# Area of region lying inside r(t) = sin(2t) and r(t) = cos(2t)

• Nov 7th 2010, 06:12 PM
scherz0
Area of region lying inside r(t) = sin(2t) and r(t) = cos(2t)
Dear all,

I am having trouble with the following exercise and would like to ask if the step underlined in red could be please explained.

I understand everything up to that point but don't understand why only $\sin 2\theta$ is considered in the integral. What about $\cos 2\theta$?

Also, what does the $8 \cdot 2\int^{\pi/8}_{0} \frac{1}{2} \sin^2 2\theta d\theta$ mean? Does it represent the fact that the wedge (highlighted in light green below in my graph) is multiplied by 16 times?

Thank you very much.

---

Question:
Find the area of the region that lies inside both $r = \sin 2\theta$ and $r = \cos 2\theta$.

Given solution: http://img830.imageshack.us/img830/9516/is2js10431.jpg

My graph: http://img838.imageshack.us/img838/4710/isjs10431.jpg

• Nov 7th 2010, 06:34 PM
skeeter
note the attached graph ...

$r = \sin(2t)$ in red from $0$ to $\frac{\pi}{8}$

$r = \cos(2t)$ in blue from $\frac{\pi}{8}$ to $\frac{\pi}{4}$

area swept out by $r = \sin(2t)$ from $0$ to $\frac{\pi}{8}$ equals the area swept out by $\cos(2t)$ from $\frac{\pi}{8}$ to $\frac{\pi}{4}$

so, taking advantage of the symmetry of the graph, 16 times one of those single areas is the desired area defined by the problem.

$\displaystyle 16 \int_0^{\frac{\pi}{8}} \frac{\sin^2(2t)}{2} \, dt = 16 \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \frac{\cos^2(2t)}{2} \, dt$

either one of the above will calculate the total area.
• Nov 9th 2010, 01:47 AM
tonio
Quote:

Originally Posted by skeeter
note the attached graph ...

$r = \sin(2t)$ in red from $0$ to $\frac{\pi}{8}$

$r = \cos(2t)$ in blue from $\frac{\pi}{8}$ to $\frac{\pi}{4}$

area swept out by $r = \sin(2t)$ from $0$ to $\frac{\pi}{8}$ equals the area swept out by $\cos(2t)$ from $\frac{\pi}{8}$ to $\frac{\pi}{4}$

so, taking advantage of the symmetry of the graph, 16 times one of those single areas is the desired area defined by the problem.

$\displaystyle 16 \int_0^{\frac{\pi}{8}} \frac{\sin^2(2t)}{2} \, dt = 16 \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \frac{\cos^2(2t)}{2} \, dt$

either one of the above will calculate the total area.

Pretty nice, but I think the OP's problem, and mine as well, is to justify why we integrate $\frac{\sin^22t}{2}}$ to

obtain that area, since that function doesn't appear at all in the given data...

Tonio
• Nov 9th 2010, 07:07 AM
skeeter
Quote:

Originally Posted by tonio
Pretty nice, but I think the OP's problem, and mine as well, is to justify why we integrate $\frac{\sin^22t}{2}}$ to

obtain that area, since that function doesn't appear at all in the given data...

Tonio

Since this was not a simple/beginner type area problem, I made the assumption that the OP was familiar with the basics of finding the area enclosed by a polar curve ...

$\displaystyle A = \int_{\theta_1}^{\theta_2} \frac{[r(\theta)]^2}{2} \, d\theta$
• Nov 9th 2010, 05:02 PM
scherz0
Thank you for your response, skeeter.

I can see by inspection that

Quote:

Originally Posted by skeeter
...

the area swept out by $r = \sin(2t)$ from $0$ to $\frac{\pi}{8}$ equals the area swept out by $\cos(2t)$ from $\frac{\pi}{8}$ to $\frac{\pi}{4}$

so, taking advantage of the symmetry of the graph, 16 times one of those single areas is the desired area defined by the problem ...

.

However, if I were not sure about the symmetry of your graph, what computation could I perform to verify this symmetry? Must I compute $\displaystyle \int_{0}^{\frac{\pi}{8}} \sin(2t) dt$ and $\displaystyle \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \cos(2t) dt$ or is there some easier method?
• Nov 9th 2010, 05:15 PM
skeeter
Quote:

Originally Posted by scherz0
Thank you for your response, skeeter.

I can see by inspection that

.

However, if I were not sure about the symmetry of your graph, what computation could I perform to verify this symmetry? Must I compute $\displaystyle \int_{0}^{\frac{\pi}{8}} \sin(2t) dt$ and $\displaystyle \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} \cos(2t) dt$ or is there some easier method?

$\sin(2t)$ and $\cos(2t)$ are not the integrands ...

$\frac{\sin^2(2t)}{2}$ and $\frac{\cos^2(2t)}{2}$ are the integrands for the area.