2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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

enter image description here

$\endgroup$
5
  • $\begingroup$ Hi, thank you for your response. I appreciate your solution to the problem; however, ultimately my question was to understand my method didn't work. If you could take a look at my solution I would greatly appreciate it. $\endgroup$
    – steven
    Jan 19 '18 at 21:51
  • $\begingroup$ @steven ok, I'will take a look! $\endgroup$
    – user
    Jan 19 '18 at 22:07
  • $\begingroup$ @steven I think your method should work, you only should add $\frac12$ in front to the integral! $\endgroup$
    – user
    Jan 19 '18 at 22:25
  • $\begingroup$ Thank you so much! :) $\endgroup$
    – steven
    Jan 20 '18 at 21:03
  • $\begingroup$ @steven You are welcome, Bye! $\endgroup$
    – user
    Jan 20 '18 at 21:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.