# Two very irritating area problems. (polar, and restricted area)

#### reaz

hi guys

1. A curve is given in polar coordinates of r = cos ^ 2 fi , 0 <= fi <= 2 pi
How big area does the place/area have which is enclosed by the curve?.

2. calculate the area of the restricted area which is located between the curves y = x^3 and y = x^4.

Ok as far as I am concerned you don't necesserily need to draw this in x and y axis do I? the thing is how to get to know the limits of integration analytically ? I am very bad at finding stuff like this in graphical solutions

#### galactus

MHF Hall of Honor
hi guys

1. A curve is given in polar coordinates of $$\displaystyle r = cos ^ {2}({\phi}) , \;\ 0 <= {\phi} <= 2 {\pi}$$
How big area does the place/area have which is enclosed by the curve?.
That is 'phi'(the Greek letter). Not 'fi', as in "fee-fi-fo-fum, I smell the blood of an....".

Wait, before I finish that, you aren't an Englishman, are you?. (Rofl)

Anyway, Just use the polar integration thingy.

$$\displaystyle \frac{1}{2}\int_{0}^{2\pi}cos^{4}({\phi})d{\phi}$$

#### AllanCuz

hi guys

1. A curve is given in polar coordinates of r = cos ^ 2 fi , 0 <= fi <= 2 pi
How big area does the place/area have which is enclosed by the curve?.

2. calculate the area of the restricted area which is located between the curves y = x^3 and y = x^4.

Ok as far as I am concerned you don't necesserily need to draw this in x and y axis do I? the thing is how to get to know the limits of integration analytically ? I am very bad at finding stuff like this in graphical solutions
For the second question,

$$\displaystyle \int_0^1 dx \int_{x^4}^{x^3} dy$$

Where the dx domain comes from equating $$\displaystyle x^3 = x^4$$

These are equal at 0 and 1. But notice on the interval $$\displaystyle 0 \le x \le 1$$ that $$\displaystyle x^4 \le x^3$$

So our dy domain is from $$\displaystyle x^4$$ to $$\displaystyle x^3$$

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#### reaz

... Englishman, Be he alive, or be he dead. I'll have his bones to grind my bread ^^

Anyway thanks for the hints I will try to proceed from what's been given by u guys.