Question

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

Short answer

The final component \(\sqrt {\rm{5}} \frac{{{{{\rm{(2\pi )}}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ + 2\pi }}\)

Step-by-step solution

Would you like to alter the line integral to this format.

\(\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}} \right){\rm{ds = f(x(t) + y(t) + z(t))}}\sqrt {{{\left( {\frac{{{\rm{dx}}}}{{{\rm{dt}}}}} \right)}^{\rm{2}}}{\rm{ + }}{{\left( {\frac{{{\rm{dy}}}}{{{\rm{dt}}}}} \right)}^{\rm{2}}}{\rm{ + }}{{\left( {\frac{{{\rm{dz}}}}{{{\rm{dt}}}}} \right)}^{\rm{2}}}{\rm{dt}}} \)

With regard to, take the derivative of \({\rm{x,y}}\)and \({\rm{z}}\)\(\sqrt {\rm{5}} \frac{{{{{\rm{(2\pi )}}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ + 2\pi }}\)

\(\frac{{{\rm{dx}}}}{{{\rm{dt}}}}{\rm{ = 1,}}\frac{{{\rm{dy}}}}{{{\rm{dt}}}}{\rm{ = - 2sin(2t),}}\frac{{{\rm{dz}}}}{{{\rm{dt}}}}{\rm{ = 2cos(2t)}}\)

Substitute into the equation.

\(\begin{array}{c}\sqrt {{{\frac{{{\rm{dx}}}}{{{\rm{dt}}}}}^{\rm{2}}}{\rm{ + }}{{\frac{{{\rm{dy}}}}{{{\rm{dt}}}}}^{\rm{2}}}{\rm{ + }}\frac{{{\rm{d}}{{\rm{z}}^{\rm{2}}}}}{{{\rm{dt}}}}} {\rm{ = }}\sqrt {{\rm{1 + 4si}}{{\rm{n}}^{\rm{2}}}{\rm{(2t) + 4co}}{{\rm{s}}^{\rm{2}}}{\rm{(2t)}}} \\{\rm{ = }}\sqrt {\rm{5}} \end{array}\)

Substitute into Step \({\rm{1}}\)equation and remove the constant.

\({c_c}\left( {{x^2} + {y^2} + {z^2}} \right)ds = \sqrt 5 _0^{2\pi }\left( {{t^2} + {{\cos }^2}(2t) + {{\sin }^2}(2t)} \right)dt\)

\({\rm{ = }}\sqrt {\rm{5}} _{\rm{0}}^{{\rm{2\pi }}}\left( {{{\rm{t}}^{\rm{2}}}{\rm{ + 1}}} \right){\rm{dt = }}\sqrt {\rm{5}} \frac{{{{{\rm{(2\pi )}}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ + 2\pi }}\)

Assess and simplify \(\sqrt {\rm{5}} \frac{{{{{\rm{(2\pi )}}}^{\rm{3}}}}}{{\rm{3}}}{\rm{ + 2\pi }}\).