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Section 16.2 : Line Integrals - Part I

For problems 1 – 7 evaluate the given line integral. Follow the direction of \(C\) as given in the problem statement.

  1. Evaluate \( \displaystyle \int\limits_{C}{{3{x^2} - 2y,円ds}}\) where \(C\) is the line segment from \(\left( {3,6} \right)\) to \(\left( {1, - 1} \right)\). Solution
  2. Evaluate \( \displaystyle \int\limits_{C}{{2y{x^2} - 4x,円ds}}\) where \(C\) is the lower half of the circle centered at the origin of radius 3 with clockwise rotation. Solution
  3. Evaluate \( \displaystyle \int\limits_{C}{{6x,円ds}}\) where \(C\) is the portion of \(y = {x^2}\) from \(x = - 1\) to \(x = 2\). The direction of \(C\) is in the direction of increasing \(x\). Solution
  4. Evaluate \( \displaystyle \int\limits_{C}{{xy - 4z,円ds}}\) where \(C\) is the line segment from \(\left( {1,1,0} \right)\) to \(\left( {2,3, - 2} \right)\). Solution
  5. Evaluate \( \displaystyle \int\limits_{C}{{{x^2}{y^2},円ds}}\) where \(C\) is the circle centered at the origin of radius 2 centered on the \(y\)-axis at \(y = 4\). See the sketches below for orientation. Note the "odd" axis orientation on the 2D circle is intentionally that way to match the 3D axis the direction.
    Solution
  6. Evaluate \( \displaystyle \int\limits_{C}{{16{y^5},円ds}}\) where \(C\) is the portion of \(x = {y^4}\) from \(y = 0\) to \(y = 1\) followed by the line segment from \(\left( {1,1} \right)\) to \(\left( {1, - 2} \right)\) which in turn is followed by the line segment from \(\left( {1, - 2} \right)\) to \(\left( {2,0} \right)\). See the sketch below for the direction. Solution
  7. Evaluate \( \displaystyle \int\limits_{C}{{4y - x,円ds}}\) where \(C\) is the upper portion of the circle centered at the origin of radius 3 from \(\displaystyle\left( {\frac{3}{{\sqrt 2 }},\frac{3}{{\sqrt 2 }}} \right)\) to \(\displaystyle\left( { - \frac{3}{{\sqrt 2 }}, - \frac{3}{{\sqrt 2 }}} \right)\) in the counter clockwise rotation followed by the line segment from \(\displaystyle\left( { - \frac{3}{{\sqrt 2 }}, - \frac{3}{{\sqrt 2 }}} \right)\) to \(\displaystyle\left( {4, - \frac{3}{{\sqrt 2 }}} \right)\) which in turn is followed by the line segment from \(\displaystyle\left( {4, - \frac{3}{{\sqrt 2 }}} \right)\) to \(\left( {4,4} \right)\). See the sketch below for the direction. Solution
  8. Evaluate \( \displaystyle \int\limits_{C}{{{y^3} - {x^2},円ds}}\) for each of the following curves.
    1. \(C\) is the line segment from \(\left( {3,6} \right)\) to \(\left( {0,0} \right)\) followed by the line segment from \(\left( {0,0} \right)\) to \(\left( {3, - 6} \right)\).
    2. \(C\) is the line segment from \(\left( {3,6} \right)\) to \(\left( {3, - 6} \right)\).
    Solution
  9. Evaluate \( \displaystyle \int\limits_{C}{{4{x^2},円ds}}\) for each of the following curves.
    1. \(C\) is the portion of the circle centered at the origin of radius 2 in the 1st quadrant rotating in the clockwise direction.
    2. \(C\) is the line segment from \(\left( {0,2} \right)\) to \(\left( {2,0} \right)\).
    Solution
  10. Evaluate \( \displaystyle \int\limits_{C}{{2{x^3},円ds}}\) for each of the following curves.
    1. \(C\) is the portion \(y = {x^3}\) from \(x = - 1\) to \(x = 2\).
    2. \(C\) is the portion \(y = {x^3}\) from \(x = 2\) to \(x = - 1\).
    Solution
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