I’m evaluating the double integral $$\int_{0}^{1/2} \int_{(\sqrt{3})y}^{\sqrt{1-y^{2}}} 1 dx dy$$
The outer limit stops at $y = \frac12$ because the curves $$x = (\sqrt{3})y$$ $$x = \sqrt{1-y^{2}}$$ intersect when $y = \frac12$, so beyond that value, the intended region between the curves no longer exists.
My question is: What would it mean to integrate past $y = \frac12$, for example, $$\int_{0}^{1} \int_{(\sqrt{3})y}^{\sqrt{1-y^{2}}} 1 dx dy$$ Is there a geometric interpretation?
Notice that \begin{equation} \int_0^1 \int_{\sqrt{3}y}^\sqrt{1-y^2} \mathrm{d}x\mathrm{d}y = \int_0^{\frac{1}{2}} \int_{\sqrt{3}y}^\sqrt{1-y^2} \mathrm{d}x\mathrm{d}y - \int_{\frac{1}{2}}^1 \int_\sqrt{1-y^2}^{\sqrt{3}y} \mathrm{d}x\mathrm{d}y \end{equation}
The first integral represents the area between $x=\sqrt{3}y$ and $x=\sqrt{1-y^2}$ starting from the x-axis up until the intersection point at $y=1/2$. (Blue in the attached graph below)
The second integral represents the area between $x = \sqrt{1-y^2}$ and $x=\sqrt{3}y$ starting from the intersection point up until $y = 1$. (Red in the attached graph below)
Notice that the second area is counted negatively towards the original integral, because we had to swap the bounds $\sqrt{3}y$ and $\sqrt{1-y^2}$.
Here is a link to a screenshot of a graph showing the geometrical interpretation of this integral. The integral calculates (blue area) - (red area)
