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Section 16.3 : Line Integrals - Part II
For problems 1 – 5 evaluate the given line integral. Follow the direction of \(C\) as given in the problem statement.
- Evaluate \( \displaystyle \int\limits_{C}{{\sqrt {1 + y} \,dy}}\) where \(C\) is the portion of \(y = {{\bf{e}}^{2x}}\) from \(x = 0\) to \(x = 2\). Solution
- Evaluate \( \displaystyle \int\limits_{C}{{2y\,dx + \left( {1 - x} \right)\,dy}}\) where \(C\) is portion of \(y = 1 - {x^3}\) from \(x = - 1\) to \(x = 2\). Solution
- Evaluate \( \displaystyle \int\limits_{C}{{{x^2}\,dy - yz\,dz}}\) where \(C\) is the line segment from \(\left( {4, - 1,2} \right)\) to \(\left( {1,7, - 1} \right)\). Solution
- Evaluate \( \displaystyle \int\limits_{C}{{1 + {x^3}\,dx}}\) where \(C\) is the right half of the circle of radius 2 with counter clockwise rotation followed by the line segment from \(\left( {0,2} \right)\) to \(\left( { - 3, - 4} \right)\). See the sketch below for the direction.
Solution
- Evaluate \( \displaystyle \int\limits_{C}{{2{x^2}\,dy - xy\,dx}}\) where \(C\) is the line segment from \(\left( {1, - 5} \right)\) to \(\left( { - 2, - 3} \right)\) followed by the portion of \(y = 1 - {x^2}\) from \(x = - 2\) to \(x = 2\) which in turn is followed by the line segment from \(\left( {2, - 3} \right)\) to \(\left( {4, - 3} \right)\). See the sketch below for the direction.
Solution
- Evaluate \( \displaystyle \int\limits_{C}{{\left( {x - y} \right)\,dx - y{x^2}\,dy}}\) for each of the following curves.
- \(C\) is the portion of the circle of radius 6 in the 1st, 2nd and 3rd quadrant with clockwise rotation.
- \(C\) is the line segment from \(\left( {0, - 6} \right)\) to \(\left( {6,0} \right)\).
- Evaluate \( \displaystyle \int\limits_{C}{{{x^3}\,dy - \left( {y + 1} \right)\,dx}}\) for each of the following curves.
- \(C\) is the line segment from \(\left( {1,7} \right)\) to \(\left( { - 2,4} \right)\).
- \(C\) is the line segment from \(\left( { - 2,4} \right)\) to \(\left( {1,7} \right)\).