I need some help with this problem. I started with graphing the polar equation: The table of values are: 0 -> 5, $[\pi]

The table of values generated by calculator are: $0 \to 5$, $\pi/2 \to 2$, $\pi \to -1$, and $3\pi/2 \to 2$

I know that the angle of $-1$ is $\pi$. So, that is my lower bound. Now, I need to determine the angle at the pole by letting $r = 0$.

$r = 2 + 3\cos\theta$

$0 = 2 + 3\cos\Theta $

$-2 = 3\cos\Theta $

$\Theta =\cos^{-1}\left(\frac{-2}{3}\right)$

Then, I inputted the following in my calculator: $A = 2(\frac{1}{2})\int_{\cos^{-1}(\frac{-2}{3}) }^{\pi}(2+3\cos\Theta )^{2}d\Theta $

I put $2$ in the front because the area to be solved is just the half of the inner loop. To get the whole area, I multiplied it to two. In doing so, it returned an answer of approximately $0.44$ units squared.

Is this correct?


1 Answer 1


Yes, it is correct!

I would just write the calculations a bit clearer..

The origin is given by $r=0$ and parametrized on the curve by $\theta=\arccos(-\frac{2}{3})$. The turning point in the inner circle is given by $\theta=\pi$. Thus, one can compute the area as follows

\begin{align} A &= 2 \int_{\arccos(-\frac{2}{3})}^\pi \int_0^{2+3\cos \theta} r \, dr d\theta = 2 \int_{\arccos(-\frac{2}{3})}^\pi \frac{1}{2} (2+3\cos \theta)^2 \, d\theta \\ &= -3 \sqrt{5} + \frac{17}{2} \arccos{\frac{2}{3}} \approx 0.4409 \end{align}


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