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I can graph this polar equation but using the polar curve integral formula, I can't seem to be able to find the answer. Also an additional question, what type of curve does this fall under? I can't seem to find much information on this.

enter image description here

AndroidV11
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2 Answers2

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The area of a region in polar coordinates defined by the equation $r=f(θ)$ with $α≤θ≤β$ is given by the integral $$\mathcal A=\frac 12∫^β_α[f(θ)]^2\,dθ$$

Now we have $$\mathcal A'=\frac 12∫^{\pi/2}_{0}\left[2 \tan \left(\frac{\theta}2\right)\right]^2\,dθ=\left[4\left(\tan\left(\dfrac{{\theta}}{2}\right)-\dfrac{{\theta}}{2}\right)\right]^{\pi/2}_{0}=4 − π$$ The figure has symmetry in relation to the $y$-axis then the area $\mathcal A=\color{magenta}{2}\mathcal A'=8-2\pi.$

Sebastiano
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    How did you arrive with pi over 2 and 0 as the limits? – AndroidV11 Jan 18 '23 at 13:48
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    In polar coordinate $\theta\in [0,2\pi[$. When $r=0$, so $2\tan(\theta/2)=0\iff \theta =0$. When $r=2$ (in the top), then $2\tan(\theta/2)=2\iff \theta=\pi/2$. Thus, the polar region is ${(r,\theta): 0\leqslant \theta\leqslant \pi/2, 0\leqslant r\leqslant 2\tan(\theta/2)}$. Anyway, that is one sheet, but its figure has symmetry, so the answer should be $2(4-\pi)$. – A. P. Jan 18 '23 at 14:32
  • @AndroidV11 Looking the picture. – Sebastiano Jan 18 '23 at 19:41
  • @A.P. You're right. I had forgotten the symmetry of the question. – Sebastiano Jan 18 '23 at 19:41
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    Is there a way to find the limits without having to worry too much about the graph? That's my concern since I was given the question during an exam and I really had no idea that the graph would look like that. The only thing I knew was to equate r = 0. Which led to my initial confusion since I thought the limits were 0 and pi as they were the two values that occurred when r = 0.

    Is the test for symmetry a good idea to use since you can perform it without looking at the graph? My idea is that if you are able to prove symmetry, you are able to reduce the limits to use. I think that is doable.

     – AndroidV11 Jan 18 '23 at 21:29
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    Additional question, why doesn't the test for symmetry with respect to the y-axis seem to work? I tried substituting theta with 180 minus theta and I can't seem to bring out the original equation to prove the test for symmetry. – AndroidV11 Jan 18 '23 at 21:45
  • @AndroidV11 You could easily see from the graph the limits of r and the tangent and the symmetry with respect to y. The operation you have to execute is the same as the one done by the user A. P. – Sebastiano Jan 19 '23 at 20:01
  • But is it possible to obtain the limits without the graph and through solving it? – AndroidV11 Jan 21 '23 at 02:46
  • @AndroidV11 The graph it is important.However the limit s are In polar coordinate $θ∈[0,2π[$ . When $r=0$ , so $2\tan(θ/2)=0⟺θ=0$ . When $r=2$ (in the top), then $2\tan(θ/2)=2⟺θ=π/2$ . Thus, the polar region is $(r,θ):0⩽θ⩽π/2,0⩽r⩽2\tan(θ/2)$. – Sebastiano Jan 21 '23 at 11:45
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Since we have the classic substitution of $r = 2\tan{\frac{\theta}{2}}$, we can express $x, y$ in terms of $r$ and integrate.

If $y = r\sin{\theta} \\ x= r\cos{\theta}$, and we have $\sin{\theta} = \frac{2\tan{\frac{\theta}{2}}}{1+\tan^2{\frac{\theta}{2}}} = \frac{r}{1+\frac{r^2}{4}} = \frac{4r}{4 + r^2} \\ \cos{\theta} = \frac{1-\tan^2{\frac{\theta}{2}}}{1+\tan^2{\frac{\theta}{2}}} = \frac{4 - r^2}{4 + r^2}$

then, $y(r) = \frac{4r^2}{4 + r^2} \\ x(r) = \frac{4r - r^3}{4 + r^2}$

It is easier to integrate w.r.t $y$, with one complete integral and, since $y = r$ at the intercepts, the bounds remain unchanged. That is,

$$2\int_{0}^{2}{x(y)dy} = 2\int_{0}^{2}{x(r)\frac{dy}{dr}dr} = 2\int_{0}^{2}{x(r)\frac{dy}{dr}dr} = 8 - 2\pi$$

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