# Science:Math Exam Resources/Courses/MATH200/December 2013/Question 05 (a)/Solution 1

We have to start by drawing the following curves to see the domain of integration

${\displaystyle (1)\quad y=0}$

${\displaystyle (2)\quad y=1}$

${\displaystyle (3)\quad x={\sqrt {3}}y}$

${\displaystyle (4)\quad x={\sqrt {4-y^{2}}}}$

${\displaystyle (1)}$ and ${\displaystyle (2)}$ are just horizontal lines at ${\displaystyle y=0}$ and ${\displaystyle y=1}$ respectively (shown below).

${\displaystyle (3)}$ represents a line ${\displaystyle y={\frac {x}{\sqrt {3}}}}$

${\displaystyle (4)}$ represents half a circle of radius ${\displaystyle 2}$ defined for ${\displaystyle x>0}$. In our case, only the part of the half circle in the first quadrant is necessary because of ${\displaystyle (1)}$ and ${\displaystyle (2)}$.

As a result, the domain of integration is the area bounded by these four curves.