abondantQ

2021-05-11

Evaluate the line integral, where C is the given curve
C xy ds
C: $x={t}^{2},y=2t,0\le t\le 5$

Dora

Integral solution:

Jeffrey Jordon

To evaluate the line integral ${\int }_{C}xyds$, where $C$ is the curve $C:x={t}^{2},y=2t,0\le t\le 5$, we first need to parameterize the curve in terms of a single parameter.
We can parameterize the curve by letting $x={t}^{2}$ and $y=2t$, which gives us the parametric equations $x={t}^{2}$ and $y=2t$, with $0\le t\le 5$.
Next, we need to express $ds$ in terms of $dt$. We have
$ds=\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt.$
Substituting $x={t}^{2}$ and $y=2t$ into this equation, we get
$ds=\sqrt{\left(2t{\right)}^{2}+\left(2{\right)}^{2}}dt=\sqrt{4{t}^{2}+4}dt=2\sqrt{{t}^{2}+1}dt.$
Now we can evaluate the line integral:
${\int }_{C}xyds={\int }_{0}^{5}\left({t}^{2}\right)\left(2t\right)\left(2\sqrt{{t}^{2}+1}\right)dt=4{\int }_{0}^{5}{t}^{3}\sqrt{{t}^{2}+1}dt.$
To evaluate this integral, we can use the substitution $u={t}^{2}+1$, which gives us $du=2tdt$. Substituting this into the integral, we get
${\int }_{0}^{5}{t}^{3}\sqrt{{t}^{2}+1}dt=\frac{1}{2}{\int }_{1}^{26}\left(u-1{\right)}^{3/2}du.$
Using the power rule for integration, we can evaluate this integral to get
$\frac{1}{2}{\int }_{1}^{26}\left(u-1{\right)}^{3/2}du=\frac{1}{2}·\frac{2}{5}\left(u-1{\right)}^{5/2}{|}_{1}^{26}=\frac{1}{5}\left(26\sqrt{677}-8\sqrt{2}-5\right).$
Therefore, the value of the line integral is $\overline{)\frac{1}{5}\left(26\sqrt{677}-8\sqrt{2}-5\right)}$.

RizerMix

The given line integral can be expressed as:

where $C$ is the curve defined by $x={t}^{2}$, $y=2t$, and $0\le t\le 5$.
To evaluate the line integral, we need to parameterize the curve in terms of a single variable. We can choose $t$ as the parameter, so that $x={t}^{2}$ and $y=2t$.
Then, the differential element of arc length $ds$ is given by:

Substituting for $x$, $y$, and $ds$ in the integral, we get:

Simplifying the integrand, we get:

To evaluate this integral, we can make the substitution $u={t}^{2}+1$, so that and ${t}^{2}=u-1$. Then, the integral becomes:

Evaluating the definite integral, we get:

Therefore, the value of the line integral is:

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