Question #314036

Evaluate ∫c š¹. š‘‘š‘Ÿ,where š¹ = š‘‹^2 āˆ’ š‘Œ^2š‘– + š‘„š‘¦š‘— and curve š¶ is the arc of the curve š‘¦ = š‘‹^3 from (0,0) to (2,8).


Expert's answer

∫CFdr=[y=x3dy=3x2dxx:0→2F1=x2āˆ’y2=x2āˆ’x6F2=xy=x4]=∫02(x2āˆ’x6)dx+x4ā‹…3x2dx==(x33+2x77)∣02=82421\int_C{Fdr}=\left[ \begin{array}{c} y=x^3\\ dy=3x^2dx\\ x:0\rightarrow 2\\ F_1=x^2-y^2=x^2-x^6\\ F_2=xy=x^4\\\end{array} \right] =\int_0^2{\left( x^2-x^6 \right) dx+x^4\cdot 3x^2dx}=\\=\left( \frac{x^3}{3}+\frac{2x^7}{7} \right) |_{0}^{2}=\frac{824}{21}


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