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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.
- 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
- 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
- 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
- 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
- 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.
- 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
- 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
- Evaluate \( \displaystyle \int\limits_{C}{{{y^3} - {x^2}\,ds}}\) for each of the following curves.
- \(C\) 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 \(\left( {3, - 6} \right)\).
- \(C\) is the line segment from \(\left( {3,6} \right)\) to \(\left( {3, - 6} \right)\).
- Evaluate \( \displaystyle \int\limits_{C}{{4{x^2}\,ds}}\) for each of the following curves.
- \(C\) is the portion of the circle centered at the origin of radius 2 in the 1st quadrant rotating in the clockwise direction.
- \(C\) is the line segment from \(\left( {0,2} \right)\) to \(\left( {2,0} \right)\).
- Evaluate \( \displaystyle \int\limits_{C}{{2{x^3}\,ds}}\) for each of the following curves.
- \(C\) is the portion \(y = {x^3}\) from \(x = - 1\) to \(x = 2\).
- \(C\) is the portion \(y = {x^3}\) from \(x = 2\) to \(x = - 1\).