Question

Line integral calculus. Help!!

Answer 1

The path
`C_2` is parameterized by

`\mathbf r(t)=(x(t),y(t))=((\cos t,\sin t))/(1+e^t)`

so that substituting
`x(t),y(t)` into
`\mathbf F` gives

`\mathbf F(x(t),y(t))=\frac{(\cos t,\sin t)}{(1+e^t)\sqrt{1-\frac1{(1-e^t)^2}}}`

Compute the differential of
`\mathbf r(t)`:

`\mathrm d\mathbf r=((-\sin t-e^t(\cos t+\sin t),\cos t+e^t(\cos t-\sin t)))/((1+e^t)^2)\,\mathrm dt`

The line integral then reduces to

`\displaystyle\int_(C_2)\mathbf F\cdot\mathrm d\mathbf r=\int_0^\infty-\frac{e^t}{(1+e^t)^3\sqrt{1-\frac1{(1+e^t)^2}}}\,\mathrm dt=-\int_0^\infty(e^t)/((1+e^t)^2√((1+e^t)^2-1))\,\mathrm dt`

To compute the integral, substitute
`s=1+e^t` and
`\mathrm ds=e^t\,\mathrm dt`, so the integration domain changes to the interval
`[2,\infty)`.

`\displaystyle\int_(C_2)\mathbf F\cdot\mathrm d\mathbf r=-\int_2^\infty(\mathrm ds)/(s^2√(s^2-1))`

Then substitute
`s=\sec r` and
`\mathrm ds=\sec r\tan r\,\mathrm dr`. Note that in order for this substitution to be reversible, we restrict
`r` to be between -π/2 and π/2, so that the integration domain changes to
`\left[\frac\pi3,\frac\pi2\right)`.

`\displaystyle\int_(C_2)\mathbf F\cdot\mathrm d\mathbf r=-\int_(\pi/3)^(\pi/3)(\sec r\tan r)/(\sec^2r√(\sec^2r-1))\,\mathrm dr=-\int_(\pi/3)^(\pi2)\cos r\,\mathrm dr=\boxed{-1+\frac{\sqrt3}2}`