Determine the area enclosed by the curve y = cos ^2 fi

**Riazy** · March 6th 2011, 09:56 AM

Hi guys you can see my problem attaced to the pictuse + my attempt to solve it.
and my question is how I can go further from there.

Determine the area enclosed by the curve y = cos ^2 fi-eul-001.jpg

**AllanCuz** · March 6th 2011, 10:02 AM

Originally Posted by Riazy

Hi guys you can see my problem attaced to the pictuse + my attempt to solve it.
and my question is how I can go further from there.

Click image for larger version.

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If you know what the curve looks like it becomes a bit easiar to evaluate the area. Look at cos^2 x - Wolfram|Alpha and let us know how you want to approach this now.

**Riazy** · March 6th 2011, 10:35 AM

If you can see my attempt, i would like if someone could complete and show me the euler step there, I want to learn how to solve this analytically. Thanks

**Soroban** · March 6th 2011, 10:45 AM

Hello, Riazy!

$\text{A polar curve is given: }\:r \,=\,\cos^2\!\theta,\;0 \le \theta \le 2\pi$

$\text{Calculate the area enclosed by the curve.}$

$\displaystyle \text{Formula: }\:A \;=\;\tfrac{1}{2}\!\!\int^{\beta}_{\alpha}\!\! r^2\,d\theta$

$\displaystyle \text{We have: }\;A \;=\;\tfrac{1}{2}\!\!\int^{2\pi}_0\!\! \cos^4\!\theta\,d\theta$

Why involve complex variables? . . . Why not standard Trig identities?

$\displaystyle \tfrac{1}{2}\cos^4\!\theta \;=\;\tfrac{1}{2}(\cos^2\!\theta)^2 \;=\;\tfrac{1}{2}\left(\frac{1+\cos2\theta}{2}\rig ht)^2 \;=\;\tfrac{1}{8}(1 + 2\cos2\theta + \cos^2\!2\theta)$

. . $=\;\frac{1}{8}\left(1 + 2\cos2\theta + \dfrac{1 + \cos4\theta}{2}\right) \;=\;\frac{1}{8}\left(\frac{3}{2} + 2\cos2\theta + \frac{1}{2}\cos4\theta\right)$

. . $=\;\frac{1}{16}(3 + 4\cos2\theta + \cos4\theta)$

$\displaystyle \text{We have: }\:A \;=\;\tfrac{1}{16}\!\int^{2\pi}_0 \!\!\left(3 + 4\cos2\theta + \cos4\theta \right)\,d\theta$

. . $=\;\frac{1}{16}\!\bigg[3\theta + 2\sin2\theta + \frac{1}{4}\sin4\theta\bigg]^{2\pi}_0$

. . $=\;\frac{1}{16}\bigg[\left(6\pi + 2\sin4\pi + \frac{1}{4}\sin8\pi\right) - \left(0 + 2\sin0 + \frac{1}{4}\sin0\right)\bigg]$

. . $=\;\;\dfrac{3\pi}{8}$

**Riazy** · March 6th 2011, 10:48 AM

Really helpfull, Soroban, Thanks a lot.

**Riazy** · March 7th 2011, 09:54 AM

Okay there is one thing I didnt understand, sorry but i dont consider this to be a bump as I have got my answer, but I would like to ask

how you got 1/8 into 1/16 I can see that you multiplied everything within the brackets with 2. but what about 1/16?

Thanks for answers!

**SammyS** · March 7th 2011, 08:34 PM

What Soroban did was to change 2 to 4/2, then factored out a 1/2 .

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