Given the curve $\gamma(t)=\left( \frac{1}{2}t^2, \; \frac{1}{3}t^3 \right), \;\; t \in [0, \sqrt{3}]$ evaluate its length as well as the line integral
$$\oint_{\gamma} f(x, y)\,{\rm d}s$$
where $f(x, y)=1+2x$.
Solution
The length of the curve is given by the integral representation:
$$\mathcal{L}=\int_{0}^{\sqrt{3}}\sqrt{t^2+t^4}\, {\rm d}t$$
So we need to evaluate the above integral.However:
$$\begin{align*}
\int_{0}^{\sqrt{3}}\sqrt{t^2+t^4}\, {\rm d}t &=\int_{0}^{\sqrt{3}}t\sqrt{t^2+1}\, {\rm d}t \\
&= \frac{1}{2}\int_{0}^{\sqrt{3}}\left ( t^2+1 \right )' \sqrt{t^2+1}\, {\rm d}t\\
&= \frac{1}{2}\left [ \frac{2}{3}\left ( 1+t^2 \right )^{3/2} \right ]_0^{\sqrt{3}}\\
&= \frac{7}{3}
\end{align*} $$
For the evaluation of the line integral we are applying the formula. Hence:
$$\begin{aligned}
\oint_{\gamma}f(x, y)\, {\rm d}(x, y) &=\int_{0}^{\sqrt{3}}f\left ( \gamma(t) \right )\left \| \gamma'(t) \right \|\, {\rm d}t \\
&= \int_{0}^{\sqrt{3}}\left ( 1+t^2 \right )\sqrt{t^2 +t^4}\, {\rm d}t\\
&= \cdots \\
&= \frac{31}{5}
\end{aligned}$$
The last integral where the dots are presented can be dealt in a similar manner.
$$\oint_{\gamma} f(x, y)\,{\rm d}s$$
where $f(x, y)=1+2x$.
Solution
The length of the curve is given by the integral representation:
$$\mathcal{L}=\int_{0}^{\sqrt{3}}\sqrt{t^2+t^4}\, {\rm d}t$$
So we need to evaluate the above integral.However:
$$\begin{align*}
\int_{0}^{\sqrt{3}}\sqrt{t^2+t^4}\, {\rm d}t &=\int_{0}^{\sqrt{3}}t\sqrt{t^2+1}\, {\rm d}t \\
&= \frac{1}{2}\int_{0}^{\sqrt{3}}\left ( t^2+1 \right )' \sqrt{t^2+1}\, {\rm d}t\\
&= \frac{1}{2}\left [ \frac{2}{3}\left ( 1+t^2 \right )^{3/2} \right ]_0^{\sqrt{3}}\\
&= \frac{7}{3}
\end{align*} $$
For the evaluation of the line integral we are applying the formula. Hence:
$$\begin{aligned}
\oint_{\gamma}f(x, y)\, {\rm d}(x, y) &=\int_{0}^{\sqrt{3}}f\left ( \gamma(t) \right )\left \| \gamma'(t) \right \|\, {\rm d}t \\
&= \int_{0}^{\sqrt{3}}\left ( 1+t^2 \right )\sqrt{t^2 +t^4}\, {\rm d}t\\
&= \cdots \\
&= \frac{31}{5}
\end{aligned}$$
The last integral where the dots are presented can be dealt in a similar manner.
Let's deal with the integral
ReplyDelete\(\displaystyle{I=\int_{0}^{\sqrt{3}}\left(1+t^2\right)\,\sqrt{t^2+t^4}\,\mathrm{d}t}\) .
We have that :
\(\displaystyle{\begin{aligned}I&=\int_{0}^{\sqrt{3}}\left(1+t^2\right)\,\sqrt{t^2+t^4}\,\mathrm{d}t \\&=\int_{0}^{\sqrt{3}}t\,\left(1+t^2\right)\,\sqrt{1+t^2}\,\mathrm{d}t\\&=\int_{0}^{\sqrt{3}}t\,\left(1+t^2\right)^{3/2}\,\mathrm{d}t\\&=\dfrac{1}{2}\,\int_{0}^{\sqrt{3}}\left(1+t^2\right)^{3/2}\,\mathrm{d}(1+t^2)\\&=\left[\dfrac{1}{5}\,\left(1+t^2\right)^{5/2}\right]_{0}^{\sqrt{3}}\\&=\dfrac{1}{5}\,\left[(1+3)^{5/2}-1^{5/2}\right]\\&=\dfrac{1}{5}\,\left[\sqrt{4^{5}}-1\right]\\&=\dfrac{31}{5}\end{aligned}}\)
:D
ReplyDeleteThank you Vaggelis for presenting the calculations of the last integral.