Science:Math Exam Resources/Courses/MATH101/April 2010/Question 02 (b)

MATH101 April 2010

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Question 02 (b)
Full-Solution Problem. Justify your answer and show all your work. Simplification of the answer is not required. Find the area of the region enclosed by one loop of the polar curve $r=4\cos(2\theta ).$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Can you sketch the polar curve for 0 ≤ θ ≤ 2π ? The question asks you to find the area of one loop, do you see it on your sketch?

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We can consider one loop of the polar curve to be defined by the values of $\theta$ in between successive points where $r=0$. Consider that $\cos(z)=0$, where $z=-\pi /2$ and $\pi /2$, so we can take $\theta$ to be vary in between $\theta =-\pi /4$ and $\pi /4$ to define a loop. Now consider that the area of a wedge with angle $\Delta \theta$ is equal to $r^{2}\Delta \theta /2$ where $r$ is the 'radius' of the wedge (i.e: from the point to the rounded boundary). To compute the total area of the loop we simply integrate in between the appropriate angles. ${\begin{aligned}A&=\int _{-\pi /4}^{\pi /4}{\frac {1}{2}}r(\theta )^{2}\,d\theta \\&=8\int _{-\pi /4}^{\pi /4}\cos ^{2}(2\theta )\,d\theta \end{aligned}}$ Using a trigonometric identity, we can simplify this integral so we are not dealing with the square of a trig function. ${\begin{aligned}A&=8\int _{-\pi /4}^{\pi /4}\cos ^{2}(2\theta )\,d\theta \\&=4\int _{-\pi /4}^{\pi /4}1+\cos(4\theta )\,d\theta \\&=2\pi \end{aligned}}$ Therefore, the area of one loop of the polar curve $r=4\cos(2\theta )$ is $2\pi$.

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