1. ## find the area

I neeeed help ... important

SCHAUMS OUTLINES
THEORY AND PROBLEMS OF DIFFERENTIAL AND INTEGRAL
CHAPTER 34 - PLANE AREAS BY INTEGRAL
PROBLEM 15 -T

here .... the picture

find the area bounded by the given curve

2. $A = \int_0^\pi \left(x(t)^2 + y(t)^2\right) \, dt$

note that this cardioid can also be represented by the equation

$r = 2-2\cos{t}$

and the area of this cardioid is the same ...

$A = \int_0^\pi r(t)^2 \, dt$

... which leads me to believe that if you take the expressions for $x(t)$ and $y(t)$, square each, and find their sum, you'll get $r^2$.

I'll leave it for you to complete the necessary algebra/trig grunt work for confirmation.

3. thank you sir
I complete what you mention in the picture .. and I get the correct answer 6pi ,but to tell you the truth I havent study this way by using polar coordinates (if the teacher ask me why from 0 to pi !!! I have no idea).

but if you dont mind Ill show you what I did and give me your help ... 30 minutes later.

I am sorry about my English ..It isnt my native language ... I am studying it

4. let me start with the black area in the previos picture

could you please correct what I did
Is it right or wrong?

5. Originally Posted by rqeeb
thank you sir
I complete what you mention in the picture .. and I get the correct answer 6pi ,but to tell you the truth I havent study this way by using polar coordinates (if the teacher ask me why from 0 to pi !!! I have no idea).

$A = \int_0^{2\pi} \frac{r^2}{2} \, dt$

using symmetry (upper half of the cardioid has the same area as the lower half), so integrate from $0$ to $\pi$ and double the area ...

$A = 2\int_0^{\pi} \frac{r^2}{2} \, dt = \int_0^{\pi} r^2 \, dt$