How do you find the area between the loop of #r=1+2costheta#?

Question

Ezekiel Alexander · Accepted Answer

Please see the explanation.

This reference Area with Polar Coordinates does a very similar exercise. The equation from the reference is: #Area = int_alpha^beta 1/2r^2 d theta# We know #r(theta)# but we need to find the value of #alpha and beta# The sample problem tells us that the loop starts at: #theta = (2pi)/3# and it ends at #theta = (4pi)/3# The integral for the area of the loop is #Area = int_((2pi)/3)^((4pi)/3) 1/2(1 + 2cos(theta))^2d theta# #Area = int_((2pi)/3)^((4pi)/3) 1/2(1 + 4cos(theta) + 4cos^2(theta))d theta# #Area = int_((2pi)/3)^((4pi)/3) 1/2(3 + 4cos(theta) + 2cos(2theta))d theta# #Area = 1/2(3theta + 4sin(theta) + sin(2theta))|_((2pi)/3)^((4pi)/3)# I am going to let you do the evaluation.

Luke Alvarez · Answer

undefined To find the area between the loop of the polar curve $r = 1 + 2\cos(\theta)$, you need to calculate the definite integral of $r$ squared with respect to $\theta$ over the appropriate interval. Since the loop repeats every $\pi$ radians, you would integrate from $\theta = 0$ to $\theta = \pi$. The formula for finding the area between a polar curve and the origin is:

$$A = \frac{1}{2} \int_{\alpha}^{\beta} [r(\theta)]^2 d\theta$$

Where $\alpha$ and $\beta$ are the initial and final angles, respectively, and $r(\theta)$ is the polar function.

So, for $r = 1 + 2\cos(\theta)$, the area would be:

$$A = \frac{1}{2} \int_{0}^{\pi} (1 + 2\cos(\theta))^2 d\theta$$

You would then evaluate this integral to find the area between the loop of the curve.