# area under parametric curve

I have difficulty on how to eliminate parameter especially the equation involved trigonometry equation.

The question is asking for the area bounded by the curve , the 2 axes and the line $$y=1$$.

$$x=4 \sin t$$

$$y=\cot t$$

where $$t$$ is in the range $$(0, \pi)$$. I have tried to use the trigonometric formula $$1+ \cot^2 t =\csc^2 t$$

but I got the wrong answer.The answer given is 4ln[(square root 2)+1].

• can you draw the curve? – jimjim Mar 1 '19 at 5:47
• The curve is not given by the question. – Lee Jason Mar 1 '19 at 6:01
• yes it is, it is parametrically given, can you draw it or not? – jimjim Mar 1 '19 at 6:03
• no, I don't know how to draw – Lee Jason Mar 1 '19 at 6:06
• put t=0 and find x and y, put t = 0.1 and find x and y, do a few steps or use wolfram alpha to draw it for you, it is fun. – jimjim Mar 1 '19 at 6:15

The Desmos plot of the relevant part of the curve looks like this:

The desired area is simply the area "under" (to the left of) the curve from $$y=0$$ to $$y=1$$.

We now express $$x$$ in terms of $$y$$. Since $$1+\cot^2t=\csc^2t=\frac1{\sin^2t}$$ and thus $$\frac1{1+\cot^2t}=\sin^2t$$, we have $$x=4\sin t=\frac4{\sqrt{1+\cot^2t}}=\frac4{\sqrt{1+y^2}}$$ So the desired area is $$\int_0^1\frac4{\sqrt{1+y^2}}\,dy=4[\sinh^{-1}y]_0^1=4\sinh^{-1}1=4\ln(1+\sqrt2)$$ The last equality is well-known; see e.g. OEIS A091648.