- #1
kostoglotov
- 231
- 6
EDIT: I figured out my mistake...no option to delete silly post. Oh well.
1. Homework Statement
The problem is: use iterated integrals in polar form to find the area of one leaf of the rose-shaped curve r = cos(3*theta).
My setup agrees exactly with the solutions manual...but then something weird happens.
The solution manual and I both get to the same point, but then the solutions manual solves it with a trig identity, and I use the integral tables in the back...two different answers are arrived at. After looking through it again and again, I cannot find a calculation error.
I will begin where my solution and the solution manual's solutions diverge.
\int_{-\pi/6}^{\pi/6} \frac{1}{2}\cos^2{3\theta} \ d\theta = 2 \int_{0}^{\pi/6} \frac{1}{2}\left(\frac{1+\cos{6\theta}}{2}\right) d\theta;]<br /> <br /> 2 \int_{0}^{\pi/6} \frac{1}{2}\left(\frac{1+\cos{6\theta}}{2}\right) d\theta = \frac{1}{2}\left[\theta + \frac{1}{6}\sin{6\theta}\right]_{0}^{\pi/6} = \frac{\pi}{12};]<br /> <br /> No problem, I can see how all that works.<br /> <br /> <h2>Homework Equations</h2><br /> <br /> I used the integral table in the back that gives<br /> <br /> \int \cos^2{u} \ du = \frac{1}{2}u + \frac{1}{4}\sin{2u} + C ;]&lt;br /&gt; &lt;br /&gt; &lt;h2&gt;The Attempt at a Solution&lt;/h2&gt;&lt;br /&gt; &lt;br /&gt; Using this I get&lt;br /&gt; &lt;br /&gt; \int_{-\pi/6}^{\pi/6} \frac{1}{2}\cos^2{3\theta} \ d\theta = \frac{1}{2} \left[\frac{3\theta}{2} + \frac{1}{4}\sin{6\theta}\right]_{-\pi/6}^{\pi/6}&lt;br /&gt; &lt;br /&gt; \frac{1}{2} \left[\frac{3\theta}{2} + \frac{1}{4}\sin{6\theta}\right]_{-\pi/6}^{\pi/6} = \frac{1}{2}\left[\frac{\pi}{4} + \frac{1}{4}\sin{\pi} - \frac{-\pi}{4} + \frac{1}{4}\sin{\pi}\right] = \frac{1}{2}\left[\frac{\pi}{4}+\frac{\pi}{4}\right]&lt;br /&gt; &lt;br /&gt; \frac{1}{2}\left[\frac{\pi}{4}+\frac{\pi}{4}\right] = \frac{\pi}{4}&lt;br /&gt; &lt;br /&gt; \frac{\pi}{12} \neq \frac{\pi}{4}&lt;br /&gt; &lt;br /&gt; What is going on? Am I just too tired to see where I&amp;#039;ve made some minor mistake??
1. Homework Statement
The problem is: use iterated integrals in polar form to find the area of one leaf of the rose-shaped curve r = cos(3*theta).
My setup agrees exactly with the solutions manual...but then something weird happens.
The solution manual and I both get to the same point, but then the solutions manual solves it with a trig identity, and I use the integral tables in the back...two different answers are arrived at. After looking through it again and again, I cannot find a calculation error.
I will begin where my solution and the solution manual's solutions diverge.
\int_{-\pi/6}^{\pi/6} \frac{1}{2}\cos^2{3\theta} \ d\theta = 2 \int_{0}^{\pi/6} \frac{1}{2}\left(\frac{1+\cos{6\theta}}{2}\right) d\theta;]<br /> <br /> 2 \int_{0}^{\pi/6} \frac{1}{2}\left(\frac{1+\cos{6\theta}}{2}\right) d\theta = \frac{1}{2}\left[\theta + \frac{1}{6}\sin{6\theta}\right]_{0}^{\pi/6} = \frac{\pi}{12};]<br /> <br /> No problem, I can see how all that works.<br /> <br /> <h2>Homework Equations</h2><br /> <br /> I used the integral table in the back that gives<br /> <br /> \int \cos^2{u} \ du = \frac{1}{2}u + \frac{1}{4}\sin{2u} + C ;]&lt;br /&gt; &lt;br /&gt; &lt;h2&gt;The Attempt at a Solution&lt;/h2&gt;&lt;br /&gt; &lt;br /&gt; Using this I get&lt;br /&gt; &lt;br /&gt; \int_{-\pi/6}^{\pi/6} \frac{1}{2}\cos^2{3\theta} \ d\theta = \frac{1}{2} \left[\frac{3\theta}{2} + \frac{1}{4}\sin{6\theta}\right]_{-\pi/6}^{\pi/6}&lt;br /&gt; &lt;br /&gt; \frac{1}{2} \left[\frac{3\theta}{2} + \frac{1}{4}\sin{6\theta}\right]_{-\pi/6}^{\pi/6} = \frac{1}{2}\left[\frac{\pi}{4} + \frac{1}{4}\sin{\pi} - \frac{-\pi}{4} + \frac{1}{4}\sin{\pi}\right] = \frac{1}{2}\left[\frac{\pi}{4}+\frac{\pi}{4}\right]&lt;br /&gt; &lt;br /&gt; \frac{1}{2}\left[\frac{\pi}{4}+\frac{\pi}{4}\right] = \frac{\pi}{4}&lt;br /&gt; &lt;br /&gt; \frac{\pi}{12} \neq \frac{\pi}{4}&lt;br /&gt; &lt;br /&gt; What is going on? Am I just too tired to see where I&amp;#039;ve made some minor mistake??
Last edited: