Question 752371
Your idea is good, but to convince yourself, you need a drawing or two (or a very good imagination). Here are my drawings.
{{{drawing(300,300,-2,8,-2,8,
grid(0),green(line(-3,-2,9,6)),
locate(6,6.5,green(y=(2/3)x)),
triangle(0,0,6,4,6,0),
locate(3.5,1.5,R)
)}}}
Let's loox at the x-y plane, and consider points in region R that at part of that cross section at {{{x=a}}}.
They form segment AP, that goes from the x-axis to that green line.
Points A and P have {{{x=a}}}. The radius of the circle is AP {{{drawing(300,300,-2,8,-2,8,
rectangle(5.8,0,6,0.2),
red(line(-2,0,8,0)),red(line(0,-2,0,8)),
red(arrow(5,0,8,0)),red(arrow(0,5,0,8)),
locate(7.6,-0.2,x),locate(-0.5,7.9,y),
green(line(-3,-2,9,6)),
locate(6,6.5,green(y=(2/3)x)),
blue(line(6,3.8,6,0.2)),
locate(7.5,2.5,R),
line(6,-0.2,6,0.2),locate(5.9,-0.3,a),locate(6.1,0.5,A),
circle(6,4,0.2),locate(6.1,4.6,P)
)}}}
Since P is on the line, its y-coordinate is {{{(2/3)a}}}, and that is the distance AP, the radius of the circle.
So {{{A=pi*(2a/3)^2}}} --> {{{A=(4a^2/9)*pi}}} or {{{A=4a^2*pi/9}}}