Flux of a vector field across a plane curve, ∫_C F · n ds = ∫ P dy − Q dx. 35 problems with worked solutions; in 35 of them every
equation is proved by a computer algebra system.
4 worked problems
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
Find the flux of \( \displaystyle \mathbf F = \left\langle x, y \right\rangle \) across the segment from (0, 0) to (1, 1).
With r(t) = <t, t>, the normal n ds = ⟨y′, −x′⟩ dt (to the right of the direction of travel), so flux = ∫ (P y′ − Q x′) dt.
Reviewed
\[ \frac{d}{d t} t \left. x \right|_{\substack{ x=t\\ y=t }} - \frac{d}{d t} t \left. y \right|_{\substack{ x=t\\ y=t }} = 0 \]
P y′ − Q x′.✓ Proved
\[ \int\limits_{0}^{1} 0\, dt = 0 \]
Integrate.✓ Proved
Answer \( 0 \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the flux of \( \displaystyle \mathbf F = \left\langle x + y, x - y \right\rangle \) 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.
Reviewed
\[ - \frac{d}{d t} t \left. x - y \right|_{\substack{ x=t\\ y=3 t }} + \frac{d}{d t} 3 t \left. x + y \right|_{\substack{ x=t\\ y=3 t }} = 14 t \]
P y′ − Q x′.✓ Proved
\[ \int\limits_{0}^{1} 14 t\, dt = 7 \]
Integrate.✓ Proved
Answer \( 7 \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find the flux of \( \displaystyle \mathbf F = \left\langle x y, y^{2} \right\rangle \) 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.
Reviewed
\[ - \frac{d}{d t} t \left. y^{2} \right|_{\substack{ x=t\\ y=3 t }} + \frac{d}{d t} 3 t \left. x y \right|_{\substack{ x=t\\ y=3 t }} = 0 \]
P y′ − Q x′.✓ Proved
\[ \int\limits_{0}^{1} 0\, dt = 0 \]
Integrate.✓ Proved
Answer \( 0 \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify