Problem
$r = 4 + 6\cdot cos(\theta)$
$$A =0.5 \cdot\int r^2 d\theta$$
Find the area in between the inner and outer loop.
My Solution
$$\int\limits_0^{2\pi} \big((4+6\cdot cos(\theta)\big)\cdot |4+6\cdot cos(\theta)| \, d\theta $$
I think this should work because when theta is in the inner loop, the radius is going to be negative. By putting the absolute value function, we can ensure that when theta is in the loop, the area will instead be subtracted instead of double counted if we computer the area from 0->2pi. I think my method should work.
Question
I calculated the value using a TI-Nspire, but I got the value that represents the area from 0->2pi, meaning it double counted the inner loop. Can someone point out why my method dosen't work? Or does the calculator simply not know how to compute a definite integral with an absolute value?
Your method is very clever and correct, you only need to add the factor $\frac12$ in front to the integral to obtain
$$\frac12\int\limits_0^{2\pi} \big((4+6\cdot cos(\theta)\big)\cdot |4+6\cdot cos(\theta)| \, d\theta \approx 103.3$$
Integral evaluation
As an alternative, we have that
$$r = 4 + 6\cdot cos \theta=0 \implies \cos \theta = -\frac23 \implies\theta_1=\arccos \left(-\frac23\right), \,\theta_2=2\pi-\arccos \left(-\frac23\right)$$
$$r(0)=10, r(\pi/2)=4, r(\theta_1)=0, r(\pi)=2$$
thus the area between the inner and outer loop can be calculated by
$$S=2(A_1-A_2)\approx 103.3$$
$$A_1=\frac12 \cdot\int_0^{\theta1} r^2 d\theta \quad A_2=\frac12 \cdot\int_{\theta1}^{\pi} r^2 d\theta$$
Parametric plot
Integrals evaluation