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Line integrals of scalar functions practice problems

Line integrals of a scalar function with respect to arc length, ∫_C f ds. 36 problems with worked solutions; in 36 of them every equation is proved by a computer algebra system.

20 worked problems

Try each one before opening the solution. Every line is labeled with what checked it. How we verify

Problem 12.310

medium
Evaluate \( \displaystyle \int_C x y + z \, ds \), where \( \displaystyle C \) is one turn of the helix r(t) = ⟨cos t, sin t, t⟩.
  1. Parametrize: r(t) = <cos(t), sin(t), t>, 0 ≤ t ≤ 2*pi; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} t\right)^{2} + \left(\frac{d}{d t} \sin{\left(t \right)}\right)^{2} + \left(\frac{d}{d t} \cos{\left(t \right)}\right)^{2}} = \sqrt{2} \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{2 \pi} \sqrt{2} \left. x y + z \right|_{\substack{ x=\cos{\left(t \right)}\\ y=\sin{\left(t \right)}\\ z=t }}\, dt = 2 \sqrt{2} \pi^{2} \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( 2 \sqrt{2} \pi^{2} \)

✓ 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

Problem 12.311

medium
Evaluate \( \displaystyle \int_C x + y + z \, ds \), where \( \displaystyle C \) is one turn of the helix r(t) = ⟨cos t, sin t, t⟩.
  1. Parametrize: r(t) = <cos(t), sin(t), t>, 0 ≤ t ≤ 2*pi; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} t\right)^{2} + \left(\frac{d}{d t} \sin{\left(t \right)}\right)^{2} + \left(\frac{d}{d t} \cos{\left(t \right)}\right)^{2}} = \sqrt{2} \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{2 \pi} \sqrt{2} \left. x + y + z \right|_{\substack{ x=\cos{\left(t \right)}\\ y=\sin{\left(t \right)}\\ z=t }}\, dt = 2 \sqrt{2} \pi^{2} \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( 2 \sqrt{2} \pi^{2} \)

✓ 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

Problem 12.312

medium
Evaluate \( \displaystyle \int_C x^{2} + y^{2} + z \, ds \), where \( \displaystyle C \) is the segment from the origin to (1, 2, 2).
  1. Parametrize: r(t) = <t, 2*t, 2*t>, 0 ≤ t ≤ 1; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} t\right)^{2} + 2 \left(\frac{d}{d t} 2 t\right)^{2}} = 3 \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{1} \left. x^{2} + y^{2} + z \right|_{\substack{ x=t\\ y=2 t\\ z=2 t }} 3\, dt = 8 \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( 8 \)

✓ 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

Problem 12.313

medium
Evaluate \( \displaystyle \int_C 3 \, ds \), where \( \displaystyle C \) is the segment from (0, 1, 0) to (1, 0, 0).
  1. Parametrize: r(t) = <t, 1 - t, 0>, 0 ≤ t ≤ 1; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} 0\right)^{2} + \left(\frac{d}{d t} t\right)^{2} + \left(\frac{d}{d t} \left(1 - t\right)\right)^{2}} = \sqrt{2} \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{1} \sqrt{2} \left. 3 \right|_{\substack{ x=t\\ y=1 - t\\ z=0 }}\, dt = 3 \sqrt{2} \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( 3 \sqrt{2} \)

✓ 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

Problem 12.314

medium
Evaluate \( \displaystyle \int_C x^{2} + y^{2} + z \, ds \), where \( \displaystyle C \) is the segment from (0, 1, 0) to (1, 0, 0).
  1. Parametrize: r(t) = <t, 1 - t, 0>, 0 ≤ t ≤ 1; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} 0\right)^{2} + \left(\frac{d}{d t} t\right)^{2} + \left(\frac{d}{d t} \left(1 - t\right)\right)^{2}} = \sqrt{2} \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{1} \sqrt{2} \left. x^{2} + y^{2} + z \right|_{\substack{ x=t\\ y=1 - t\\ z=0 }}\, dt = \frac{2 \sqrt{2}}{3} \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( \frac{2 \sqrt{2}}{3} \)

✓ 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

Problem 12.315

medium
Evaluate \( \displaystyle \int_C x^{2} + y^{2} + z \, ds \), where \( \displaystyle C \) is the quarter circle of radius 1 in the xy-plane, from (1, 0, 0) to (0, 1, 0).
  1. Parametrize: r(t) = <cos(t), sin(t), 0>, 0 ≤ t ≤ pi/2; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} 0\right)^{2} + \left(\frac{d}{d t} \sin{\left(t \right)}\right)^{2} + \left(\frac{d}{d t} \cos{\left(t \right)}\right)^{2}} = 1 \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{\frac{\pi}{2}} \left. x^{2} + y^{2} + z \right|_{\substack{ x=\cos{\left(t \right)}\\ y=\sin{\left(t \right)}\\ z=0 }} \cdot 1\, dt = \frac{\pi}{2} \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( \frac{\pi}{2} \)

✓ 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

Problem 12.316

medium
Evaluate \( \displaystyle \int_C z \, ds \), where \( \displaystyle C \) is the quarter circle of radius 3 in the xy-plane, from (3, 0, 0) to (0, 3, 0).
  1. Parametrize: r(t) = <3*cos(t), 3*sin(t), 0>, 0 ≤ t ≤ pi/2; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} 0\right)^{2} + \left(\frac{d}{d t} 3 \sin{\left(t \right)}\right)^{2} + \left(\frac{d}{d t} 3 \cos{\left(t \right)}\right)^{2}} = 3 \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{\frac{\pi}{2}} \left. z \right|_{\substack{ x=3 \cos{\left(t \right)}\\ y=3 \sin{\left(t \right)}\\ z=0 }} \cdot 3\, dt = 0 \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ 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

Problem 12.317

medium
Evaluate \( \displaystyle \int_C x + y + z \, ds \), where \( \displaystyle C \) is the segment from (0, 1, 0) to (1, 0, 0).
  1. Parametrize: r(t) = <t, 1 - t, 0>, 0 ≤ t ≤ 1; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} 0\right)^{2} + \left(\frac{d}{d t} t\right)^{2} + \left(\frac{d}{d t} \left(1 - t\right)\right)^{2}} = \sqrt{2} \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{1} \sqrt{2} \left. x + y + z \right|_{\substack{ x=t\\ y=1 - t\\ z=0 }}\, dt = \sqrt{2} \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( \sqrt{2} \)

✓ 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

Problem 12.318

medium
Evaluate \( \displaystyle \int_C z \, ds \), where \( \displaystyle C \) is the segment from (0, 1, 0) to (1, 0, 0).
  1. Parametrize: r(t) = <t, 1 - t, 0>, 0 ≤ t ≤ 1; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} 0\right)^{2} + \left(\frac{d}{d t} t\right)^{2} + \left(\frac{d}{d t} \left(1 - t\right)\right)^{2}} = \sqrt{2} \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{1} \sqrt{2} \left. z \right|_{\substack{ x=t\\ y=1 - t\\ z=0 }}\, dt = 0 \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ 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

Problem 12.319

medium
Evaluate \( \displaystyle \int_C 3 \, ds \), where \( \displaystyle C \) is the segment from the origin to (1, 2, 2).
  1. Parametrize: r(t) = <t, 2*t, 2*t>, 0 ≤ t ≤ 1; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} t\right)^{2} + 2 \left(\frac{d}{d t} 2 t\right)^{2}} = 3 \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{1} \left. 3 \right|_{\substack{ x=t\\ y=2 t\\ z=2 t }} 3\, dt = 9 \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( 9 \)

✓ 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

Problem 12.320

medium
Evaluate \( \displaystyle \int_C x + y + z \, ds \), where \( \displaystyle C \) is the segment from the origin to (1, 2, 2).
  1. Parametrize: r(t) = <t, 2*t, 2*t>, 0 ≤ t ≤ 1; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} t\right)^{2} + 2 \left(\frac{d}{d t} 2 t\right)^{2}} = 3 \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{1} \left. x + y + z \right|_{\substack{ x=t\\ y=2 t\\ z=2 t }} 3\, dt = \frac{15}{2} \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( \frac{15}{2} \)

✓ 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

Problem 12.321

medium
Evaluate \( \displaystyle \int_C x y + z \, ds \), where \( \displaystyle C \) is the quarter circle of radius 3 in the xy-plane, from (3, 0, 0) to (0, 3, 0).
  1. Parametrize: r(t) = <3*cos(t), 3*sin(t), 0>, 0 ≤ t ≤ pi/2; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} 0\right)^{2} + \left(\frac{d}{d t} 3 \sin{\left(t \right)}\right)^{2} + \left(\frac{d}{d t} 3 \cos{\left(t \right)}\right)^{2}} = 3 \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{\frac{\pi}{2}} \left. x y + z \right|_{\substack{ x=3 \cos{\left(t \right)}\\ y=3 \sin{\left(t \right)}\\ z=0 }} \cdot 3\, dt = \frac{27}{2} \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( \frac{27}{2} \)

✓ 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

Problem 12.322

medium
Evaluate \( \displaystyle \int_C z \, ds \), where \( \displaystyle C \) is one turn of the helix r(t) = ⟨cos t, sin t, t⟩.
  1. Parametrize: r(t) = <cos(t), sin(t), t>, 0 ≤ t ≤ 2*pi; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} t\right)^{2} + \left(\frac{d}{d t} \sin{\left(t \right)}\right)^{2} + \left(\frac{d}{d t} \cos{\left(t \right)}\right)^{2}} = \sqrt{2} \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{2 \pi} \sqrt{2} \left. z \right|_{\substack{ x=\cos{\left(t \right)}\\ y=\sin{\left(t \right)}\\ z=t }}\, dt = 2 \sqrt{2} \pi^{2} \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( 2 \sqrt{2} \pi^{2} \)

✓ 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

Problem 12.323

medium
Evaluate \( \displaystyle \int_C x y + z \, ds \), where \( \displaystyle C \) is the segment from the origin to (1, 2, 2).
  1. Parametrize: r(t) = <t, 2*t, 2*t>, 0 ≤ t ≤ 1; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} t\right)^{2} + 2 \left(\frac{d}{d t} 2 t\right)^{2}} = 3 \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{1} \left. x y + z \right|_{\substack{ x=t\\ y=2 t\\ z=2 t }} 3\, dt = 5 \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( 5 \)

✓ 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

Problem 12.324

medium
Evaluate \( \displaystyle \int_C x + y \, ds \), where \( \displaystyle C \) is one turn of the helix r(t) = ⟨cos t, sin t, t⟩.
  1. Parametrize: r(t) = <cos(t), sin(t), t>, 0 ≤ t ≤ 2*pi; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} t\right)^{2} + \left(\frac{d}{d t} \sin{\left(t \right)}\right)^{2} + \left(\frac{d}{d t} \cos{\left(t \right)}\right)^{2}} = \sqrt{2} \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{2 \pi} \sqrt{2} \left. x + y \right|_{\substack{ x=\cos{\left(t \right)}\\ y=\sin{\left(t \right)}\\ z=t }}\, dt = 0 \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ 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

Problem 12.405

medium
Evaluate \( \displaystyle \int_C 3 \, ds \), where \( \displaystyle C \) is the quarter circle of radius 2 in the xy-plane, from (2, 0, 0) to (0, 2, 0).
  1. Parametrize: r(t) = <2*cos(t), 2*sin(t), 0>, 0 ≤ t ≤ pi/2; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} 0\right)^{2} + \left(\frac{d}{d t} 2 \sin{\left(t \right)}\right)^{2} + \left(\frac{d}{d t} 2 \cos{\left(t \right)}\right)^{2}} = 2 \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{\frac{\pi}{2}} \left. 3 \right|_{\substack{ x=2 \cos{\left(t \right)}\\ y=2 \sin{\left(t \right)}\\ z=0 }} \cdot 2\, dt = 3 \pi \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( 3 \pi \)

✓ 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

Problem 12.410

medium
Evaluate \( \displaystyle \int_C x + y \, ds \), where \( \displaystyle C \) is the quarter circle of radius 2 in the xy-plane, from (2, 0, 0) to (0, 2, 0).
  1. Parametrize: r(t) = <2*cos(t), 2*sin(t), 0>, 0 ≤ t ≤ pi/2; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} 0\right)^{2} + \left(\frac{d}{d t} 2 \sin{\left(t \right)}\right)^{2} + \left(\frac{d}{d t} 2 \cos{\left(t \right)}\right)^{2}} = 2 \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{\frac{\pi}{2}} \left. x + y \right|_{\substack{ x=2 \cos{\left(t \right)}\\ y=2 \sin{\left(t \right)}\\ z=0 }} \cdot 2\, dt = 8 \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( 8 \)

✓ 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

Problem 12.411

medium
Evaluate \( \displaystyle \int_C x y + z \, ds \), where \( \displaystyle C \) is the segment from (0, 1, 0) to (1, 0, 0).
  1. Parametrize: r(t) = <t, 1 - t, 0>, 0 ≤ t ≤ 1; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} 0\right)^{2} + \left(\frac{d}{d t} t\right)^{2} + \left(\frac{d}{d t} \left(1 - t\right)\right)^{2}} = \sqrt{2} \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{1} \sqrt{2} \left. x y + z \right|_{\substack{ x=t\\ y=1 - t\\ z=0 }}\, dt = \frac{\sqrt{2}}{6} \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( \frac{\sqrt{2}}{6} \)

✓ 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

Problem 12.412

medium
Evaluate \( \displaystyle \int_C x + y + z \, ds \), where \( \displaystyle C \) is the quarter circle of radius 3 in the xy-plane, from (3, 0, 0) to (0, 3, 0).
  1. Parametrize: r(t) = <3*cos(t), 3*sin(t), 0>, 0 ≤ t ≤ pi/2; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} 0\right)^{2} + \left(\frac{d}{d t} 3 \sin{\left(t \right)}\right)^{2} + \left(\frac{d}{d t} 3 \cos{\left(t \right)}\right)^{2}} = 3 \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{\frac{\pi}{2}} \left. x + y + z \right|_{\substack{ x=3 \cos{\left(t \right)}\\ y=3 \sin{\left(t \right)}\\ z=0 }} \cdot 3\, dt = 18 \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( 18 \)

✓ 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

Problem 12.413

medium
Evaluate \( \displaystyle \int_C x + y \, ds \), where \( \displaystyle C \) is the segment from the origin to (1, 2, 2).
  1. Parametrize: r(t) = <t, 2*t, 2*t>, 0 ≤ t ≤ 1; ds = ‖r′(t)‖ dt.
    Reviewed
  2. \[ \sqrt{\left(\frac{d}{d t} t\right)^{2} + 2 \left(\frac{d}{d t} 2 t\right)^{2}} = 3 \]
    ‖r′(t)‖.✓ Proved
  3. \[ \int\limits_{0}^{1} \left. x + y \right|_{\substack{ x=t\\ y=2 t\\ z=2 t }} 3\, dt = \frac{9}{2} \]
    ∫ f(r(t)) ‖r′(t)‖ dt.✓ Proved
Answer \( \frac{9}{2} \)

✓ 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

More line integrals of scalar functions practice

Evaluate \( \displaystyle \int_C x + y \, ds \), where \( \displaystyle C \) is the quarter circle of radius 1 in the xy-plane, from (1, 0, 0) to (0, 1, 0).
Problem 12.406medium✓ Every equation proved
Evaluate \( \displaystyle \int_C z \, ds \), where \( \displaystyle C \) is the quarter circle of radius 2 in the xy-plane, from (2, 0, 0) to (0, 2, 0).
Problem 12.407medium✓ Nihil obstat
Evaluate \( \displaystyle \int_C x + y + z \, ds \), where \( \displaystyle C \) is the quarter circle of radius 1 in the xy-plane, from (1, 0, 0) to (0, 1, 0).
Problem 12.408medium✓ Every equation proved
Evaluate \( \displaystyle \int_C 3 \, ds \), where \( \displaystyle C \) is the quarter circle of radius 3 in the xy-plane, from (3, 0, 0) to (0, 3, 0).
Problem 12.409medium✓ Nihil obstat
Evaluate \( \displaystyle \int_C x y + z \, ds \), where \( \displaystyle C \) is the quarter circle of radius 2 in the xy-plane, from (2, 0, 0) to (0, 2, 0).
Problem 12.414medium✓ Nihil obstat
Evaluate \( \displaystyle \int_C 3 \, ds \), where \( \displaystyle C \) is the quarter circle of radius 1 in the xy-plane, from (1, 0, 0) to (0, 1, 0).
Problem 12.415medium✓ Every equation proved
Evaluate \( \displaystyle \int_C x^{2} + y^{2} + z \, ds \), where \( \displaystyle C \) is the quarter circle of radius 3 in the xy-plane, from (3, 0, 0) to (0, 3, 0).
Problem 12.416medium✓ Nihil obstat
Evaluate \( \displaystyle \int_C x y + z \, ds \), where \( \displaystyle C \) is the quarter circle of radius 1 in the xy-plane, from (1, 0, 0) to (0, 1, 0).
Problem 12.417medium✓ Nihil obstat
Evaluate \( \displaystyle \int_C z \, ds \), where \( \displaystyle C \) is the segment from the origin to (1, 2, 2).
Problem 12.418medium✓ Nihil obstat
Evaluate \( \displaystyle \int_C x + y \, ds \), where \( \displaystyle C \) is the segment from (0, 1, 0) to (1, 0, 0).
Problem 12.419medium✓ Nihil obstat
Evaluate \( \displaystyle \int_C 3 \, ds \), where \( \displaystyle C \) is one turn of the helix r(t) = ⟨cos t, sin t, t⟩.
Problem 12.420medium✓ Nihil obstat
Evaluate \( \displaystyle \int_C z \, ds \), where \( \displaystyle C \) is the quarter circle of radius 1 in the xy-plane, from (1, 0, 0) to (0, 1, 0).
Problem 12.421medium✓ Nihil obstat
Evaluate \( \displaystyle \int_C x^{2} + y^{2} + z \, ds \), where \( \displaystyle C \) is one turn of the helix r(t) = ⟨cos t, sin t, t⟩.
Problem 12.422medium✓ Nihil obstat
Evaluate \( \displaystyle \int_C x + y + z \, ds \), where \( \displaystyle C \) is the quarter circle of radius 2 in the xy-plane, from (2, 0, 0) to (0, 2, 0).
Problem 12.465medium✓ Nihil obstat
Evaluate \( \displaystyle \int_C x^{2} + y^{2} + z \, ds \), where \( \displaystyle C \) is the quarter circle of radius 2 in the xy-plane, from (2, 0, 0) to (0, 2, 0).
Problem 12.475medium✓ Nihil obstat
Evaluate \( \displaystyle \int_C x + y \, ds \), where \( \displaystyle C \) is the quarter circle of radius 3 in the xy-plane, from (3, 0, 0) to (0, 3, 0).
Problem 12.476medium✓ Nihil obstat