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This is exercise 7, of the book Engineering Mathematics by Stroud, Chapter 24, Further Problems section.

Here's a graph i made of the figure as i see it: enter image description here

It gives the answer as $π+8$. The integral i constructed is the following: $$ A=2\int_{0}^{π/2} \int_{0}^{2}r\ drdθ +2\int_{π/2}^{π} \int_{0}^{2+2cosθ} r\ drdθ $$

The first $r$ is the equation of the circle $r=2$, which dominates from $0$ to $\frac{π}{2}$ and the second $r$ is the equation of the cardioid $r=2+2cosθ$. I'm using the double integral method.. The answer i get by solving the expression above is $5π-8$.

What am i missing? What is wrong? Thanks in advance.

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  • $\begingroup$ Judging from the drawing, I wouldn't immediately think double integral, but rather an integral of the form $\frac12\int_{-\pi/2}^{3\pi/2}r(\theta)\,d\theta$. I would find it more intuitive to set up, although you should, for all intents and purposes, end up with the same integral ultimately. $\endgroup$
    – Arthur
    Jul 23, 2016 at 15:46

2 Answers 2

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Given the formula for the area in polar coordinates and the symmetry of the configuration, the area you want to compute is just: $$ 2\pi + \int_{\pi/2}^{\pi}\left[2\left(1+\cos\theta\right)\right]^2\,d\theta =\color{red}{5\pi-8}.$$

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    $\begingroup$ How can the area be just between $π/2$ and $π$? From what i see the area enclosed is symmetric about the x axis. So what i did is 2* the area from 0 to pi/2 + 2* the area from pi/2 to pi.. Isn't that logical? What am i missing here? $\endgroup$
    – KeyC0de
    Jul 23, 2016 at 16:05
  • $\begingroup$ I considered the sum between the area of the half-circle and twice the area of the cardioid sector enclosed by $\theta\in\left[\frac{\pi}{2},\pi\right]$. $\endgroup$ Jul 23, 2016 at 16:07
  • $\begingroup$ Hmm one more thing. You say you consider twice the area of the cardioid enclosed in $[π/2,π]$. But you square it.. you don't multiply it by 2.. $\endgroup$
    – KeyC0de
    Jul 23, 2016 at 16:28
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    $\begingroup$ The area of the cardiod sector enclosed by $\theta\in\left[\frac{\pi}{2},\pi\right]$ is given by $$\frac{1}{2}\int_{\pi/2}^{\pi}r(\theta)^2\,d\theta $$ with $r(\theta)=2(1+\cos\theta)$. That is the formula for the area in polar coordinates. $\endgroup$ Jul 23, 2016 at 16:30
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It is not referring to the area you show. By definition of "area enclosed" in polar coordinates, it should be the area enclosed for the same range of $\theta$, in other word I think the author is referring to the area on the right hand side:

$$A=\int_{-\pi/2}^{\pi/2}\left[\frac{1}{2}(2(1+\cos\theta))^2-\frac{1}{2}2^2\right] d\theta$$ $$=\int_{-\pi/2}^{\pi/2}\frac{1}{2}\left(4(1+\cos\theta)^2-4\right)d\theta$$ $$=\int_{-\pi/2}^{\pi/2}\frac{1}{2}(8\cos\theta+4\cos^2\theta)d\theta$$ $$=4[\sin\theta]_{-\pi/2}^{\pi/2}+2\int_{-\pi/2}^{\pi/2}\frac{1+\cos 2\theta}{2}d\theta$$ $$=8+\pi$$

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  • $\begingroup$ Oh, it's not the area i drew with brown color? Isn't that the "interesting" area between the two graphs? You say it is the area on the right hand side? Why is that? Can you please explain a bit more? $\endgroup$
    – KeyC0de
    Jul 23, 2016 at 16:02
  • $\begingroup$ I am not sure, does the author say from $-\pi/2$ to $\pi/2$? $\endgroup$
    – velut luna
    Jul 23, 2016 at 16:03
  • $\begingroup$ No it doesn't. The exercise is stated exactly as in the title of my question. $\endgroup$
    – KeyC0de
    Jul 23, 2016 at 16:04
  • $\begingroup$ Then I am not sure. According to my understanding, it should include the area at the left as well. $\endgroup$
    – velut luna
    Jul 23, 2016 at 16:09
  • $\begingroup$ Yes. Exactly as i thought. See my comments on the answer below. Is there a way to do it using double integrals? $\endgroup$
    – KeyC0de
    Jul 23, 2016 at 16:10

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