Question

# Evaluate the line integral, \int_c xyds, Where c is the given curve, c:x=t^2,y=2t,0\leq t\leq5

Integrals
Evaluate the line integral
$$\displaystyle\int_{{c}}{x}{y}{d}{s}$$
Where c is the given curve
$$\displaystyle{c}:{x}={t}^{{2}},{y}={2}{t},{0}\leq{t}\leq{5}$$

## Expert Answers (1)

2021-09-17
Explanation
Line integrals can be used to eavaluate integrals of two or three dimensioal curves
The curve c is expressed by the following parametric equations given by
$$\displaystyle{x}={t}^{{2}}$$
and
$$\displaystyle{y}={2}{t}$$
Thus, to evaluate the following integral given by
$$\displaystyle{I}=\int_{{c}}{x}{y}{d}{s}$$
we use the following line integral formula given by
$$\displaystyle\int_{{c}}{f{{\left({x},{y}\right)}}}{d}{s}={\int_{{a}}^{{b}}}{f{{\left({x}{\left({t}\right)},{y}{\left({t}\right)}\right)}}}{\left|{r}'\right|}{\left.{d}{t}\right.}$$
$$\displaystyle={\int_{{a}}^{{b}}}{f{{\left({x}{\left({t}\right)},{y}{\left({t}\right)}\right)}}}{\left|{\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}\right|}{\left.{d}{t}\right.}$$
$$\displaystyle={\int_{{a}}^{{b}}}{f{{\left({x}{\left({t}\right)},{y}{\left({t}\right)}\right)}}}\sqrt{{{\left({\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}\right)}{2}+{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}\right)}^{{2}}}}{\left.{d}{t}\right.}$$
Since $$\displaystyle{x}{\left({t}\right)}={t}^{{2}}$$ and $$\displaystyle{y}{\left({t}\right)}={2}{t}$$
We want to evaluate $$\displaystyle\int_{{c}}{x}{y}{d}{s}$$ as follows
Firstly, we must find the integrand $$\displaystyle{f{{\left({x}{\left({t}\right)},{y}{\left({t}\right)}\right)}}}$$ as follows
$$\displaystyle{f{{\left({x}{\left({t}\right)},{y}{\left({t}\right)}\right)}}}={x}{y}$$
$$\displaystyle={t}^{{2}}{\left({2}{t}\right)}$$
$$\displaystyle={2}{t}^{{3}}$$
Also, the derivatives of x(f) and y(t) must be
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{x}{\left({t}\right)}={\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({t}^{{2}}\right)}={2}{t}$$
and
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{y}{\left({t}\right)}={\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({2}{t}\right)}={2}$$
Since $$\displaystyle{0}\leq{t}\leq{5}$$, so that the upper and lower limits of the integral must be
$$\displaystyle{a}={0}$$
and
$$\displaystyle{b}={5}$$
Thus, using the line integral formula, we get
$$\displaystyle\int_{{c}}{x}{y}{d}{s}={\int_{{a}}^{{b}}}{f{{\left({x}{\left({t}\right)}\right)}}}\sqrt{{{\left({\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}\right)}^{{2}}+{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}\right)}^{{2}}}}{\left.{d}{t}\right.}$$
$$\displaystyle={\int_{{0}}^{{5}}}{\left({2}{t}^{{3}}\right)}\sqrt{{{\left({2}{t}\right)}^{{2}}+{\left({2}\right)}^{{2}}}}{\left.{d}{t}\right.}$$
$$\displaystyle={2}{\int_{{0}}^{{5}}}{t}^{{3}}\sqrt{{{4}{t}^{{2}}+{4}}}{\left.{d}{t}\right.}={2}{\int_{{0}}^{{5}}}{t}^{{3}}\sqrt{{{4}{\left({t}^{{2}}+{1}\right)}}}{\left.{d}{t}\right.}$$
$$\displaystyle={2}{\int_{{0}}^{{5}}}{2}{t}^{{3}}\sqrt{{{\left({t}^{{2}}+{1}\right)}}}{\left.{d}{t}\right.}$$
$$\displaystyle={4}{\int_{{0}}^{{5}}}{t}^{{3}}\sqrt{{{\left({t}^{{2}}+{1}\right)}}}{\left.{d}{t}\right.}$$
To evaluate the following integral $$\displaystyle{\int_{{0}}^{{5}}}{t}^{{3}}\sqrt{{{\left({t}^{{2}}+{1}\right)}}}{\left.{d}{t}\right.}$$, we use the method of substitution as follows
Let $$\displaystyle{u}={t}^{{2}}+{1}$$, so that
Differentiate both sides of the following equation $$\displaystyle{u}={t}^{{2}}+{1}$$ implies that
$$\displaystyle{d}{u}={2}{t}{\left.{d}{t}\right.}$$
and thus
$$\displaystyle{t}{\left.{d}{t}\right.}={\frac{{{d}{u}}}{{{2}}}}$$
Also, we have
$$\displaystyle{t}^{{2}}={u}-{1}$$
Also, we find that the upper and lower limits of the integral must be
$$\displaystyle{t}={0}\Rightarrow{u}={0}^{{2}}+{1}={1}$$
and
$$\displaystyle{t}={5}\Rightarrow{u}={5}^{{2}}+{1}={26}$$
Thus, the line integral becomes
$$\displaystyle\int_{{c}}{x}{y}{d}{s}={4}{\int_{{{t}={0}}}^{{5}}}{t}^{{2}}\sqrt{{{t}^{{2}}+{1}}}{t}{\left.{d}{t}\right.}$$
$$\displaystyle={4}{\int_{{{u}={1}}}^{{{26}}}}{\left({u}-{1}\right)}\sqrt{{{u}}}{\left({\frac{{{d}{u}}}{{{2}}}}\right)}$$
$$\displaystyle={\frac{{{4}}}{{{2}}}}{\int_{{{u}={1}}}^{{{26}}}}{\left({u}-{1}\right)}\sqrt{{{u}}}{d}{u}$$
$$\displaystyle={2}{\int_{{{u}={1}}}^{{{26}}}}{\left({u}^{{{\frac{{{3}}}{{{2}}}}}}-{u}^{{\frac{{{1}}}{{{2}}}}}\right)}{d}{u}$$
Thus, evaluating the following line integral implies that
$$\displaystyle\int_{{c}}{x}{y}{d}{s}={2}{{\left[{\frac{{{u}^{{{\frac{{{5}}}{{{2}}}}}}}}{{{\frac{{{5}}}{{{2}}}}}}}-{\frac{{{u}^{{{\frac{{{3}}}{{{2}}}}}}}}{{{\frac{{{3}}}{{{2}}}}}}}\right]}_{{{u}={1}}}^{{{26}}}}$$
$$\displaystyle={2}{\left[{\frac{{{2}{u}^{{{\frac{{{5}}}{{{2}}}}}}}}{{{5}}}}-{\frac{{{2}{u}^{{{\frac{{{3}}}{{{2}}}}}}}}{{{3}}}}\right]}$$
$$\displaystyle={2}{\left[{\left({\frac{{{2}{\left({26}\right)}^{{\frac{{{5}}}{{{2}}}}}}}{{{5}}}}-{\frac{{{2}{\left({26}\right)}^{{\frac{{{3}}}{{{2}}}}}}}{{{3}}}}\right)}-{\left({\frac{{{2}{\left({1}\right)}^{{\frac{{{5}}}{{{2}}}}}}}{{{5}}}}-{\frac{{{2}{\left({1}\right)}^{{\frac{{{3}}}{{{2}}}}}}}{{{3}}}}\right)}\right]}$$
$$\displaystyle={\frac{{{8}}}{{{15}}}}{\left({949}\sqrt{{{26}}}+{1}\right)}$$
Result:
We have shown that the following line integral implies that
$$\displaystyle\int_{{c}}{x}{y}{d}{s}={\frac{{{8}}}{{{15}}}}{\left({949}\sqrt{{{26}}}+{1}\right)}$$
for the curve c expressed by the following parametric equations given by
$$\displaystyle{x}={t}^{{2}}$$
and $$\displaystyle{y}={2}{t}$$