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

• May 24th 2010, 06:43 AM
reaz
Two very irritating area problems. (polar, and restricted area)
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
• May 24th 2010, 12:53 PM
galactus
Quote:

Originally Posted by reaz
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.

$\displaystyle \frac{1}{2}\int_{0}^{2\pi}cos^{4}({\phi})d{\phi}$
• May 24th 2010, 02:32 PM
AllanCuz
Quote:

Originally Posted by 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

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 $\displaystyle x^3$
• May 25th 2010, 12:01 AM
reaz
... Englishman, Be he alive, or be he dead. I'll have his bones to grind my bread :D ^^

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