Question

Evaluate the line integral, where C is the given curve. \(\) \(\int_{\rm{C}} {\left( {{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{3}}}{\rm{ - }}\sqrt {\rm{x}} } \right)} {\rm{dy,}}\) C is the arc of the curve \({\rm{y = }}\sqrt {\rm{x}} \)from \({\rm{(1,1)}}\) to \({\rm{(4,2)}}\).

Short answer

The line integral \(\int_{\rm{C}} {\left( {{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{3}}}{\rm{ - }}\sqrt {\rm{x}} } \right)} {\rm{dy = }}\frac{{{\rm{243}}}}{{\rm{8}}}\).

Step-by-step solution

Explanation of solution.

\(\int_{\rm{C}} {\left( {{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{3}}}{\rm{ - }}\sqrt {\rm{x}} } \right)} {\rm{dy}}\)

Given \({\rm{y = }}\sqrt {\rm{x}} \) Therefore \({\rm{x = }}{{\rm{y}}^{\rm{2}}}\)

\({\rm{ = }}\int_{\rm{C}} {\left( {{{\left( {{{\rm{y}}^{\rm{2}}}} \right)}^{\rm{2}}}{{\rm{y}}^{\rm{3}}}{\rm{ - y}}} \right)} {\rm{dy}}\)

Calculation.

Move from \({\rm{(1,1) }}\)to \({\rm{(4,2)}}\), y from rises \({\rm{1}}\)to \({\rm{2}}\).

As a result, from \({\rm{1}}\) to\({\rm{2}}\), integrate with respect to y.

\(\begin{array}{l}{\rm{ = }}\int_{\rm{1}}^{\rm{2}} {{{\rm{y}}^{\rm{7}}}} {\rm{ - y dy}}\\{\rm{ = }}\left( {\frac{{{{\rm{y}}^{\rm{8}}}}}{{\rm{8}}}{\rm{ - }}\frac{{{{\rm{y}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{\rm{1}}^{\rm{2}}\\{\rm{ = }}\left( {\frac{{{{\rm{2}}^{\rm{8}}}}}{{\rm{8}}}{\rm{ - }}\frac{{{{\rm{2}}^{\rm{2}}}}}{{\rm{2}}}} \right){\rm{ - }}\left( {\frac{{\rm{1}}}{{\rm{8}}}{\rm{ - }}\frac{{\rm{1}}}{{\rm{2}}}} \right)\\{\rm{ = }}\left( {\frac{{{\rm{256}}}}{{\rm{8}}}{\rm{ - }}\frac{{{\rm{4 \times 4}}}}{{\rm{8}}}} \right){\rm{ - }}\left( {\frac{{\rm{1}}}{{\rm{8}}}{\rm{ - }}\frac{{\rm{4}}}{{\rm{8}}}} \right)\\{\rm{ = }}\frac{{{\rm{256 - 16 - 1 + 4}}}}{{\rm{8}}}{\rm{ = }}\frac{{{\rm{243}}}}{{\rm{8}}}{\rm{.}}\end{array}\)

The line integral \(\int_{\rm{C}} {\left( {{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{3}}}{\rm{ - }}\sqrt {\rm{x}} } \right)} {\rm{dy = }}\frac{{{\rm{243}}}}{{\rm{8}}}\).