# Thread: 243.11.5.9 Area of intersection cardioid and circle

1. OK just seeing if this is setup OK
before I pursue all the steps
I thot adding areas would be easier

2. I see it like so ...

$\displaystyle \int_0^\pi \dfrac{8^2}{2} \, d\theta + \int_\pi^{2\pi} \dfrac{[8(1+\sin{\theta})]^2}{2} \, d\theta$

... note there are also opportunities to use symmetry.

Originally Posted by skeeter
I see it like so ...

$\displaystyle \int_0^\pi \dfrac{8^2}{2} \, d\theta + \int_\pi^{2\pi} \dfrac{[8(1+\sin{\theta})]^2}{2} \, d\theta$

... note there are also opportunities to use symmetry.
I'll try the symmetry .... half the limits mult by 2

$\displaystyle A=2\left[\int_0^{\pi/2} 64\, d\theta + \int_{3\pi/2}^{2\pi} [8(1+\sin{\theta})]^2 \, d\theta\right]$

$128\left[\displaystyle \left[\theta\right]_0^{\pi/2} +\left[\theta-2cos\theta-\dfrac{\sin\left(2x\right)-2x}{4}\right]_{3\pi/2}^{2\pi}\right]$

sorry I just can't get this the bk ans is $16(5\pi - 8)$

4. I would use symmetry and known the area of a semicircle to write:

$\displaystyle A=\frac{8^2}{2}\left(\pi+2\int_{-\frac{\pi}{2}}^{0}\left(1+\sin(\theta)\right)^2\,d\theta\right)=32\left(\pi+\frac{3}{2}\pi-4\right)=16(5\pi-8)$

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