Find the volume of the solid - a frustum of a right circular cone

Find the volume of the solid S:

a frustum of a right circular cone with height h, lower base radius R, and top radius r.

This is literally all the problem statement gives. I don't understand this at all. What I have taken from this problem (if at all correct) is:


V = pi * integral (x=0 to h) [ y*(r-R)/h + R]^2 dy
My teacher hinted that the shape should be tilted so that it is along the x-axis but the hint my tutor gave (although she said she had no idea how to do it) was based on the equation above and revolving around the y-axis. I am not entirely sure how to set this problem up. Thank you for your help!

You almost had the set-up correct except for some notation.

You could do it along either axis, but you need to be consistent in your notation for the integral. If x goes from 0 to h, then integrate the radius function of x dx. On the other hand, if you integrate the radius function of y dy, make y go from 0 to h.

So either do pi * integral (x=0 to h) [ x*(r-R)/h + R]^2 dx or
pi * integral (y=0 to h) [ y*(r-R)/h + R]^2 dy

which of course gives the same answer either way, because changing the name of a variable has no effect on the value of a definite integral.

Lord bless you today!