# Area of the region

Find the area of the region bounded by: $$r^2 = 128*cos(2\theta)$$

$$\int 1/2*(\sqrt{128*cos(2\theta)})^2$$

that's my integral, but i dont know what the bounds are. i tried typing it into my calculator, but i could only find one bound(.78539816) and im not even sure if this is even correct. can someone help me find the bounds?

AKG
Homework Helper
You know that cosine is periodic in 2pi, so theta can go from 0 to 2pi. Just by looking at it, you should be able to see that it will be a four leaf clover, and that each of the leaves will have the same area, so the total area will be 4 x the area of one. You should also be able to see that, since cosine is zero at pi/2 and 3pi/2, that one of the clover leafs will be from pi/4 to 3pi/4. And don't worry about r being negative, since the leaf created by taking r negative and theta going from pi/4 to 3pi/4 is just the leaf taking r positive going form 5pi/4 to 7pi/4, which you're already counting when taking the area of the one leaf and multiplying it by 4.

So your answer will be 4 x area of one leaf, where r is positive and theta ranges from pi/4 to 3pi/4.

ehild
Homework Helper
ILoveBaseball said:
Find the area of the region bounded by: $$r^2 = 128*cos(2\theta)$$

$$\int 1/2*(\sqrt{128*cos(2\theta)})^2$$

that's my integral, but i dont know what the bounds are.

r^2 can not be negative, so

$$-\pi/4\le \theta \le \pi/4$$ and $$3\pi/4\le \theta \le 5\pi/4$$

ehild

Last edited:
AKG
Homework Helper
How could you possibly get -64? I don't remember the formula for finding the area bounded by a curve given in polar co-ordinates, but if the formula you're using is correct, your integrand is strictly positive (it's a squared number (positive) times 1/2 (positive)).

Oh, I can't believe I didn't notice this. Something is wrong (maybe) since cosine will take on negative values on that region, but r² is positive, so I suppose it is implied that theta will only range over values where the equation makes sense. Cosine is non-negative from 3pi/2 to pi/2, so integrate from 3pi/4 to pi/4 (and not the other way around. Now, I'm guessing that if you're given r² = 128*cos(theta), maybe they want you to take the positive and negative values of r, so you'll get a 2-leafed clover. The area you need is then 2 x the area from 3pi/4 to pi/4 (of course, use 5pi/4, not pi/4), and that area is just 1/2r² d(theta), and r² is given to you, you don't have to write it as the square of a square root. You need to compute:

$$2\int _{\frac{3\pi }{4}} ^{\frac{5\pi }{4}} \frac{1}{2}r^2\, d\theta$$

$$= \int _{\frac{3\pi }{4}} ^{\frac{5\pi }{4}} 128\cos (2\theta)\, d\theta$$

$$= 64\sin (2\theta)|_ {\theta = \frac{3\pi }{4}} ^{\theta = \frac{5\pi }{4}}$$

$$= 64\left (\sin \left (\frac{pi}{2}\right ) - \sin \left (\frac{3\pi }{2}\right )\right )$$

$$= 128$$