Let 
be defined as
(i) Sketch the graph of .
(ii) Evaluate the line integrals:
Solution
(i) The graph of the curve is depicted below:
(ii)
(i) For the first integral we have successively:
![\begin{align*} \oint \limits_{\gamma} \left(x^2+y^2\right)\,\mathrm{d}s&=\int_{0}^{\infty}\left(x^2(t)+y^2(t)\right)\,\||\gamma^\prime(t)||\,\mathrm{d}t\\ &=\int_{0}^{\infty}e^{-2\,t}\,\sqrt{2}\,e^{-t}\,\mathrm{d}t\\ &=\sqrt{2}\,\int_{0}^{\infty}e^{-3\,t}\,\mathrm{d}t\\ &=\left[-\dfrac{\sqrt{2}}{3}\,e^{-3\,t}\right]_{0}^{\infty}\\ &=\dfrac{\sqrt{2}}{3} \end{align*}](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-a69954bcfd193cd923ce1ccc556c7b8a_l3.png "Rendered by QuickLaTeX.com")
(ii) For the second integral we have successively:
![\begin{align*} \oint \limits_{\gamma} \left(-y,x\right)\cdot d(x,y)&=\int_{0}^{\infty}\left(-y(t),x(t)\right)\cdot \gamma^\prime(t)\,\mathrm{d}t\\ &=\int_{0}^{\infty}-e^{-t}\cdot (-e^{-t})\,\mathrm{d}t\\ &=\int_{0}^{\infty}e^{-2\,t}\,\mathrm{d}t\\ &=\left[-\dfrac{e^{-2\,t}}{2}\right]_{0}^{\infty}\\ &=\dfrac{1}{2} \end{align*}](https://www.math.tolaso.com.gr/wp-content/ql-cache/quicklatex.com-44be5bc7399f7333f043e861eccbeb62_l3.png "Rendered by QuickLaTeX.com")
This exercise was an exam’s question somewhere in Greece.