I neeeed help ... important
SCHAUM`S 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
I neeeed help ... important
SCHAUM`S 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
$\displaystyle A = \int_0^\pi \left(x(t)^2 + y(t)^2\right) \, dt$
note that this cardioid can also be represented by the equation
$\displaystyle r = 2-2\cos{t}$
and the area of this cardioid is the same ...
$\displaystyle A = \int_0^\pi r(t)^2 \, dt$
... which leads me to believe that if you take the expressions for $\displaystyle x(t)$ and $\displaystyle y(t)$, square each, and find their sum, you'll get $\displaystyle r^2$.
I'll leave it for you to complete the necessary algebra/trig grunt work for confirmation.
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 haven`t study this way by using polar coordinates (if the teacher ask me why from 0 to pi !!! I have no idea).
but if you don`t mind I`ll show you what I did and give me your help ... 30 minutes later.
I am sorry about my English ..It isn`t my native language ... I am studying it
$\displaystyle A = \int_0^{2\pi} \frac{r^2}{2} \, dt$
using symmetry (upper half of the cardioid has the same area as the lower half), so integrate from $\displaystyle 0$ to $\displaystyle \pi$ and double the area ...
$\displaystyle A = 2\int_0^{\pi} \frac{r^2}{2} \, dt = \int_0^{\pi} r^2 \, dt$