Question

Evaluate the line integral, where C is the given curve.

\(\) \(\int_{\rm{C}} {\rm{x}} {\rm{ sin y ds}}\), C is the line segment from \({\rm{(0,}}\mid {\rm{3) to (4,6)}}\).

Short answer

The line integral \(\frac{{{\rm{20 sin 6 - 60 cos 6 - 20 sin 3}}}}{{\rm{9}}}\).

Step-by-step solution

Explanation of solution.

Remember howmade the section between the two points was parameterized. To start, determine the vector determined by the two points given

\({\rm{(4,6) - (0,3) = (4,3)}}\)

The starting point is \(\left( {{\rm{0,3}}} \right)\)The segment may now be parameterized as follows

\({\rm{\gamma (l) = (0,3) + (4l,3l) = (4l,3l + 3),l}} \in {\rm{(0,1)}}\)

Find the parametrization's derivative next

\({\gamma ^\prime }{\rm{(t) = (4,3)}} \Rightarrow \left\| {{\gamma ^\prime }{\rm{(t)}}} \right\|{\rm{ = }}\sqrt {{{\rm{4}}^{\rm{2}}}{\rm{ + }}{{\rm{3}}^{\rm{2}}}} {\rm{ = 5}}\).

Calculation.

Finally, calculate the integral

\(\begin{array}{c}\int_{\rm{0}}^{\rm{1}} {\rm{4}} {\rm{t sin(3 + 3t)}} \cdot {\rm{5dt = 20}}\int_{\rm{0}}^{\rm{1}} {\rm{t}} {\rm{ sin(3 + 3t)dt}}\\{\rm{ = }}\frac{{{\rm{20 sin 6 - 60 cos 6 - 20 sin 3}}}}{{\rm{9}}}.\end{array}\)

The line integral \(\frac{{{\rm{20 sin 6 - 60 cos 6 - 20 sin 3}}}}{{\rm{9}}}\).