What is the approach for integrating 2θsin2θ in a polar equation problem?

[rjs123]

Homework Statement

r = \theta + sin(2\theta) for 0 \le \theta \le \pi

a. Find the area bounded by the curve and the x-axis

b. Find the angle \theta that corresponds to the point on the curve with x-coordinate -2

c. For \frac{\pi}{3} < \theta <\frac{2\pi}{3} , \frac{dr}{d\theta} is negative. What does this fact say about r? What does this fact say about the curve?

d. Find the value of \theta in the interval 0 \le \theta \le \frac{\pi}{2} that corresponds to the point on the curve in the first quadrant with the greatest distance from the origin.

The Attempt at a Solution

a. 1/2\int_{0}^{\pi }(\theta + sin2\theta)^2 d\theta}


When I foil I end up with \theta^2 + 2\theta\sin2\theta + sin^2(2\theta)

this is the part where I believe I made the mistake. I know the area is 4.9348 by the use of my calculator...if it is indeed correct...I can easily integrate the 1st and 3rd term...but i think the second term needs to be done by parts if I am not mistaken.

 

[blather]

Parts should be ok to solve that. It's pretty straight forward. If you're familiar with the "ultra-violet voodoo" pneumonic, then let the \theta term be the "doo" (du) part.

 

[rjs123]

i seem to be having trouble doing the integration by parts

2\theta\sin2\theta

using uv - \int vdu

u = 2\theta dv = sin2\theta
du = 2d\theta v = \frac{-1}{2}cos2\theta

-\theta\cos2\theta + 1/2sin2\theta



the final integral from \theta^2 + 2\theta\sin2\theta + sin^2(2\theta) becomes:


\frac{1}{3}\theta^3-\theta\cos2\theta+ 1/2sin2\theta + \frac{1 - cos4\theta}{2}

from 0 to \pi

 

[blather]

Try

u=2\theta
\Rightarrow du=2d\theta
v=\sin 2\theta
\Rightarrow dv=-2\cos 2\theta

Then plug into your expression of

uv-\int v du

Don't forget to evaluate that first part with limits.

 

[rjs123]

Try

u=2\theta
\Rightarrow du=2d\theta
v=\sin 2\theta
\Rightarrow dv=-2\cos 2\theta

Then plug into your expression of

uv-\int v du

Don't forget to evaluate that first part with limits.

i'm pretty sure dv = sin2\theta

and its integral v = -1/2cos2\theta

 

[blather]

Yes, you're correct. But it is of no consequence because we don't need the differential of v.

 

[blather]

Wait. Hang on. No.

dv=-(1/2)\cos 2\theta d\theta

 

[rjs123]

Wait. Hang on. No.

dv=-(1/2)\cos 2\theta d\theta

you set u to one part, and dv to another

du is the differential of u,

v is the integral of dv.

I'll work on this tomorrow...ahhhh so frustrating.

 

[rjs123]

any help at integrating this?


2\theta\sin2\theta