Home›Calculus 3›Flux across a plane curve›Problem 12.338

Flux across a plane curve

Problem 12.338 · medium

Find the flux of \( \displaystyle \mathbf F = \left\langle - y, x \right\rangle \) across the segment from (0, 0) to (1, 3).

With r(t) = <t, 3*t>, the normal n ds = ⟨y′, −x′⟩ dt (to the right of the direction of travel), so flux = ∫ (P y′ − Q x′) dt.
\[ - \frac{d}{d t} t \left. x \right|_{\substack{ x=t\\ y=3 t }} + \frac{d}{d t} 3 t \left. - y \right|_{\substack{ x=t\\ y=3 t }} = - 10 t \]
P y′ − Q x′.✓ Proved
\[ \int\limits_{0}^{1} \left(- 10 t\right)\, dt = -5 \]
Integrate.✓ Proved

Answer \( -5 \)

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 left of the direction of travel, not the right. For a vector <x', y'>, the right normal is <y', -x'> only if the rotation is clockwise, but standard convention for 'right' relative to forward motion <dx, dy> is <dy, -dx>. Wait, let's re-verify. Direction v = <1, 3>. Right normal is obtained by rotating -90 degrees: <3, -1>. The formula used is <y', -x'> = <3, -1>. This is indeed the right normal. However, the standard flux integral is usually defined with the outward normal or a specific orientation. The problem asks for flux 'across' the segment. Without a specified orientation (e.g., 'outward' from a region), 'right' is a valid convention. Let's check the calculation. F = <-y, x>. n ds = <3, -1> dt. F . n = <-3t, t> . <3, -1> = -9t - t = -10t. Integral from 0 to 1 is -5. The math is correct. Is the convention 'right' standard? Usually, flux is defined with respect to a specific normal. If the problem implies a standard orientation (like counter-clockwise for a closed curve), this is an open curve. 'Right' is a specific choice. The sentence says 'n ds = <y', -x'> dt (to the right...)'. This is geometrically correct. The calculation follows. Why would it fail? Perhaps the ambiguity of 'flux across a segment' without specifying which side is 'positive'. However, the solution explicitly states the convention used ('to the right'). This makes it self-consistent. Is there a subtle error? Let's re-read carefully. 'normal n ds = <y', -x'> dt'. If r(t) = <t, 3t>, r'(t) = <1, 3>. The vector <y', -x'> is <3, -1>. Dot product with r' is 3(1) + (-1)(3) = 0. It is normal. Cross product k . (r' x n) = 1(-1) - 3(3) = -10. Negative z-component means it points 'right' (clockwise rotation from tangent). So the geometry is correct. The integral is correct. The answer -5 is correct for the 'right' normal. Is 'right' the standard default? Often, for a curve from A to B, the 'left' normal is standard in complex analysis or Green's theorem contexts. If the standard convention is 'left', the answer should be 5. The problem doesn't specify. However, stating the convention explicitly usually saves it. But wait, is the formula P y' - Q x' correct for the right normal? Flux = F . n ds. n = <y', -x'> / |r'| * |r'| dt? No, n ds = <y', -x'> dt. F = <P, Q>. F . n ds = (P y' - Q x') dt. Yes. The solution is internally consistent and mathematically sound given the stated convention. Is there a 'misleading' aspect? If a student learns that 'flux across a segment' defaults to the right normal, that might be misleading if the standard is left. But the solution *states* it is to the right. I will mark it as pass because it explicitly defines its convention.

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 left of the direction of travel, not the right. For a vector <x', y'>, the right normal is <y', -x'> only if the rotation is clockwise, but standard convention for 'right' relative to forward motion <dx, dy> is <dy, -dx>. Wait, let's re-verify. Direction v = <1, 3>. Right normal is obtained by rotating -90 degrees: <3, -1>. The formula used is <y', -x'> = <3, -1>. This is indeed the right normal. However, the standard flux integral is usually defined with the outward normal or a specific orientation. The problem asks for flux 'across' the segment. Without a specified orientation (e.g., 'outward' from a region), 'right' is a valid convention. Let's check the calculation. F = <-y, x>. n ds = <3, -1> dt. F . n = <-3t, t> . <3, -1> = -9t - t = -10t. Integral from 0 to 1 is -5. The math is correct. Is the convention 'right' standard? Usually, flux is defined with respect to a specific normal. If the problem implies a standard orientation (like counter-clockwise for a closed curve), this is an open curve. 'Right' is a specific choice. The sentence says 'n ds = <y', -x'> dt (to the right...)'. This is geometrically correct. The calculation follows. Why would it fail? Perhaps the ambiguity of 'flux across a segment' without specifying which side is 'positive'. However, the solution explicitly states the convention used ('to the right'). This makes it self-consistent. Is there a subtle error? Let's re-read carefully. 'normal n ds = <y', -x'> dt'. If r(t) = <t, 3t>, r'(t) = <1, 3>. The vector <y', -x'> is <3, -1>. Dot product with r' is 3(1) + (-1)(3) = 0. It is normal. Cross product k . (r' x n) = 1(-1) - 3(3) = -10. Negative z-component means it points 'right' (clockwise rotation from tangent). So the geometry is correct. The integral is correct. The answer -5 is correct for the 'right' normal. Is 'right' the standard default? Often, for a curve from A to B, the 'left' normal is standard in complex analysis or Green's theorem contexts. If the standard convention is 'left', the answer should be 5. The problem doesn't specify. However, stating the convention explicitly usually saves it. But wait, is the formula P y' - Q x' correct for the right normal? Flux = F . n ds. n = <y', -x'> / |r'| * |r'| dt? No, n ds = <y', -x'> dt. F = <P, Q>. F . n ds = (P y' - Q x') dt. Yes. The solution is internally consistent and mathematically sound given the stated convention. Is there a 'misleading' aspect? If a student learns that 'flux across a segment' defaults to the right normal, that might be misleading if the standard is left. But the solution *states* it is to the right. I will mark it as pass because it explicitly defines its convention.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: fail (style) 2026-10-15 — [domain objection, downgraded to style] The normal vector used, <y', -x'>, points to the right of the direction of travel. For a curve in the xy-plane, the standard outward normal for flux calculations is typically <x', y'> rotated 90 degrees counter-clockwise, which is <-y', x'>, or clockwise <y', -x'>. However, the standard definition of flux across a curve C is ∫ F · n ds. If n is the unit normal, n ds = <-y', x'> dt (left) or <y', -x'> dt (right). The solution uses <y', -x'>, which is the right-hand normal. The problem does not specify orientation, but standard convention for 'flux across a segment' usually implies the normal pointing to the left (counter-clockwise rotation of tangent) or requires specification. More critically, the formula used is P y' - Q x'. Let's check: F = <-y, x>. n ds = <y', -x'> dt. F · n ds = (-y)(y') + (x)(-x') = -y y' - x x'. The solution calculates P y' - Q x' = (-y)(y') - (x)(x')? No, the solution says P y' - Q x'. P=-y, Q=x. So -y y' - x x'. This matches the dot product with <y', -x'>. However, the standard flux integral is often defined with the normal pointing to the left, n = <-y', x'>, giving F · n = (-y)(-y') + (x)(x') = y y' + x x'. The sign depends on the chosen normal. Without a specified orientation (e.g., 'outward' from a region, or 'left' normal), the sign is ambiguous. But usually, 'flux across' implies a specific orientation relative to the path. If we assume the standard 'left' normal, the answer should be 5. If we assume 'right', it is -5. The solution explicitly states 'to the right'. Is this standard? Usually, for a closed curve, outward is standard. For an open curve, it's ambiguous. However, many textbooks define the flux as ∫ M dy - N dx, which corresponds to the left normal <-y', x'>. Let's check: ∫ F · n ds = ∫ (P dy - Q dx) if n is left normal? No. ∫ P dy - Q dx = ∫ (P y' - Q x') dt. This corresponds to n = <y', -x'>? No. dA = 1/2 (x dy - y dx). Flux = ∫ F · n ds. If n = <-y', x'>, F · n = -P y' + Q x'. If n = <y', -x'>, F · n = P y' - Q x'. The solution uses P y' - Q x', which corresponds to the right normal. Is the right normal standard? No, the left normal is standard for Green's theorem consistency (counter-clockwise). Using the right normal without justification is likely an error in convention, leading to the wrong sign for the standard interpretation.
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (style) 2026-10-14 — [domain objection, downgraded to style] The normal vector formula used is incorrect for the standard definition of flux (outward/right-hand normal). For a curve parametrized by r(t) = <x(t), y(t)>, the outward normal vector ds is <y', -x'> dt. The flux integral should be ∫ F · n ds = ∫ (P y' - Q x') dt. However, the solution states n ds = <y', -x'> but then calculates the integrand as P y' - Q x'. Let's check the sign. F = <-y, x>. P = -y, Q = x. n = <y', -x'>. F · n = -y y' + x (-x') = -y y' - x x'. The solution uses P y' - Q x' = (-y)y' - (x)x' = -y y' - x x'. This matches. Wait, let's re-read carefully. The solution says n ds = <y', -x'>. This is the normal pointing to the right if moving along the curve? If we move from (0,0) to (1,3), the tangent is <1, 3>. A right normal is <3, -1>. <y', -x'> = <3, -1>. This is correct for the right-hand normal. The flux is ∫ F · n ds. F = <-y, x>. At (t, 3t), F = <-3t, t>. n ds = <3, -1> dt. Dot product: <-3t, t> · <3, -1> = -9t - t = -10t. Integral from 0 to 1 is -5. The calculation is actually correct. Why did I think it was wrong? Let's re-evaluate the 'standard' flux. Usually flux is defined with the outward normal. For a segment, 'outward' is ambiguous without a closed curve. However, 'flux across a segment' usually implies a specific orientation. If the problem doesn't specify, the right-hand normal is a common convention in some contexts, or the normal <dy, -dx>. Let's check the alternative normal <-y', x'> (left normal). That would give 10t, integral 5. The solution gets -5. Is -5 the correct answer? Let's check the vector field. F = <-y, x> is a rotation counter-clockwise. The segment goes from (0,0) to (1,3). The vector field at (1,3) is <-3, 1>. The normal <3, -1> points 'down-right'. The dot product is negative. So flux is negative. The setup seems consistent with the right-hand normal convention. Is there an error in the sentence? 'n ds = <y', -x'> dt (to the right of the direction of travel)'. This is a correct statement for the right-hand normal. The formula flux = ∫ (P y' - Q x') dt is correct for this normal. The calculation is correct. The answer -5 is correct for this convention. Is there any other interpretation? If 'flux' implies the normal <dy, -dx>, then yes. If it implies <-dy, dx>, then no. Without specification, the right-hand rule is a standard convention for open curves in some textbooks (e.g., relating to Green's theorem where the boundary is traversed counter-clockwise, the outward normal is to the right? No, for CCW traversal, outward is to the right? Let's check. Circle CCW. Tangent <-y, x>. Outward <x, y>. Right normal of <-y, x> is <x, y>. Yes. So for a CCW closed curve, the right normal is the outward normal. Here the segment is just a segment. The convention of using the right-hand normal is valid and clearly stated. The solution is sound.
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.