Home›Calculus 3›Flux across a plane curve›Problem 12.326

Flux across a plane curve

Problem 12.326 · medium

Find the flux of \( \displaystyle \mathbf F = \left\langle x, y \right\rangle \) across the parabola y = x² from (0, 0) to (1, 1).

With r(t) = <t, t**2>, the normal n ds = ⟨y′, −x′⟩ dt (to the right of the direction of travel), so flux = ∫ (P y′ − Q x′) dt.
\[ - \frac{d}{d t} t \left. y \right|_{\substack{ x=t\\ y=t^{2} }} + \frac{d}{d t} t^{2} \left. x \right|_{\substack{ x=t\\ y=t^{2} }} = t^{2} \]
P y′ − Q x′.✓ Proved
\[ \int\limits_{0}^{1} t^{2}\, dt = \frac{1}{3} \]
Integrate.✓ Proved

Answer \( \frac{1}{3} \)

Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Not checked	—	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
answer, a second way	✓ Checked independently	sympy 1.14.0 + mpmath 1.3.0	F · n summed over a 20,000-segment polyline

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The normal vector used, <y', -x'>, points to the right of the direction of travel. For the curve y=x^2 from (0,0) to (1,1), 'right' corresponds to the inward/downward normal, whereas standard flux problems typically imply the outward/upward normal or require specifying the orientation. More critically, the formula P y' - Q x' corresponds to the normal <y', -x'>. If the standard outward normal <y', -x'> is intended for a closed curve, it's ambiguous here. However, the primary error is likely the sign convention: usually, flux across a curve parameterized by t is defined with a specific normal. If the problem implies the standard 'outward' or 'upward' normal for the region below the curve, the normal should be <-y', x'> or similar, leading to a sign change. Let's look closer. The vector <y', -x'> for r=<t, t^2> is <2t, -1>. At t=0.5, this is <1, -1>, pointing down-right. The 'standard' normal for flux often assumes the normal pointing to the left of the tangent vector for counter-clockwise orientation, or simply specifies the normal. Without a specified normal, 'flux across' is ambiguous. However, in many calculus contexts, if not specified, one might assume the normal pointing away from the origin or 'up'. The solution arbitrarily chooses 'right'. If the standard convention for 'flux across a curve' in 2D without specified orientation often defaults to the normal obtained by rotating the tangent 90 degrees counter-clockwise (left), which would be <-y', x'> = <-2t, 1>. The solution uses <y', -x'> (right). This is a sign error relative to the standard CCW convention, or at least an unjustified choice that contradicts the typical 'outward' normal for the region bounded by the parabola and axes if that were the context. Given the ambiguity, stating 'to the right' is a specific choice, but is it the *correct* one for the problem? Usually, problems imply the normal pointing into the first quadrant or 'up'. The vector <2t, -1> points into the fourth quadrant (mostly). This is likely the wrong normal direction for the intended physical interpretation (e.g., flow out of the region under the curve).

Every verdict on record (6)

qwen3.6:27b-mlx: fail (style) 2026-10-17 — [domain objection, downgraded to style] The normal vector used, <y', -x'>, points to the right of the direction of travel. For the curve y=x^2 from (0,0) to (1,1), 'right' corresponds to the inward/downward normal, whereas standard flux problems typically imply the outward/upward normal or require specifying the orientation. More critically, the formula P y' - Q x' corresponds to the normal <y', -x'>. If the standard outward normal <y', -x'> is intended for a closed curve, it's ambiguous here. However, the primary error is likely the sign convention: usually, flux across a curve parameterized by t is defined with a specific normal. If the problem implies the standard 'outward' or 'upward' normal for the region below the curve, the normal should be <-y', x'> or similar, leading to a sign change. Let's look closer. The vector <y', -x'> for r=<t, t^2> is <2t, -1>. At t=0.5, this is <1, -1>, pointing down-right. The 'standard' normal for flux often assumes the normal pointing to the left of the tangent vector for counter-clockwise orientation, or simply specifies the normal. Without a specified normal, 'flux across' is ambiguous. However, in many calculus contexts, if not specified, one might assume the normal pointing away from the origin or 'up'. The solution arbitrarily chooses 'right'. If the standard convention for 'flux across a curve' in 2D without specified orientation often defaults to the normal obtained by rotating the tangent 90 degrees counter-clockwise (left), which would be <-y', x'> = <-2t, 1>. The solution uses <y', -x'> (right). This is a sign error relative to the standard CCW convention, or at least an unjustified choice that contradicts the typical 'outward' normal for the region bounded by the parabola and axes if that were the context. Given the ambiguity, stating 'to the right' is a specific choice, but is it the *correct* one for the problem? Usually, problems imply the normal pointing into the first quadrant or 'up'. The vector <2t, -1> points into the fourth quadrant (mostly). This is likely the wrong normal direction for the intended physical interpretation (e.g., flow out of the region under the curve).
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: fail (style) 2026-10-15 — [domain objection, downgraded to style] The normal vector formula used, <y', -x'>, points to the right of the direction of travel. For the curve y=x^2 from (0,0) to (1,1), 'right' corresponds to the inward/downward normal, whereas standard flux problems typically require the outward/upward normal (or specify orientation). More critically, the standard flux integral formula for a vector field <P, Q> across a curve is ∫(P dy - Q dx) or ∫(P n_x + Q n_y) ds. The solution uses P y' - Q x', which corresponds to the normal <y', -x'>. However, for the standard outward normal (pointing away from the x-axis for this parabola in the first quadrant), the normal should be <-y', x'> or <1, -2t> normalized? No, for y=x^2, the tangent is <1, 2t>. The outward normal (pointing generally down/right? No, usually 'outward' from the region bounded by the curve and axes is ambiguous without a closed loop. But typically, for an open curve, one must specify the side. The formula P y' - Q x' gives the flux across the normal <y', -x'>. Let's check the sign. Tangent T = <1, 2t>. Normal N_right = <2t, -1>. Dot product F.N = <t, t^2> . <2t, -1> = 2t^2 - t^2 = t^2. Integral is 1/3. If the intended normal was the other side <-2t, 1>, the flux would be -1/3. Without specification, 'flux across' is ambiguous, but usually implies a specific orientation or the problem implies a standard convention. However, the bigger issue is often the definition of 'flux across a curve'. If it means the flow across the curve from one side to the other, the sign matters. The solution assumes the 'right' normal. Is this standard? Often, for a graph y=f(x), the upward normal is <-f', 1>. The solution uses <y', -x'> which is <2t, -1>, pointing down/right. This is the 'downward' normal. If the problem implies the standard orientation (counter-clockwise for closed, or upward for graphs), this is likely the wrong sign. Given the answer is 1/3, and the integral of t^2 is 1/3, the calculation is consistent with the chosen normal. But is the chosen normal the correct one for 'flux across the parabola'? Usually, if not specified, one might assume the normal pointing away from the origin or upward. The normal <2t, -1> points into the fourth quadrant relative to the tangent. At t=1, normal is <2, -1>, pointing down and right. The vector field is <1, 1>. The dot product is positive. If we used the upward normal <-2, 1>, the dot product is -1, integral -1/3. Most textbooks define flux across a curve y=f(x) from a to b using the normal < -f'(x), 1 > (upward) or < f'(x), -1 > (downward). There is no universal default. However, the sentence claims 'n ds = <y', -x'> dt (to the right)'. This is a correct statement about the geometry of that specific normal. The error is likely that the problem implies a specific orientation (e.g., outward from the region bounded by y=x^2, y=1, x=0?) or simply that the sign is arbitrary without specification. But wait, looking at similar problems online, 'flux across' often defaults to the normal pointing in the positive y-direction or outward from a region. If this is an open curve, the problem is ill-posed without specifying the normal direction. However, if we assume the standard convention for line integrals of the second kind or flux, often the 'positive' normal is considered. Let's look for a clearer error. The formula for flux is ∫ F · n ds. If n = <y', -x'>/|r'|, then n ds = <y', -x'> dt. F = <x, y>. F · n ds = (x y' - y x') dt. The solution calculates P y' - Q x'. Here P=x, Q=y. So x y' - y x'. This matches. The result is 1/3. Is there a sign error in the standard convention? Often, for a curve traversed from left to right, the 'positive' normal is upward <-y', x'>. The solution uses the downward/right normal. If the standard answer is -1/3, then this is an error. If the standard answer is 1/3, it's fine. Without the problem statement specifying 'outward' or 'upward', it's ambiguous. BUT, look at the vector field F=<x,y>. It points radially outward. The parabola y=x^2. The normal <2t, -1> points somewhat 'inward' towards the x-axis? No, at (1,1), <2,-1> points away from the y-axis and down. The vector <1,1> points up and right. They are somewhat opposed in the y-component. Let's check the other normal <-2t, 1>. Dot product <-2t, 1> · <t, t^2> = -2t^2 + t^2 = -t^2. Integral -1/3. Which is the 'standard' flux? Usually, flux is defined with respect to a specified orientation. If none is specified, it's a bad problem. However, in many contexts (like Green's theorem applications), the normal is outward. For an open curve, this is undefined. Let's assume the question implies the normal pointing away from the 'inside' of the parabola (the region y > x^2)? That would be the normal pointing towards the x-axis? No, the region y > x^2 is 'inside' the cup. The normal pointing into the cup is <-2t, 1> (upward/leftish). The normal pointing out of the cup is <2t, -1> (downward/rightish). If 'flux across' means flow from inside to outside, it's 1/3. If it means outside to inside, it's -1/3. Or if it refers to the region bounded by the curve and the axes, the outward normal on the curve part would point away from the region (down/right), so 1/3. This seems plausible. Is there a more definitive error? The sentence says 'n ds = <y', -x'> dt'. This is correct for the right-hand normal. The calculation is correct for that normal. The ambiguity is the main issue, but is it an 'error'? In a strict adversarial review, if the problem doesn't specify, and the solution picks one and gets a positive answer, it might be 'pass' if that's a common convention. But wait, look at the vector field again. F = <x, y>. Divergence is 2. If we close the curve with x=0 and y=1, we can use Green's theorem. Region D: 0<=x<=1, x^2<=y<=1. Boundary: C1 (parabola), C2 (top y=1), C3 (left x=0). Outward normal on C1 (parabola) points away from D, i.e., down/right <2t, -1>. Flux_out = ∬ div F dA = ∬ 2 dA = 2 * Area. Area = ∫(1-x^2)dx = 1 - 1/3 = 2/3. Total Flux = 4/3. Flux on C2 (y=1, x:1->0? No, counter-clockwise boundary): C2 is y=1, x from 1 to 0? Standard CCW: (0,0)->(1,1) along parabola? No, (0,0) to (1,1) is not a closed loop. Let's close it: (0,0) to (1,1) via parabola, then (1,1) to (0,1) via y=1, then (0,1) to (0,0) via x=0. Outward normal on parabola (part of boundary) points away from region (down/right). This matches the solution's normal. So if the problem implies the region bounded by the parabola and the axes, the outward flux is 1/3. This is a very standard interpretation. So the solution is likely correct in its interpretation. Is there a subtle error? 'n ds = <y', -x'> dt'. For r(t) = <t, t^2>, r'(t) = <1, 2t>. The vector <y', -x'> is <2t, -1>. This is perpendicular to <1, 2t>. It points to the right. This is correct. The integral is correct. The answer is correct for that normal. I will mark it pass unless there is a convention violation. Actually, many sources define the normal for y=f(x) as <-f', 1> (upward). If the problem came from a context where 'flux' implies the upward normal, the answer should be -1/3. However, 'flux across' is ambiguous. Let's look for a clearer error. Did the user make a sign error in the formula? P y' - Q x'. F=<P,Q>. n=<y', -x'>. F.n = P y' - Q x'. Correct. I will give it a pass, noting the ambiguity is resolved by the likely 'outward from region' interpretation.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The formula for flux is ∫(P dy - Q dx) = ∫(P y' - Q x') dt. The solution sets up this formula correctly in text, but the calculation in line 2 computes -x'*y + y'*x (or similar mix-up) resulting in t^2, whereas P y' - Q x' for F=<x,y> is x(2t) - y(1) = t(2t) - t^2 = t^2. Wait, let's re-evaluate. P=x, Q=y. x=t, y=t^2. x'=1, y'=2t. P y' - Q x' = t(2t) - t^2(1) = 2t^2 - t^2 = t^2. The calculation is actually correct. However, the text says "normal n ds = <y', -x'> dt (to the right)". For a curve traversed from (0,0) to (1,1), the tangent is <1, 2t>. The normal <y', -x'> is <2t, -1>. Dot product with tangent is 2t - 2t = 0. Is it to the right? At t=0.5, tangent is <1, 1>. Right normal is <1, -1>. <2t, -1> is <1, -1>. Yes, it is the right normal. The flux integral is ∫ F · n ds. F = <t, t^2>. n ds = <2t, -1> dt. Dot product: 2t^2 - t^2 = t^2. Integral 0 to 1 is 1/3. The solution is actually correct. Why did I think it was wrong? Let me re-read carefully. "normal n ds = <y', -x'> dt". This is the outward normal for a closed curve? No, just a normal. The standard flux integral across a curve C is ∫_C F · n ds. If n is the unit normal, ds is arc length. Here n ds = <dy, -dx> or <-dy, dx>. The vector <y', -x'> dt corresponds to <dy, -dx>. This is the normal pointing to the right of the direction of travel. The problem asks for flux across the parabola. Usually, "flux across a curve" implies a specific orientation. If not specified, it's ambiguous. However, the solution explicitly chooses the "right" normal. Is this the standard convention? Often, for open curves, the normal is not uniquely defined without context (e.g., boundary of a region). But if we assume the standard 2D flux definition ∫ P dy - Q dx, this corresponds to the normal <dy, -dx> which is indeed the right-hand normal. The calculation is correct. The answer 1/3 is correct. So why fail? Maybe the prompt implies a specific normal? "Flux across the parabola" is ambiguous. But the solution states its choice. Let's look for a subtle error. Line 1: "normal n ds = <y', -x'> dt". This vector has magnitude ds? |<y', -x'>| = sqrt(y'^2 + x'^2) = ds/dt. So n ds = <y', -x'> dt is correct. The text says "to the right". This is a valid choice. The calculation is correct. The answer is correct. Is there a style issue? "Subs(y, (x, y), (t, t**2))" is weird notation but the result is correct. I will pass it.
gpt-oss:20b: pass 2026-10-14

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/flux_plane, checked 2026-10-17 with SymPy 1.14.0.