double integral of a circle question

Question

Consider a circle of radius $R$, with centre at the origin. Express the area of the circle as an appropriate double integral over Cartesian coordinates, and evaluate this integral.

So a circle that can be described as $x^2+y^2=R^2$.

The answer starts with $$A=\int_{-R}^R\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}1\ \mathrm{d}y\mathrm{d}x$$

Where does the $1$ come from here? I assumed it would be $x^2+y^2$ or $R^2$, not $1$.

Answer 1

If the area of the circle is $A$, the double integral expression of $A$ over Cartesian coordinates is: $$A=\int_{-R}^R\int_{-\sqrt{R^2-y^2}}^{\sqrt{R^2-y^2}}\mathrm dx\mathrm dy= \int_{-R}^R2\sqrt{R^2-y^2}\mathrm dy$$

Answer 2

The area differential is dxdy and there is no function involved, so the evaluation is defined by the integration limits only.