For a curve given by: $x=e^{-t}\cos{2t}$, $y=\sin t$

R is the region bounding this curve, the x axis and the y axis (y-intercept is point a and x-intercept is point b).

  1. Find the exact coordinates of a and b

  2. By considering $\int_0^ax~dy$, show that the exact area of region R is $\frac{3}{10}[\sqrt2e^{\frac{-3\pi}{4}}-e^{-\pi}]$

Alright, sadly I am extremely unfamiliar with parametric equations, so I don't even know how to do part 1. I assumed that for the y-intercept you just let $\sin t =0$ but that seems to give me the wrong answer. I resorted to typing it into a graphical calculator, and I did get the answers a=0.707 and b=1. 0.707 appears to be $\frac{\sqrt2}{2}$. I only did this so that I could proceed with the question, this is obviously not a viable answer.

On to the next part. Firstly, I rewrote the equation in terms of x and y, using the fact that $\cos(2\arcsin y) = 1-2y^2$

$y=\arcsin t \implies x=e^{-\arcsin y}-2y^2e^{-\arcsin y}$

Since we need the area of region R, I then considered

$\int_0^{\frac{\sqrt2}{2}}e^{-\arcsin y}-2y^2e^{-\arcsin y}~dy$

I then tried a u-substitution, where $u=\arcsin y$. Hence $y=\sin u$ and $dy=\cos u du$

The limits (which as I have mentioned earlier I am not sure about) then just become $\frac{\pi}{4}$ and 0. Hence we have:

$\int_0^{\frac{\pi}{4}}[e^{-u}-2e^{-u}\sin ^2u]\cos u ~du$. Which I have no idea how to integrate.

In conclusion: If indeed this is the right method (i doubt it), how do I proceed? Otherwise, I am really hoping that there is a much simpler method along the way, perhaps avoiding the $\arcsin y$.

Thank you very much!

UPDATE I know understand how to find the x and y intercepts, which still leaves the integral.

  • $\begingroup$ I dont know why putting $y=\sin t=0$ gives the wrong answer. In fact, x-intercept is $1$ (you get it by putting $y=0$) and y-intercept is $\frac1{\sqrt2}$. (by putting $x=0$) $\endgroup$
    – Sayan
    Commented Oct 30, 2014 at 2:17
  • $\begingroup$ Ahhh I see thank you very much! Will update the question $\endgroup$ Commented Oct 30, 2014 at 2:39
  • $\begingroup$ The number you say should be the area of the region is certainly not correct. If you graph the region, you see that it is roughly a triangle with base 1 and height $\sqrt{2}/2$, so its area is about 0.35, but the value you indicate is about 0.02. $\endgroup$ Commented Oct 30, 2014 at 2:56

1 Answer 1


The following figure shows a part of the curve in question:

enter image description here

As you can see there are many regions "bounded by the curve and the axes" (not all of them visible in the figure). Most likely the triangular region $R$ with corners $(0,0)$, $(1,0)$, and $(0,{1\over\sqrt{2}})$ is meant.

When a region is given by parametric representations of its boundary arcs the simplest way to compute its area is to use one of Green's formulas $${\rm area}(R)=\left\{\eqalign{&\int_{\partial R} x\>dy,\cr &-\int_{\partial R}y\>dx,\cr &{1\over2}\int_{\partial R}(x\>dy-y\>dx)\ .\cr}\right.$$ Note that we have to integrate counterclockwise along the full boundary of $R$. Since $\partial R$ consists of two segments on the axes and the arc $$\gamma:\quad t\mapsto\bigl(e^{-t}\cos(2t),\sin t\bigr)\qquad \biggl(0\leq t\leq{\pi\over4}\biggr)$$ of your curve we use the first of the above formulas, because along the axis segments we have either $x(t)\equiv0$ or $\dot y(t)\equiv0$, so that these segments give no contribution to the total boundary integral. It then follows that $${\rm area}(R)=\int_\gamma x\>dy=\int_0^{\pi/4}x(t)\>\dot y(t)\ dt=\int_0^{\pi/4}e^{-t}\cos(2t)\> \cos t \ dt={1\over10}\bigl(3+e^{-\pi/4}\bigr)\ .$$ While we are at it we can compute the area of the little loop $L$ to the left of the $x$-axis. This loop is bounded by the single arc $$\gamma':\quad t\mapsto\bigl(e^{-t}\cos(2t),\sin t\bigr)\qquad \biggl({\pi\over 4}\leq t\leq{3\pi\over4}\biggr)\ ,$$ but $\gamma'$ goes the wrong way around. Therefore we obtain $${\rm area}(L)=\int_{\partial L}x\>dy=-\int_{\gamma'}x\>dy=-\int_{\pi/4}^{3\pi/4}e^{-t}\cos(2t)\> \cos t \ dt={e^{-3\pi/4}\over 5\sqrt{2}}\bigl(e^{\pi/2}-3\bigr)\ .$$


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