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Evaluate the line integral\(\int_{\rm{C}} {\rm{F}} {\rm{ \times dr}}\) where \({\rm{C}}\)is given by the vector function\({\rm{r(t)}}\).

\({\rm{F(x,y) = xyi + 3}}{{\rm{y}}^{\rm{2}}}{\rm{j}}\)

Short Answer

Expert verified

The value of the line integral is \({\rm{45}}\).

Step by step solution

01

Finding  Derivative.

\({\rm{r'(t) = 44}}{{\rm{t}}^{\rm{3}}}{\rm{i + 3}}{{\rm{t}}^{\rm{2}}}{\rm{j}}\)

02

Integrating the Vector field.

\(\int_{\rm{C}} {\rm{F}} \cdot {\rm{ds = }}\int_{\rm{C}} {\rm{F}} {\rm{(r(t))}} \cdot {\rm{r'(t)dt}}\)

\(\begin{aligned}\int\limits_{\rm{0}}^{\rm{1}} {\left( {{\rm{11}}{{\rm{t}}^{\rm{4}}} \cdot {{\rm{t}}^{\rm{3}}}{\rm{,3}}{{\left( {{{\rm{t}}^{\rm{3}}}} \right)}^{\rm{2}}}} \right) \cdot \left( {{\rm{44}}{{\rm{t}}^{\rm{3}}}{\rm{,3}}{{\rm{t}}^{\rm{2}}}} \right)} &{\rm{ = }}\int_{\rm{0}}^{\rm{1}} {\rm{1}} {\rm{1}}{{\rm{t}}^{\rm{7}}} \cdot {\rm{44}}{{\rm{t}}^{\rm{3}}}{\rm{ + 3}}{{\rm{t}}^{\rm{6}}} \cdot {\rm{3}}{{\rm{t}}^{\rm{2}}}{\rm{dt}}\\&{\rm{ = }}\int_{\rm{0}}^{\rm{1}} {\rm{4}} {\rm{84}}{{\rm{t}}^{{\rm{10}}}}{\rm{ + 9}}{{\rm{t}}^{\rm{8}}}\\&{\rm{ = 45}}\end{aligned}\)

Thus, the value is\({\rm{45}}\).

A

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