# Tough definite integration

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.

• 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 $x=0$) Commented Oct 30, 2014 at 2:17
• Ahhh I see thank you very much! Will update the question Commented Oct 30, 2014 at 2:39
• 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. Commented Oct 30, 2014 at 2:56

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 $${\rm area}(L)=\int_{\partial L}x\>dy=-\int_{\gamma'}x\>dy=-\int_{\pi/4}^{3\pi/4}e^{-t}\cos(2t)\> \cos t \ dt={e^{-3\pi/4}\over 5\sqrt{2}}\bigl(e^{\pi/2}-3\bigr)\ .$$