{"version":1,"lectureId":"01M14TZG80Q03WZ8E386YW13DE","attempt":0,"publication":{"slug":"stokes-theorem-in-practice","title":"Stokes' Theorem in Practice","subject":"mathematics","summary":"One exam question, worked from the first reading to the final check. The line integral of F = -yz i + (4y + 1) j + xy k around a circle of radius 3 standing in the plane y = 4, walked clockwise as seen from the positive y-axis. Stokes' theorem trades that walk around the rim for a flux through the flat disc the rim bounds, so we compute the curl one slot at a time, settle which way the unit normal points by reading the right-hand rule off the wording of the problem rather than off a preference, and integrate a constant through a disc of known area. The answer is verified twice.","metaDescription":"Stokes' theorem on one exam question: swap the walk around the circle for the flux through the disc, and read the normal off the wording.","transcript":"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.","watch":{"version":1,"scenes":[{"title":"Stokes' Theorem in Practice","start":0,"end":281.1592083333333,"objects":{"answer":"a Math [text] that says \"$integral_C arrow(F) dot dif arrow(r) = 72 pi$\"","bowl":"a Surface [green] drawn in frame (function=<function>, u_range=(-3.0, 3.0), v_range=(-3.0, 3.0))","centre":"a Point [text] labelled \"(0, 4, 0)\" drawn in frame (location=(0.0, 4.0, 0.0))","checks_heading":"a Heading that says \"Two Independent Checks\"","components":"a Math [text] that says \"$arrow(F) = chevron.l -y z, thin 4 y + 1, thin x y chevron.r$\"","curl_gradient":"a Math [text] that says \"$nabla times nabla(2 y^2 + y) = arrow(0)$\"","curl_heading":"a Heading that says \"The Curl, One Slot at a Time\"","curl_result":"a Math [text] that says \"$nabla times arrow(F) = chevron.l x, thin -2 y, thin z chevron.r$\"","curl_work":"a Derivation [text] that says \"$(nabla times arrow(F))_1 &= (partial F_3)/(partial y) - (partial F_2)/(partial z) = x - 0 = x \\ (nabla times arrow(F))_2 &= (partial F_1)/(partial z) - (partial F_3)/(partial x) = -y - y = -2 y \\ (nabla times arrow(F))_3 &= (partial F_2)/(…$\"","direct_check":"a Derivation [text] that says \"$arrow(r)(t) = chevron.l 3 cos t, thin 4, thin 3 sin t chevron.r \\ arrow(F) dot dif arrow(r) = 36 thin dif t \\ integral_0^(2 pi) 36 thin dif t = 72 pi$\"","disc":"a Circle [blue] drawn in frame (center=(0.0, 4.0, 0.0), radius=3.0, normal_vector=(0.0, 1.0, 0.0))","dome":"a Surface [magenta] drawn in frame (function=<function>, u_range=(-3.0, 3.0), v_range=(-3.0, 3.0))","flux_heading":"a Heading that says \"The Flux Through the Disc\"","flux_work":"a Derivation [text] that says \"$nabla times arrow(F) &= chevron.l x, thin -2 y, thin z chevron.r \\ hat(n) &= chevron.l 0, thin -1, thin 0 chevron.r \\ (nabla times arrow(F)) dot hat(n) &= x dot 0 + (-2 y)(-1) + z dot 0 \\ &= 2 y = 2(4) = 8 \\ integral_C arrow(F) dot dif arr…$\"","frame":"an Axes3D (x_range=(-4.0, 4.0), y_range=(-2.0, 6.0), z_range=(-4.0, 4.0))","gradient_check":"a Math [text] that says \"$(4 y + 1) hat(j) = nabla(2 y^2 + y)$\"","mnemonic":"a Math [text] that says \"$nabla times arrow(F) = op(\"det\") mat(hat(i), hat(j), hat(k); partial_x, partial_y, partial_z; -y z, 4 y + 1, x y)$\"","normal":"a Vector [green] labelled \"hat(n)\" drawn in frame (start=(<VariableNumber normal_x = 1.25>, 4.0, <VariableNumber normal_…, end=(<VariableNumber normal_x = 1.25>, <VariableNumber normal_tip_y…)","normal_heading":"a Heading that says \"The Normal Follows the Walk\"","normal_result":"a Math [text] that says \"$hat(n) = -hat(j) = chevron.l 0, thin -1, thin 0 chevron.r$\"","normal_tip_y":"a VariableNumber (initial_value=1.65, format_spec='.2f')","normal_x":"a VariableNumber (initial_value=1.25, format_spec='.2f')","normal_z":"a VariableNumber (initial_value=1.1, format_spec='.2f')","pattern":"a Panel that says \"Each slot uses the other two components in cyclic order.\"","point":"a Point [yellow] drawn in frame (location=(3.0, 4.0, 0.0))","point_2":"a Point [yellow] drawn in frame (location=(0.0, 4.0, 3.0))","point_3":"a Point [yellow] drawn in frame (location=(0.0, 4.0, 0.0))","point_4":"a Point [yellow] drawn in frame (location=(0.0, 4.0, 0.0))","point_5":"a Point [yellow] drawn in frame (location=(0.0, 4.0, 0.0))","probe":"a Point [yellow] labelled \"y = 4.0, thin 2 y = 8\" drawn in frame (location=(<VariableNumber probe_x = -0.8>, <VariableNumber probe_y = 4.0…)","probe_x":"a VariableNumber (initial_value=-2.0, format_spec='.1f')","probe_z":"a VariableNumber (initial_value=-1.0, format_spec='.1f')","question":"a Panel that says \"Evaluate $integral_C arrow(F) dot dif arrow(r)$ for $arrow(F) = -y z thin hat(i) + (4 y + 1) thin hat(j) + x y thin hat(k)$ $C$: radius $3$, plane $y=4$, clockwise from the positive $y$-axis\"","reverse_walk":"a ParametricCurve [yellow] drawn in frame (function=<function>, t_range=(0.0, 6.283185307179586))","reverse_walk_2":"an Orientation [red] drawn in frame (path=((3.0, 4.0, -7.347880794884119e-16), (2.9855541800165906, 4.0, …, closed=True, arrows=4)","right_hand":"a Panel that says \"From the side the normal points toward, the positive boundary direction looks counterclockwise.\"","rim":"a ParametricCurve [yellow] drawn in frame (function=<function>, t_range=(0.0, 6.283185307179586))","statement":"a Math [text] that says \"$integral_C arrow(F) dot dif arrow(r) = integral.double_S (nabla times arrow(F)) dot hat(n) thin dif S$\"","theorem":"a Panel that says \"$S$ may be any oriented surface with boundary $C$, provided its normal agrees with the direction around $C$.\"","theorem_heading":"a Heading that says \"Stokes' Theorem\"","walk":"an Orientation [red] drawn in frame (path=((3.0, 4.0, 0.0), (2.9855541800165906, 4.0, 0.2940514209886818)…, closed=True, arrows=4)"},"beats":[{"start":0,"say":"We want the circulation of this vector field around one circle. Begin by reading every geometric fact in the question against that circle.","live":[],"does":[[0,"question is shown on the screen, written out."],[0,"frame is shown on the screen, written out."],[2.682,"rim is shown on the screen, drawn."]]},{"start":8.2045,"say":"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.","live":["frame","question","rim"],"does":[[9.389,"point is shown on the screen, grown."],[10.588999999999999,"point is hidden from the screen."],[12.105,"point_2 is shown on the screen, grown."],[13.174,"centre is shown on the screen, written out."],[13.305,"point_2 is hidden from the screen."],[14.078999999999999,"point_3 is shown on the screen, grown."],[15.578999999999999,"point_3 is hidden from the screen."]]},{"start":24.094499999999996,"say":"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.","live":["frame","question","rim","centre"],"does":[[28.308999999999997,"walk is shown on the screen, written out."],[33.2435,"frame moves to a new place on the board."],[33.2435,"question is hidden from the screen — left the board."]]},{"start":33.84349999999999,"say":"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.","live":["frame","rim","centre","walk"],"does":[[33.84349999999999,"theorem_heading is shown on the screen, written out."],[34.82999999999999,"statement is shown on the screen, written out."],[35.36399999999999,"statement (the \"integral_C arrow(F) dot dif arrow(r)\" part) is emphasized."],[36.40899999999999,"theorem is shown on the screen, written out."],[37.34999999999999,"statement (the \"integral.double_S (nabla times arrow(F)) dot hat(n) thin dif S\" part) is emphasized."],[37.34999999999999,"statement (the \"integral_C arrow(F) dot dif arrow(r)\" part) is no longer emphasized."]]},{"start":44.915499999999994,"say":"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.","live":["frame","rim","centre","walk","theorem","statement","theorem_heading"],"does":[[48.827999999999996,"bowl is shown on the screen, faded in."],[50.59299999999999,"bowl is hidden from the screen."],[50.59299999999999,"dome is shown on the screen, faded in."],[52.78699999999999,"dome is hidden from the screen."]]},{"start":55.67399999999999,"say":"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.","live":null,"does":[[56.15,"disc is shown on the screen, faded in."],[60.248,"point_4 is shown on the screen, grown."],[61.448,"point_4 is hidden from the screen."],[66.785,"statement is hidden from the screen — left the board."],[66.785,"theorem is hidden from the screen — left the board."],[66.785,"theorem_heading is hidden from the screen — left the board."]]},{"start":67.38499999999999,"say":"Write the field as three components: minus y z, four y plus one, and x y. These are the three entries the curl differentiates.","live":["frame","rim","centre","walk","disc"],"does":[[67.38499999999999,"curl_heading is shown on the screen, written out."],[68.929,"components is shown on the screen, written out."],[70.02,"components (the \"-y z\" part) is emphasized."],[71.518,"components (the \"-y z\" part) is no longer emphasized."],[71.518,"components (the \"4 y + 1\" part) is emphasized."],[73.306,"components (the \"4 y + 1\" part) is no longer emphasized."],[73.306,"components (the \"x y\" part) is emphasized."]]},{"start":78.109,"say":"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.","live":["frame","rim","centre","walk","disc","components","curl_heading"],"does":[[78.109,"components (the \"x y\" part) is no longer emphasized."],[81.08099999999999,"pattern is shown on the screen, written out."],[82.90399999999998,"mnemonic is shown on the screen, written out."],[82.90399999999998,"mnemonic is emphasized."],[91.53049999999999,"components moves to a new place on the board."],[91.53049999999999,"mnemonic is hidden from the screen — left the board."],[91.53049999999999,"pattern is hidden from the screen — left the board."]]},{"start":92.1305,"say":"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.","live":null,"does":[[92.1305,"curl_work is shown on the screen, written out."],[94.80099999999999,"curl_work (the \"(partial F_3)/(partial y)\" part) is emphasized."],[97.958,"curl_work (the \"(partial F_2)/(partial z)\" part) is emphasized."],[97.958,"curl_work (the \"(partial F_3)/(partial y)\" part) is no longer emphasized."],[101.64999999999999,"curl_work (the \"(partial F_2)/(partial z)\" part) is no longer emphasized."],[101.64999999999999,"curl_work (the \"x#2\" part) is emphasized."],[101.952,"curl_work (the \"0\" part) is emphasized."],[101.952,"curl_work (the \"x#2\" part) is no longer emphasized."],[103.136,"curl_work (the \"0\" part) is no longer emphasized."],[103.136,"curl_work (the \"= x\" part) is emphasized."]]},{"start":105.007,"say":"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.","live":null,"does":[[105.605,"curl_work is shown on the screen, written out."],[106.65,"curl_work (the \"= x\" part) is no longer emphasized."],[106.65,"curl_work (the \"(partial F_1)/(partial z)\" part) is emphasized."],[111.08500000000001,"curl_work (the \"(partial F_1)/(partial z)\" part) is no longer emphasized."],[111.08500000000001,"curl_work (the \"(partial F_3)/(partial x)\" part) is emphasized."],[114.069,"curl_work (the \"(partial F_3)/(partial x)\" part) is no longer emphasized."],[114.069,"curl_work (the \"-y - y\" part) is emphasized."],[116.089,"curl_work (the \"-2 y\" part) is emphasized."],[116.089,"curl_work (the \"-y - y\" part) is no longer emphasized."]]},{"start":117.6235,"say":"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.","live":null,"does":[[118.315,"curl_work is shown on the screen, written out."],[119.55699999999999,"curl_work (the \"-2 y\" part) is no longer emphasized."],[119.55699999999999,"curl_work (the \"(partial F_2)/(partial x)\" part) is emphasized."],[121.77499999999999,"curl_work (the \"(partial F_2)/(partial x)\" part) is no longer emphasized."],[121.77499999999999,"curl_work (the \"0\" part) is emphasized."],[123.725,"curl_work (the \"(partial F_1)/(partial y)\" part) is emphasized."],[123.725,"curl_work (the \"0\" part) is no longer emphasized."],[126.33699999999999,"curl_work (the \"(-z)\" part) is emphasized."],[126.33699999999999,"curl_work (the \"(partial F_1)/(partial y)\" part) is no longer emphasized."],[127.76499999999999,"curl_work (the \"(-z)\" part) is no longer emphasized."],[127.76499999999999,"curl_work (the \"= z\" part) is emphasized."]]},{"start":129.7355,"say":"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.","live":null,"does":[[130.084,"curl_result is shown on the screen, written out."],[131.88299999999998,"curl_result (the \"x\" part) is emphasized."],[132.243,"curl_result (the \"-2 y\" part) is emphasized."],[132.243,"curl_result (the \"x\" part) is no longer emphasized."],[133.21800000000002,"curl_result (the \"-2 y\" part) is no longer emphasized."],[133.21800000000002,"curl_result (the \"z\" part) is emphasized."],[134.844,"components (the \"4 y + 1\" part) is emphasized."],[144.5495,"components is hidden from the screen — left the board."],[144.5495,"curl_heading is hidden from the screen — left the board."],[144.5495,"curl_result is hidden from the screen — left the board."],[144.5495,"curl_work is hidden from the screen — left the board."]]},{"start":145.1495,"say":"Now the right-hand rule fixes the normal. From the side the normal points toward, the positive walk around the boundary must look counterclockwise.","live":["frame","rim","centre","walk","disc"],"does":[[145.1495,"normal_heading is shown on the screen, written out."],[145.82299999999998,"right_hand is shown on the screen, written out."]]},{"start":155.2695,"say":"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.","live":["frame","rim","centre","walk","disc","right_hand","normal_heading"],"does":[[156.08199999999997,"frame turns in its own slot."]]},{"start":165.96949999999998,"say":"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.","live":null,"does":[[167.89699999999996,"normal is shown on the screen, written out."],[169.55699999999996,"normal_result is shown on the screen, written out."]]},{"start":177.843,"say":"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.","live":["frame","rim","centre","walk","disc","right_hand","normal_result","normal_heading","normal"],"does":[[180.94299999999998,"walk is hidden from the screen."],[180.94299999999998,"reverse_walk_2 is shown on the screen, written out."],[183.83399999999997,"normal is redrawn as the numbers it depends on change."],[183.83399999999997,"normal_tip_y ticks to 6.35."],[185.02999999999997,"reverse_walk_2 is hidden from the screen."],[185.02999999999997,"walk is shown on the screen, written out."],[187.921,"normal is redrawn as the numbers it depends on change."],[187.921,"normal_tip_y ticks to 1.65."]]},{"start":189.5305,"say":"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.","live":null,"does":[[194.337,"normal is redrawn as the numbers it depends on change."],[194.337,"normal_x ticks to -1.15."],[195.266,"normal_z ticks to 0.65."],[198.43599999999998,"normal is redrawn as the numbers it depends on change."],[198.43599999999998,"normal_x ticks to 1.25."],[200.154,"normal is redrawn as the numbers it depends on change."],[200.154,"normal_z ticks to 1.1."],[201.05949999999999,"normal_heading is hidden from the screen — left the board."],[201.05949999999999,"normal_result is hidden from the screen — left the board."],[201.05949999999999,"right_hand is hidden from the screen — left the board."]]},{"start":201.65949999999998,"say":"Dot the curl x, minus two y, z with the normal zero, minus one, zero.","live":["frame","rim","centre","walk","disc","normal"],"does":[[201.65949999999998,"flux_heading is shown on the screen, written out."],[202.32099999999997,"flux_work is shown on the screen, written out."],[202.72799999999998,"flux_work (the \"x\" part) is emphasized."],[203.20399999999998,"flux_work (the \"-2 y\" part) is emphasized."],[203.20399999999998,"flux_work (the \"x\" part) is no longer emphasized."],[204.09799999999998,"flux_work (the \"-2 y\" part) is no longer emphasized."],[204.09799999999998,"flux_work (the \"z\" part) is emphasized."],[205.08399999999997,"flux_work is shown on the screen, written out."],[205.444,"flux_work (the \"z\" part) is no longer emphasized."],[205.444,"flux_work (the \"0\" part) is emphasized."],[205.897,"flux_work (the \"-1\" part) is emphasized."],[205.897,"flux_work (the \"0\" part) is no longer emphasized."],[206.652,"flux_work (the \"-1\" part) is no longer emphasized."],[206.652,"flux_work (the \"0#2\" part) is emphasized."]]},{"start":208.14,"say":"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.","live":["frame","rim","centre","walk","disc","normal","flux_heading"],"does":[[208.72599999999997,"flux_work is shown on the screen, written out."],[208.72599999999997,"flux_work (the \"0#2\" part) is no longer emphasized."],[208.72599999999997,"flux_work (the \"x dot 0\" part) is emphasized."],[211.21099999999998,"flux_work (the \"x dot 0\" part) is no longer emphasized."],[211.21099999999998,"flux_work (the \"z dot 0\" part) is emphasized."],[214.03199999999998,"flux_work (the \"x dot 0\" part) is slashed through — it cancels."],[214.03199999999998,"flux_work (the \"z dot 0\" part) is slashed through — it cancels."],[215.43699999999998,"flux_work (the \"(-2 y)(-1)\" part) is emphasized."],[215.43699999999998,"flux_work (the \"z dot 0\" part) is no longer emphasized."],[216.63299999999995,"flux_work is shown on the screen, written out."],[216.63299999999995,"flux_work (the \"(-2 y)(-1)\" part) is no longer emphasized."],[216.63299999999995,"flux_work (the \"2 y\" part) is emphasized."]]},{"start":220.629,"say":"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.","live":null,"does":[[220.629,"probe is shown on the screen, written out."],[222.85199999999995,"point_5 is shown on the screen, grown."],[223.85199999999995,"point_5 is hidden from the screen."],[224.62899999999996,"probe is redrawn as the numbers it depends on change."],[224.62899999999996,"probe_x ticks to 1.9."],[225.18599999999995,"probe_z ticks to 0.8."],[229.02899999999997,"flux_work (the \"2 y\" part) is no longer emphasized."],[229.02899999999997,"flux_work (the \"2(4) = 8\" part) is emphasized."],[229.53999999999996,"probe is redrawn as the numbers it depends on change."],[229.53999999999996,"probe_x ticks to -0.8."],[229.88799999999995,"probe_z ticks to 1.8."],[232.75549999999998,"probe is hidden from the screen."]]},{"start":233.35549999999998,"say":"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.","live":null,"does":[[233.35549999999998,"flux_work (the \"2(4) = 8\" part) is no longer emphasized."],[234.04099999999994,"flux_work is shown on the screen, written out."],[235.18999999999994,"flux_work (the \"8\" part) is emphasized."],[235.90999999999994,"flux_work is shown on the screen, written out."],[238.89399999999995,"flux_work (the \"8\" part) is no longer emphasized."],[238.89399999999995,"flux_work (the \"pi dot 3^2\" part) is emphasized."],[241.59899999999996,"flux_work (the \"72 pi\" part) is emphasized."],[241.59899999999996,"flux_work (the \"pi dot 3^2\" part) is no longer emphasized."]]},{"start":243.49349999999998,"say":"Therefore the requested circulation is seventy-two pi.","live":null,"does":[[245.71699999999996,"answer is shown on the screen, written out."],[247.07449999999997,"answer moves to a new place on the board."],[247.07449999999997,"flux_heading is hidden from the screen — left the board."],[247.07449999999997,"flux_work is hidden from the screen — left the board."],[247.07449999999997,"A box is drawn around answer."]]},{"start":247.6745,"say":"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.","live":["frame","rim","centre","walk","disc","normal","answer"],"does":[[247.6745,"checks_heading is shown on the screen, written out."],[251.80899999999997,"gradient_check is shown on the screen, written out."],[251.80899999999997,"gradient_check (the \"4 y + 1\" part) is emphasized."],[254.43199999999996,"gradient_check (the \"4 y + 1\" part) is no longer emphasized."],[254.43199999999996,"gradient_check (the \"nabla(2 y^2 + y)\" part) is emphasized."],[258.0079999999999,"curl_gradient is shown on the screen, written out."]]},{"start":259.746,"say":"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.","live":["frame","rim","centre","walk","disc","normal","answer","gradient_check","curl_gradient","checks_heading"],"does":[[259.746,"gradient_check (the \"nabla(2 y^2 + y)\" part) is no longer emphasized."],[261.929,"direct_check is shown on the screen, written out."],[263.42599999999993,"direct_check (the \"3 cos t\" part) is emphasized."],[264.645,"direct_check (the \"3 cos t\" part) is no longer emphasized."],[264.645,"direct_check (the \"4\" part) is emphasized."],[265.55099999999993,"direct_check (the \"3 sin t\" part) is emphasized."],[265.55099999999993,"direct_check (the \"4\" part) is no longer emphasized."],[267.16499999999996,"direct_check is shown on the screen, written out."],[269.18499999999995,"direct_check (the \"3 sin t\" part) is no longer emphasized."],[269.18499999999995,"direct_check (the \"36\" part) is emphasized."]]},{"start":271.4455,"say":"Integrating thirty-six from zero to two pi gives seventy-two pi again. The boundary calculation and the flux calculation agree.","live":null,"does":[[271.4455,"direct_check (the \"36\" part) is no longer emphasized."],[271.7939999999999,"direct_check is shown on the screen, written out."],[274.893,"direct_check (the \"72 pi\" part) is emphasized."],[279.42099999999994,"answer (the \"72 pi\" part) is emphasized."],[280.1175416666666,"answer is hidden from the screen — left the board."],[280.1175416666666,"checks_heading is hidden from the screen — left the board."],[280.1175416666666,"curl_gradient is hidden from the screen — left the board."],[280.1175416666666,"direct_check is hidden from the screen — left the board."],[280.1175416666666,"frame is hidden from the screen — left the board."],[280.1175416666666,"rim is hidden from the screen — frame left the board."],[280.1175416666666,"centre is hidden from the screen — frame left the board."],[280.1175416666666,"walk is hidden from the screen — frame left the board."],[280.1175416666666,"disc is hidden from the screen — frame left the board."],[280.1175416666666,"normal is hidden from the screen — frame left the board."],[280.1175416666666,"gradient_check is hidden from the screen — left the board."]]}]}]},"durationSeconds":281,"chapters":[{"title":"Stokes' Theorem in Practice","startSeconds":0,"narration":"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."}]}}