Curve and line integrals

Let $\gamma$ be defined as $\gamma(t) = e^{-t} (\cos t, \sin t ), \;\; t \geq 0$.

a) Sketch the graph of the curve.
b) Evaluate the line integrals:

$$ \begin{matrix} &  \displaystyle ({\rm i})\; \oint_{\gamma}\left ( x^2 +y^2 \right )\, {\rm d}s &  & ({\rm ii}) \displaystyle \oint_{\gamma} (-y, x)\cdot {\rm d}(x, y) \end{matrix}$$

Solution



a) The graph of the curve is part of the log. spiral and is depicted at the image below.

 To get a better view of why this is the graph we can re-write the curve as $\displaystyle \gamma(t)=\left \| \gamma(t) \right \|e^{it}$ where $\left \| \gamma(t) \right \|= e^{-t}$. As $t \rightarrow +\infty$ the "measure" of the curve tends to $0$. Hence we get the desired graph (black in the image)

b)
(i) We have that:

$$ \begin{aligned} \oint_{\gamma} \left(x^2+y^2\right)\,\mathrm{d}s&=\int_{0}^{\infty}\left(x^2(t)+y^2(t)\right)\,\||\gamma^\prime(t)||\,\mathrm{d}t\\&=\int_{0}^{\infty}e^{-2\,t}\,\sqrt{2}\,e^{-t}\,\mathrm{d}t\\&=\sqrt{2}\,\int_{0}^{\infty}e^{-3\,t}\,\mathrm{d}t\\&=\left[-\dfrac{\sqrt{2}}{3}\,e^{-3\,t}\right]_{0}^{\infty}\\&=\dfrac{\sqrt{2}}{3}\end{aligned}$$

(ii) We have that:

$$ \begin{aligned} \oint_{\gamma} \left(-y,x\right)\cdot d(x,y)&=\int_{0}^{\infty}\left(-y(t),x(t)\right)\cdot \gamma^\prime(t)\,\mathrm{d}t\\&=\int_{0}^{\infty}-e^{-t}\cdot (-e^{-t})\,\mathrm{d}t\\&=\int_{0}^{\infty}e^{-2\,t}\,\mathrm{d}t\\&=\left[-\dfrac{e^{-2\,t}}{2}\right]_{0}^{\infty}\\&=\dfrac{1}{2}\end{aligned}$$