# Thread: area of polar curve

1. ## area of polar curve

find the area bounded by the spiral $\displaystyle r=ln(\theta)$ on the interval $\displaystyle \pi$ ≤ $\displaystyle \theta$ ≤ $\displaystyle 2 \pi$

(A) 2.405
(B) 2.931
(C) 3.743
(D) 4.810
(E) 7.487

2. The area of a polar curve is given by $\displaystyle A=\frac{1}{2}\int\mid{r}\mid^2\ d\theta$. You'll have to integrate by parts twice to evaluate the integral.

As a check, the area can be approximated by the area of a semicircle of radius $\displaystyle \frac{3\pi}{2}$, which is $\displaystyle \frac{1}{2}\pi\left(\frac{3\pi}{2}\right)^2=\frac{ 9}{8}\pi^3$.

Post again in this thread if you're still having trouble.

- Hollywood

3. Originally Posted by hollywood
The area of a polar curve is given by $\displaystyle A=\frac{1}{2}\int\mid{r}\mid^2\ d\theta$. You'll have to integrate by parts twice to evaluate the integral.

As a check, the area can be approximated by the area of a semicircle of radius $\displaystyle \frac{3\pi}{2}$, which is $\displaystyle \frac{1}{2}\pi\left(\frac{3\pi}{2}\right)^2=\frac{ 9}{8}\pi^3$.

Post again in this thread if you're still having trouble.

- Hollywood
$\displaystyle \frac{9}{8}\pi^3$ ≈ 34.882

problem is its not in one of the multiple choice options

4. My mistake - I left out the "ln". The correct value should be $\displaystyle \frac{1}{2}\pi\left(\ln\frac{3}{2}\pi\right)^2\app rox3.77$. But I think you're supposed to evaluate the integral:

$\displaystyle A=\frac{1}{2}\int_\pi^{2\pi}(\ln{\theta})^2\ d\theta$

which you can do by integrating by parts twice.

- Hollywood