Home›Calculus 3›Flux across a plane curve›Problem 12.328

Flux across a plane curve

Problem 12.328 · medium

Find the flux of \( \displaystyle \mathbf F = \left\langle x^{2}, y \right\rangle \) across the segment from (0, 0) to (1, 2).

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

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

Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify

The full receipt

Line	Status	Checked by	Detail
1	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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: pass

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-17
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 first quadrant going from (0,0) to (1,2), the 'right' side points into the fourth quadrant (away from the region typically bounded by such a segment if it were part of a closed loop, or simply opposite to the standard outward normal if the region is above the curve). More critically, the standard definition of flux across a curve usually implies a specific orientation (often outward from a region or consistent with the left-hand rule for counter-clockwise traversal). The problem does not specify the orientation of the normal. However, the standard convention for 'flux across a curve' without a specified region often defaults to the normal pointing to the left of the direction of travel (the standard normal < -y', x' >) or requires specifying the side. If we assume the standard 'outward' normal for a region below the curve, the normal should point down/right? No, usually for a line segment, one must specify the side. Let's look at the standard formula. Flux = int F . n ds. If n is the unit normal. The vector <y', -x'> is a normal. Is it the correct one? The problem doesn't specify 'outward' or 'left/right'. However, in many calculus contexts, if no region is specified, the 'flux across' is ambiguous. But let's look at the sign. If we use the standard normal pointing to the left of the path (standard for Green's theorem context), n = <-y', x'>. Then flux = int (-P y' + Q x') dt. The solution uses <y', -x'> which is the right normal. Without a specified region, 'right' is arbitrary. But often, 'flux across' implies a specific orientation relative to a region. If the region is the triangle with vertices (0,0), (1,0), (1,2), the outward normal on the hypotenuse points away from the triangle (down/right). The vector <2, -1> is a normal. <y', -x'> = <2, -1>. This points into the fourth quadrant. The triangle is in the first quadrant. So this normal points OUT of the triangle. If the problem implies the region bounded by the segment and the axes, this is the correct outward normal. However, the problem just says 'across the segment'. It does not define a region. In the absence of a region, the sign is ambiguous. But wait, is the calculation correct for the chosen normal? Yes. Is the choice of normal justified? The sentence says 'to the right of the direction of travel'. This is a valid choice of normal, but is it the *standard* one? Often, the standard normal is to the left. If the answer key expects the left normal, the sign would be flipped. Given the ambiguity, stating the normal is 'to the right' is a specific choice. Is it wrong? Not necessarily. But let's check the result. If we use the left normal <-2, 1>, the integrand is -2t^2 + 2t. Integral is -2/3 + 1 = 1/3. The solution gives -1/3. Which is correct? Usually, 'flux across' without qualification is ill-posed. However, many textbooks define the flux integral using the normal <y', -x'> (right) or <-y', x'> (left). There is no universal default. BUT, looking at the provided answer -1/3, it matches the 'right' normal. If the problem came from a context where 'outward' from the region under the curve is implied, the normal <2, -1> is indeed outward. So the setup is likely correct in intent. Why would it fail? Let's re-read carefully. 'n ds = <y', -x'> dt'. This is correct for the right normal. 'Flux = int (P y' - Q x') dt'. This is correct for that normal. The algebra is correct. The bounds are correct. The only potential issue is if the problem implicitly assumes the 'left' normal (standard mathematical orientation for boundaries of regions in the plane, counter-clockwise). If the region is the triangle, the boundary is traversed counter-clockwise: (0,0)->(1,0)->(1,2)->(0,0). The segment is (1,2) to (0,0) in that loop? No, the problem specifies (0,0) to (1,2). If this is part of a counter-clockwise loop enclosing the region below, the traversal should be (1,2) to (0,0). The problem specifies (0,0) to (1,2). This is clockwise relative to the region below. The outward normal for a clockwise traversal is to the left? No. Standard Green's theorem: int F . n ds = int div F dA. The normal is outward. For a counter-clockwise boundary, the outward normal is to the right of the tangent? No, for CCW, the outward normal is to the right of the tangent vector? Let's check. Circle CCW. Tangent < -y, x >. Outward normal < x, y >. Dot product? No. Tangent T = <-y, x>. Normal N = <x, y>. T rotated 90 deg clockwise is <x, y>. So for CCW traversal, the outward normal is to the RIGHT of the direction of travel. The problem specifies travel from (0,0) to (1,2). If this segment is part of the boundary of the region below it (triangle), and we want outward flux, we need to know the orientation of the boundary. If the region is the triangle, the boundary is usually CCW. The segment from (1,2) to (0,0) is part of the CCW boundary. The problem specifies (0,0) to (1,2), which is the REVERSE direction. So we are traversing the boundary clockwise. For a clockwise traversal, the outward normal is to the LEFT of the direction of travel. The solution uses the RIGHT normal. Therefore, the solution calculates the flux with the inward normal (or the normal pointing into the region below). If the question implies 'outward flux from the region bounded by the segment and the axes', the correct normal is the LEFT normal (since we are traversing clockwise relative to the region's interior if we consider the standard CCW orientation, wait. Let's stick to vectors. Region: Triangle. Boundary: (0,0)->(1,0)->(1,2)->(0,0). This is CCW. The segment in this loop is (1,2)->(0,0). The problem asks for (0,0)->(1,2). This is opposite to the CCW boundary orientation. The outward normal for the CCW boundary segment (1,2)->(0,0) is to the right of that vector. The vector is <-1, -2>. Right normal is <-2, 1>. This points into the second quadrant? No. <-1, -2> is down-left. Right normal is <-2, 1> (up-left). This points OUT of the triangle. So for the standard CCW boundary, the outward normal is <-2, 1>. The problem traverses in the opposite direction: <1, 2>. The outward normal for the region is still <-2, 1> (geometric outward). The solution uses <2, -1>. This is the INWARD normal. So the sign is wrong for outward flux. If the problem doesn't specify 'outward', it's ambiguous. But 'Find the flux... across the segment' usually implies a standard orientation. In many contexts (e.g. line integrals of vector fields), if no region is given, the 'flux' is not well-defined without a specified normal direction. However, if forced to choose, the 'left' normal is often the default 'positive' orientation in complex analysis (multiply by i). Left normal of <1, 2> is <-2, 1>. This gives +1/3. The solution gives -1/3. It is highly likely the intended answer is based on the standard 'left' normal or 'outward' from the region below (which requires careful orientation handling). Given the ambiguity, marking it 'error' might be harsh if the text allows 'right'. But wait, look at the sentence: 'n ds = <y', -x'> dt (to the right of the direction of travel)'. This explicitly chooses the right normal. Is this a valid interpretation? Yes. Is it the *standard* one? Often, no. But is it *wrong*? Only if the problem implies a specific normal. Without further context, 'flux across a curve' is ambiguous. However, in many calculus courses, the flux integral is defined as int F . n ds where n is the unit normal. If not specified, one might assume the normal pointing to the left (standard orientation). If the solution arbitrarily picks 'right' and gets a negative sign, it might be technically 'correct' for that choice, but misleading if the standard convention is left. Let's look for a clearer error. Is the formula P y' - Q x' correct for <y', -x'>? F = <P, Q>. n ds = <y', -x'> dt. F . n ds = (P y' - Q x') dt. Yes. Is the integral correct? Yes. Is the parameterization correct? Yes. The only issue is the choice of normal. If the problem is from a standard source, 'flux across' usually defaults to the normal pointing to the left of the path (or outward from the region to the left). If the region is 'under' the curve, the 'left' of the path (0,0)->(1,2) is 'above' the curve. The 'right' is 'below'. If the region is the triangle, the outward normal is 'below' (right) IF the traversal is (1,2)->(0,0). But the traversal is (0,0)->(1,2). So 'right' is 'below' (into the triangle? No, the triangle is below the line? No, the triangle with vertices (0,0), (1,0), (1,2) is below the line y=2x? No, the line is y=2x. The point (1,0) is below the line. So the region is below the line. The normal pointing 'down' (into the region) is the inward normal. The normal pointing 'up' (away from the region) is the outward normal. The vector <2, -1> points down/right. This is INTO the region (if the region is the triangle below). Wait. Line y=2x. Normal <2, -1>. At (0.5, 1), vector <2, -1> points to (2.5, 0). This is below the line. The triangle is below the line. So <2, -1> points INTO the triangle. So the solution calculates the INWARD flux. Standard convention is OUTWARD flux. Therefore, the sign is likely wrong. The solution fails to justify why the 'right' normal is the correct one for 'flux across', and likely gets the sign wrong relative to standard conventions (outward/left).
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: fail (error) 2026-10-14 — 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' or positive flux convention usually implies the normal pointing to the left (<-y', x'>) or requires explicit specification. More critically, the formula P y' - Q x' corresponds to the normal <-y', x'> (left normal). The text claims the normal is <y', -x'> (right normal) but uses the formula for the left normal, resulting in a sign error relative to the stated normal direction. If the problem implies the standard left-normal flux, the setup description is contradictory. If it implies right-normal, the formula is wrong. Given the standard convention for 2D flux integrals often defaults to the left normal (or outward for closed curves), the mismatch between the described normal vector and the integral formula is a significant conceptual error.
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.