Write an iterated integral (dy dx) for a region in the first quadrant bounded by a circle of radius 1 centered at the origin

Question

Write an iterated integral in the form dy dx for the region in the first quadrant bounded by a circle with a radius of 1 centered at the origin. So I know the equation or the circle would be $x^2$+$y^2$=1. So would I want my dx bounds to be 0 to $\sqrt{1-y^2}$ and my dy bounds to be 0 to 1?

Answer 1

Since $x^2 + y^2 = 1$, solving for $y$ gives $y = \pm\sqrt{1 - x^2}$. Observe that $\pm$ can be changed to $+$ because the area lies in Quadrant I. Also, in Quadrant I, $x \geq 0$. Therefore the value of $x$ is in the interval $[0,1]$. Similarly the value of $y$ is in the interval $\sqrt{1-x^2}$. Hence the integral is $$ \int_0^1 \int_0^{\sqrt{1-x^2}}\,dy\,dx $$

Answer 2

It is not clear if you want to integrate the function $f(x,y)=1$ over the circle or another function which may have no symetry whatsoever. In this case, the integral would be: $\int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} f(x,y)\,dy\,dx$ in which $y$ varies from the part of the circle under the $x$-axis to the part over it.