Question

Evaluate the line integral, where \({\rm{C}}\) is the given curve

\(\int_{\rm{C}} {\rm{x}} {\rm{y}}{{\rm{z}}^{\rm{2}}}{\rm{ds}}\)\({\rm{C}}\)Is the line segment from\({\rm{( - 1,5,0) to (1,6,4)}}\)

Short answer

Finally, add the integrals together.\(\frac{{{\rm{236}}\sqrt {{\rm{21}}} }}{{{\rm{15}}}}\)

Step-by-step solution

Calculate the length of the section between the two spots.

\({\rm{\gamma (t) = ( - 1 + 2t,5 + t,4t)}}\)

The parameterizations’ derivative is

\(\begin{array}{l}{{\rm{\gamma }}^{\rm{\cent}}}{\rm{(t) = (2,1,4)}}\\{\rm{ds = }}\left\| {{\rm{\gamma }}\prime {\rm{(t)}}} \right\|{\rm{ = }}\sqrt {{\rm{21}}} \end{array}\)

  Step 2: Finally, add the two integrals together.

\(\begin{array}{l}_{\rm{0}}^{\rm{1}}\sqrt {{\rm{21}}} {\rm{( - 1 + 2t)(5 + t)(4t}}{{\rm{)}}^{\rm{2}}}{\rm{dt = 0}}_{\rm{0}}^{\rm{1}}\sqrt {{\rm{21}}} \left( {{\rm{32}}{{\rm{t}}^{\rm{4}}}{\rm{ + 144}}{{\rm{t}}^{\rm{3}}}{\rm{ - 80}}{{\rm{t}}^{\rm{2}}}} \right)\\{\rm{dt = }}\frac{{{\rm{236}}\sqrt {{\rm{21}}} }}{{{\rm{15}}}}\end{array}\)

Therefore, finally, add the integrals together.\(\frac{{{\rm{236}}\sqrt {{\rm{21}}} }}{{{\rm{15}}}}\)