OK just seeing if this is setup OK
before I pursue all the steps
I thot adding areas would be easier
OK just seeing if this is setup OK
before I pursue all the steps
I thot adding areas would be easier
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.
Last edited by skeeter; September 6th, 2017 at 23:10.
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 $16(5\pi - 8)$
Last edited by karush; September 7th, 2017 at 14:58.
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)$