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I just want to ask if the polar form of the unit circle is

$$\cos^2\theta+\sin^2\theta=1$$

So, if I were to try and find the area of the circle using an integral, I would get

$$\int_0^{2\pi}\left(\cos^2\theta+\sin^2\theta-1\right)d\theta$$ But this equals $0$; therefore, you can't find the area using polar coordinates. Or am I wrong?

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    $\begingroup$ The polar form for the unit circle is $r=1$. Thus the area is $\int_{\theta = 0}^{2\pi}\int_{r=0}^1 rdr d\theta=\int_{\theta = 0}^{2\pi} \frac 12 d\theta = \pi$. $\endgroup$
    – lulu
    Mar 10, 2017 at 18:58
  • $\begingroup$ There is a slight subtlety when you create the change of coordinates from rectangular to polar. You must involve a concept called the Jacobian matrix of the change of variables: en.wikipedia.org/wiki/Jacobian_matrix_and_determinant Moreover, the unit disk which you can use to describe the region you're trying to find the area of is the region such that $0\le r \le 1$ and $0 \le \theta \le 2\pi$. $\endgroup$
    – Decaf-Math
    Mar 10, 2017 at 19:01
  • $\begingroup$ I can't use a Jacobian here, the person is in Calc 2 that I'm trying to show this to. @pyrazolam $\endgroup$
    – Heavenly96
    Mar 10, 2017 at 19:01
  • $\begingroup$ Why use polar coordinates then? Just write the upper part of the circle as the graph of $y=\sqrt {1-x^2}$ and integrate that from $-1$ to $1$ (and then multiply by $2$ to get the lower half). $\endgroup$
    – lulu
    Mar 10, 2017 at 19:03
  • $\begingroup$ Well, I know that the area is just $\pi r^2$ but the professor requires polar for this question @lulu $\endgroup$
    – Heavenly96
    Mar 10, 2017 at 19:04

2 Answers 2

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@Heavenly96, see the very first formula in teal here:

The derivation for this is mainly due to an analogy with the area of a circle. What are the areas of a full, semi, quarter, and eighth circle respectively? They are $\pi r^2, {\pi\over2}r^2, {\pi\over4}r^2$, and ${\pi\over8}r^2$. Notice something interesting here? The area of each division ends up yielding half the angle required to find such area: $$A_{\Delta\theta}\approx{\Delta\theta\over2}r^2.$$ As $\Delta \theta \to 0$, summing up each of the pieces to gain the full area gives us the formula $$A_{r(\theta)} = \int_\alpha^\beta{d\theta\over2}r^2.$$

This is how the ${1\over 2}$ comes about.

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The equation of the unit circle is polar coorinates is $r =1$

Integration in polar coordinates... If you are tutoring someone who is not in multivariate calculus and you cannot formally introduce the Jacobian, you need to introduce it in a less formal way.

When you learned the Riemann integral in Cartesian space, you divided the area under the curve into a sequence of rectangles. For a fine enough partition, the sum of the rectangles would equal your area. The base of each rectangle is $dx,$ and the height is $y(x),$ the area of each is $y\ dx$ and the area of them all is $\int_a^b y\ dx$

Moving to polar, rather than rectangles you sections of circles. The area of each is $\frac 12 r^2 d\theta.$ and the area inside a polar curve is $\int_0^{2\pi} \frac 12 r^2 dr$ (usually, the limits are $0$ to $2\pi$, but not always, and so,must be checked).

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