Different answers: integral table vs trig identity solutions

[kostoglotov]

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.

[tex] \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;]

[tex] 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};]

No problem, I can see how all that works.

Homework Equations

I used the integral table in the back that gives

[tex] \int \cos^2{u} \ du = \frac{1}{2}u + \frac{1}{4}\sin{2u} + C ;]

The Attempt at a Solution

Using this I get

$$\int_{-\pi/6}^{\pi/6} \frac{1}{2}\cos^2{3\theta} \ d\theta = \frac{1}{2} \left[\frac{3\theta}{2} + \frac{1}{4}\sin{6\theta}\right]_{-\pi/6}^{\pi/6}$$

$$\frac{1}{2} \left[\frac{3\theta}{2} + \frac{1}{4}\sin{6\theta}\right]_{-\pi/6}^{\pi/6} = \frac{1}{2}\left[\frac{\pi}{4} + \frac{1}{4}\sin{\pi} - \frac{-\pi}{4} + \frac{1}{4}\sin{\pi}\right] = \frac{1}{2}\left[\frac{\pi}{4}+\frac{\pi}{4}\right]$$

$$\frac{1}{2}\left[\frac{\pi}{4}+\frac{\pi}{4}\right] = \frac{\pi}{4}$$

$$\frac{\pi}{12} \neq \frac{\pi}{4}$$

What is going on? Am I just too tired to see where I've made some minor mistake??

 

[Ray Vickson]

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;]

[; 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};]

No problem, I can see how all that works.

Homework Equations

I used the integral table in the back that gives

[; \int \cos^2{u} \ du = \frac{1}{2}u + \frac{1}{4}\sin{2u} + C ;]

The Attempt at a Solution

Using this I get

[; \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} ;]

[; \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];]

[; \frac{1}{2}\left[\frac{\pi}{4}+\frac{\pi}{4}\right] = \frac{\pi}{4};]

[; \frac{\pi}{12} \neq \frac{\pi}{4} ;]

What is going on? Am I just too tired to see where I've made some minor mistake??

Please edit your post to make it readable; use "[t e x] ... tex-expressions.. [/t e x]" (no spaces in the word tex) instead of "[; ...tex -expressions... ;]"

For example, here is your first equation when done that way (but I did not feel like doing all the others).

$$\int_{-\pi/6}^{\pi/6} \frac{1}{2}\cos^2{3\theta} \ d\theta = 2 \int_{0}^{\pi/6} \frac{1}{2}\left(\frac{1+\cos{6\theta}}{2}\right) d\theta$$

 

[kostoglotov]

Please edit your post to make it readable; use "[t e x] ... tex-expressions.. [/t e x]" (no spaces in the word tex) instead of "[; ...tex -expressions... ;]"

For example, here is your first equation when done that way (but I did not feel like doing all the others).

$$\int_{-\pi/6}^{\pi/6} \frac{1}{2}\cos^2{3\theta} \ d\theta = 2 \int_{0}^{\pi/6} \frac{1}{2}\left(\frac{1+\cos{6\theta}}{2}\right) d\theta$$

Strange, I'm seeing normal Tex markup...must be TexTheWorld Plugin I use on Chrome. I thought I could copy paste from Reddit, but I guess not. Good to know.

I also figured out the mistake...u-sub for 3*theta...just rusty and a little weary. I was truly confused though. Thanks for taking the time.

 

[Mark44]

Strange, I'm seeing normal Tex markup...must be TexTheWorld Plugin I use on Chrome.

Before you edited your post you used [; and ;] as the delimiters. These might work with the plugin you're using, but they don't work here.

You can use [t e x] and [/t e x] as Ray suggested for standalone LaTeX, but I find it more convenient to use a pair of $ characters at the start and another pair at the end. For inline LaTeX, I use a pair of # characters at each end. This is easier than typing [i t e x] and [/ i t e x].

 

[HallsofIvy]

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.

[tex] \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;]

[tex] 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};]

No problem, I can see how all that works.

Homework Equations

I used the integral table in the back that gives

[tex] \int \cos^2{u} \ du = \frac{1}{2}u + \frac{1}{4}\sin{2u} + C ;]

The Attempt at a Solution

Using this I get

$$\int_{-\pi/6}^{\pi/6} \frac{1}{2}\cos^2{3\theta} \ d\theta = \frac{1}{2} \left[\frac{3\theta}{2} + \frac{1}{4}\sin{6\theta}\right]_{-\pi/6}^{\pi/6}$$

You are using that formula incorrectly. The formula is for $\int \cos^2(u)du$ and you have $\int\cos^2(3\theta) d\theta$. You could take $$u= 3\theta$$ but then $$du= 3 d\theta$$. So you need to "multiply and divide" the integral by 3:
$$\int cos^2(3\theta)d\theta= \frac{1}{3}\int cos^2(3\theta)(d(3\theta))$$.

Now, you can use that formula.

$$\frac{1}{2} \left[\frac{3\theta}{2} + \frac{1}{4}\sin{6\theta}\right]_{-\pi/6}^{\pi/6} = \frac{1}{2}\left[\frac{\pi}{4} + \frac{1}{4}\sin{\pi} - \frac{-\pi}{4} + \frac{1}{4}\sin{\pi}\right] = \frac{1}{2}\left[\frac{\pi}{4}+\frac{\pi}{4}\right]$$

$$\frac{1}{2}\left[\frac{\pi}{4}+\frac{\pi}{4}\right] = \frac{\pi}{4}$$

$$\frac{\pi}{12} \neq \frac{\pi}{4}$$

What is going on? Am I just too tired to see where I've made some minor mistake??