Consider the region bounded by the cardioid


and the two lines $y = x, y = −x$ to the right of the y -axis. Sketch this region. Determine the angles of the two points of intersection of the lines with the cardioid. Compute the area of the region.

Im supposed to plot all 3 in terms of polar coordinates, but I translate $y=x$ to $\sin(\theta)=\cos(\theta)$ and am confused to move further. How do I solve this type of problem?

So far, the double integral needed to be solved is



  • $\begingroup$ $y=x$ can be written as $\theta=\pi/4$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 3 '14 at 18:46
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    $\begingroup$ BTW, please notice my edits. Where you had $sin\theta$ and $cos\theta$, I changed them to $\sin\theta$ and $\cos\theta$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 3 '14 at 18:47
  • $\begingroup$ +1, added your insight to my question to further try and crank this out $\endgroup$ – user3507072 Nov 3 '14 at 18:52
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    $\begingroup$ You need $\displaystyle\int_{-\pi/4}^{\pi/4}$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 3 '14 at 18:54
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    $\begingroup$ Looking good so far, just go on integrating :) $\endgroup$ – dietervdf Nov 3 '14 at 19:04

\begin{align} & \int_{-\pi/4}^{\pi/4}\int_0^{2(1-\sin(\theta))}r\,dr\,d\theta = \int_{-\pi/4}^{\pi/4} 2(1-\sin\theta)^2\, d\theta \\[6pt] = {} & 2\int_{-\pi/4}^{\pi/4} (1 - 2\sin\theta + \sin^2\theta)\, d\theta. \end{align} If it were $\displaystyle\int_{-\pi/2}^{\pi/2} \sin^2\theta\,d\theta$ I'd use an intelligent method, but as it is (with $\pm\pi/4$ rather than $\pm\pi/2$) I'd probably just go with the half-angle formula: $\sin^2\theta = \dfrac 1 2 - \dfrac 1 2 \cos(2\theta)$ and plod through it.

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  • $\begingroup$ Yes! The half angle formula was what I was missing. $\endgroup$ – user3507072 Nov 3 '14 at 19:06
  • $\begingroup$ Intelligent method, could you tell me more? $\endgroup$ – dietervdf Nov 3 '14 at 19:08
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    $\begingroup$ By "intelligent method" I meant this: Observe that on $0\le\theta\le\pi/2$, the shapes of the graphs of $\cos^2\theta$ and $\sin^2\theta$ are identical except that left and right are reversed, and similarly on $-\pi/2\le\theta\le0$. Therefore $\displaystyle\int_{-\pi/2}^{\pi/2} \sin^2\theta\,d\theta=\int_{-\pi/2}^{\pi/2} \cos^2\theta\,d\theta$. But their sum is $\displaystyle\int_{-\pi/2}^{\pi/2} 1\,d\theta$, and that's just the area of a rectangle. So each of the two integrals is just half the area of the rectangle. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 3 '14 at 19:13

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