Thanks for adjusting it- my neck was starting to hurt!
The graph of \(\displaystyle 7y=(\sqrt{21}+ 3)x- \sqrt{21}\), which is the same as \(\displaystyle x= \frac{7y- \sqrt{21}}{\sqrt{21}+ 3}\), is a straight line. It crosses the y- axis at \(\displaystyle \left(0, \frac{\sqrt{21}}{7}\right)\) and the x-axis at \(\displaystyle \left(-\frac{21}{\sqrt{21}+ 3}, 0\right)\). The graph of \(\displaystyle 9+ y^2= 3x\), which is the same as \(\displaystyle x= 3+ \frac{y^2}{3}\), is a parabola, with horizontal axis, opening to the right, vertex at (3, 0).
They intersect where \(\displaystyle \frac{7y- \sqrt{21}}{\sqrt{21}+ 3}= 3+ \frac{y^2}{3}\). Multiply on both sides by \(\displaystyle \sqrt{21}+ 3\) to get \(\displaystyle 7y- \sqrt{21}= 3\sqrt{21}+ \frac{3}{\sqrt{21}+3}y^2\) or \(\displaystyle \frac{3}{\sqrt{21}+ 3}y^2- 7y+ 4\sqrt{21}= 0\). Solve that quadratic equation to get the y values at the points of intersection. For any y, the horizontal distance across the figure is \(\displaystyle \frac{7y- \sqrt{21}}{\sqrt{21+ 3}}- 3- \frac{y^2}{3}\). Integrate that, with respect to y, between the two y values of the intersections.
View attachment 11550