Area between curves

**nath_quam** · July 1st 2006, 04:34 AM

How would i find the area between the curves y=cosx, the y axis and the lines y =1/2 and y=srt3/2......i keep getting (pi(2-srt3)-6(-1+srt3))/12 but i doubt this is right can someone do it please thanks nath

**TD!** · July 1st 2006, 07:50 AM

Your answer is correct, apart from the sign mistake (you get approx -0.296 while it should be +0.296).

**Soroban** · July 1st 2006, 08:52 AM

Hello, Nath!

Your answrer is negative . . . otherwise, excellent work!

How would i find the area between the curves $y = \cos x$ , the $y$ -axis
and the lines $y = \frac{1}{2}$ and $y = \frac{\sqrt{3}}{2}$
i keep getting: $\frac{\pi(2 - \sqrt{3}) - 6(-1 + \sqrt{3})}{12}$ . . . but i doubt this is right.

Code:

        |
        * *
    _   |     *
   √3/2 + - - - -*- -
        |::::::::::*
    1/2 + - - - - - * - -
        |
      --+------------*---
        |           π/2

I did it "sideways" . . . with respect to $y$ .

We have: . $y = \cos x\quad\Rightarrow\quad x = \arccos y$

The area is: . $A \;= \;\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}} \arccos y\ dy$

This integral can be done by-parts . . . or from a standard formula.

We get: . $A \;= \;y\cdot\arccos y - \sqrt{1 - y^2}\:\bigg|^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}$

We have: . $\left[\frac{\sqrt{3}}{2}\cdot\arccos\frac{\sqrt{3}}{2} - \sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2} \right]$ $-$ $\left[\frac{1}{2}\cdot\arccos\frac{1}{2} - \sqrt{1 - \left(\frac{1}{2}\right)^2}\right]$

. . . $= \;\left[\frac{\sqrt{3}}{2}\cdot\frac{\pi}{6} - \frac{1}{2}\right] - \left[\frac{1}{2}\cdot\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right] \;=$ . $\frac{\pi\sqrt{3}}{12} - \frac{1}{2} - \frac{\pi}{6} + \frac{\sqrt{3}}{2}$

. . . $= \;\frac{\pi\sqrt{3}}{12} - \frac{6}{12} - \frac{2\pi}{12} + \frac{6\sqrt{3}}{12} \;=$ . $\frac{\pi\sqrt{3} - 2\pi - 6 + 6\sqrt{3}}{12}$

Therefore: . $A\;=\;\frac{\pi(\sqrt{3} - 2) + 6(\sqrt{3} - 1)}{12}$ . . . ta-DAA!

**nath_quam** · July 1st 2006, 03:18 PM

Thanks i see my mistake for some reason i switched the terminals

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