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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 \( \displaystyle \int\limits_{C}{{3{x^2} - 2y\,ds}}\) where \(C\) is the line segment from \(\left( {3,6} \right)\) to \(\left( {1, - 1} \right)\). Solution
2. Evaluate \( \displaystyle \int\limits_{C}{{2y{x^2} - 4x\,ds}}\) where \(C\) is the lower half of the circle centered at the origin of radius 3 with clockwise rotation. Solution
3. Evaluate \( \displaystyle \int\limits_{C}{{6x\,ds}}\) 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 \( \displaystyle \int\limits_{C}{{xy - 4z\,ds}}\) where \(C\) is the line segment from \(\left( {1,1,0} \right)\) to \(\left( {2,3, - 2} \right)\). Solution
5. Evaluate \( \displaystyle \int\limits_{C}{{{x^2}{y^2}\,ds}}\) 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 \( \displaystyle \int\limits_{C}{{16{y^5}\,ds}}\) 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 \( \displaystyle \int\limits_{C}{{4y - x\,ds}}\) 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 \( \displaystyle \int\limits_{C}{{{y^3} - {x^2}\,ds}}\) 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 \( \displaystyle \int\limits_{C}{{4{x^2}\,ds}}\) for each of the following curves.
   1. \(C\) is the portion of the circle centered at the origin of radius 2 in the 1^st 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 \( \displaystyle \int\limits_{C}{{2{x^3}\,ds}}\) 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