I am working on my scholarship exam practice (assume high school/pre-university math background) and I think I got half way through but I am not sure how I could continue.
Let $r$ be a positive constant. Consider the cylinder $x^2 +y^2 \leq r^2$, and let $C$ be the part of the cylinder that satisfies $0\leq z\leq y$. Consider the cross section of $C$ by the plane $x=t$ ($-r\leq t\leq r$), and express its area in terms of $r, t$.
So below is what I have got and stuck there, not sure if I am on the right track. Could you please show or hint on how I can get to the area of this cross section? The key answer $\frac {1}{2} (r^2-t^2)$ is also provided. Apologies my drawing may be amateur.
Best Answer
The cross section of C by the plane $ x = t $ is a triangle. When $ t = 0 $ $(x = 0)$ , this is an isosceles right triangle with sides r, r and $ r \sqrt 2 $: cross section of cylinder by the plane x = 0. When $ t \neq 0 $, this is an isosceles right triangle with cathetus $ \sqrt {r ^ 2-t ^ 2} $: cross section of cylinder by the plane x = t. So, area of the cross section is $ \frac{1}{2}(r^2 - t^2) $.