Area inside loop of polar equation, unsolvable problem?

Question

Is this problem solvable?

"Please find the area inside the first loop of the following equation (using polar coordinates): r = cos$(\theta)$ - sec$(\theta)$." From what I can tell, this function does not loop at all: https://www.wolframalpha.com/input/?i=r+%3D+cos%28%CE%B8%29+-+sec%28%CE%B8%29

Answer 1

It doesn't loop, and centered at the origin, the area swept out between $\theta=-\frac\pi2$ and $\theta=\frac\pi2$ would be infinite. However, the area swept out between the curve and the asymptote at $x=-1$, would be $$ \begin{align} &\int_{-\pi/2}^{\pi/2}\left[\frac12\sec^2(\theta)-\frac12\left(\cos(\theta)-\sec(\theta)\right)^2\right]\,\mathrm{d}\theta\\ &=\frac12\int_{-\pi/2}^{\pi/2}\left[2-\cos^2(\theta)\right]\,\mathrm{d}\theta\\ &=\frac12\int_{-\pi/2}^{\pi/2}\left[2-\frac{1+\cos(2\theta)}2\right]\,\mathrm{d}\theta\\ &=\frac14\int_{-\pi/2}^{\pi/2}\left[3-\cos(2\theta)\right]\,\mathrm{d}\theta\\ &=\frac{3\pi}4 \end{align} $$

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The implicit equation for this curve is $$ y^2(1+x)+x^3=0 $$ We can verify the area $$ \begin{align} 2\int_{-1}^0\sqrt{\frac{-x^3}{1+x}}\,\mathrm{d}x &=2\int_0^1\sqrt{\frac{x^3}{1-x}}\,\mathrm{d}x\\ &=2\int_0^1x^{3/2}(1-x)^{-1/2}\,\mathrm{d}x\\[9pt] &=2\,\mathrm{B}(\tfrac52,\tfrac12)\\[6pt] &=2\,\frac{\Gamma(\frac52)\Gamma(\frac12)}{\Gamma(3)}\\ &=2\,\frac{\frac34\sqrt\pi\cdot\sqrt\pi}2\\[9pt] &=\frac{3\pi}4 \end{align} $$

Answer 2

I agree there is no loop. Such a curve is a Cissoid of Diocles.