Surface area of the circle

Question

I was told to calculate the surface area of the following circle by the integration method (monte carlo)

$x^2 + y^2 = 1$

The area of this circle is determined by the following inequalities: $-1 ≤ x ≤ 1$ and $ -\sqrt{1 − x^2} ≤ y ≤ \sqrt{1 − x^2}$.

Should I change the circle equation to the polar coordinates?

(exponentiation fixed)

Answer 1

An idea with polar coordinates:

$$\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dydx=\int_0^{2\pi}\int_0^1rdrd\theta=\left.2\pi\frac{r^2}2\right|_0^1=\pi$$

Of course, the above is just the area of a circle of radius $\;1\;$ , as proposed in your question. If you want a circle with radius $\;R\;$ just change accordingly $\;-R\le x\le R\;$ .