Stokes' Theorem in Practice

We want the circulation of this vector field around one circle. Begin by reading every geometric fact in the question against that circle. Its radius is three, it lies in the plane y equals four, and its centre is the point zero, four, zero. That plane is perpendicular to the y axis, so the circle stands like a coin in a translated copy of the x z plane. Viewed from the positive y axis while looking toward the origin, the red arrows run clockwise. This direction will determine the sign of the answer. Stokes' theorem replaces circulation around the boundary with flux of the curl through a spanning surface. The boundary direction and the surface normal must be compatible. The surface may be any surface with this rim. A bowl can dip behind it, and a dome can rise in front of it. Both have exactly the same boundary. The flat disc is the useful choice. It lies in y equals four, its unit normal is constant, and its area is pi times three squared, or nine pi. Write the field as three components: minus y z, four y plus one, and x y. These are the three entries the curl differentiates. Each curl slot uses the other two components in cyclic order. The determinant mnemonic records the same pattern with i, j, k on top, the partial derivatives in the middle, and the field along the bottom. For the first slot, differentiate x y with respect to y, then subtract the z derivative of four y plus one. Those terms are x and zero, so the first slot is x. For the second slot, the z derivative of minus y z is minus y. Subtract the x derivative of x y, which is y. Minus y minus y is minus two y. For the third slot, the x derivative of four y plus one is zero. Subtract the y derivative of minus y z, which is minus z, and the result is z. Together the three slots give x, minus two y, z. Notice that four y plus one contributed nothing because both derivatives taken from it were in directions it does not depend on. Keep that term marked for the final check. Now the right-hand rule fixes the normal. From the side the normal points toward, the positive walk around the boundary must look counterclockwise. Move to the positive y viewpoint. The given red walk looks clockwise from there, so the compatible normal points away from that viewer, back toward the origin. Therefore n hat is minus j hat: zero, minus one, zero. A free vector keeps that direction when its base is moved away from the y-axis line. The orientation pair moves together. Reverse the boundary walk and the compatible normal flips to plus j. Restore the given clockwise walk and the normal returns to minus j. Because the disc is flat, that same normal belongs at every point. Slide it across the surface and only its base point changes; its direction remains minus j everywhere. Dot the curl x, minus two y, z with the normal zero, minus one, zero. The x term is multiplied by zero, and the z term is multiplied by zero. They vanish. The middle product is minus two y times minus one, leaving plus two y. Every point of this disc has y equal to four. As a point crosses the disc, its reading stays at y equals four: two y is eight at every point, even while x and z change. So the surface integral is eight times the area of a radius-three disc. The area is nine pi, and eight times nine pi is seventy-two pi. Therefore the requested circulation is seventy-two pi. The unused middle component has a structural explanation. Four y plus one in the j direction is the gradient of two y squared plus y, and every gradient has zero curl. A direct parametrization gives r of t equal to three cosine t, four, three sine t. Its dot product with the field simplifies to thirty-six d t. Integrating thirty-six from zero to two pi gives seventy-two pi again. The boundary calculation and the flux calculation agree.