SOLUTION: Ok so here's the question: i have to consider the region R = { (x,y) : x is greater than or equal to 0 and less than or equal to 6 AND y is less than or equal to (2/3)x and greater

Question 752371: Ok so here's the question: i have to consider the region R = { (x,y) : x is greater than or equal to 0 and less than or equal to 6 AND y is less than or equal to (2/3)x and greater than or equal to 0 }. Which i know is a triangle. Then i have to sketch the solid of revolution S obtained by revolving R about the x-axis. Which i did. Now here's where i'm stuck. I have to find the formula for the area of the cross-section perpendicular to the x-axis at x=a, which i know will be a circle. So i started out with the formula for area of a circle which is A= pi(r)^2. and i think that i somehow have to use (2/3)x and x=a to solve for r, and then substitute for the "r" in the equation for area so that all i have to do when given an x value later is plug it in for a, but i'm lost as to how to do that...any idea?
Answer by KMST(5398) (Show Source): You can put this solution on YOUR website!
Your idea is good, but to convince yourself, you need a drawing or two (or a very good imagination). Here are my drawings.

Let's loox at the x-y plane, and consider points in region R that at part of that cross section at .
They form segment AP, that goes from the x-axis to that green line.
Points A and P have . The radius of the circle is AP
Since P is on the line, its y-coordinate is , and that is the distance AP, the radius of the circle.
So --> or
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