty-six d t.

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Integrating thirty-six from zero
to two pi gives seventy-two pi again.

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The boundary calculation
and the flux calculation agree.