HardMathematicsHigh School (11–12)Functions and Calculus › Application of IntegralsNumericalA-Level AP IB

A wood turner shapes a solid ornament on a lathe…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A wood turner shapes a solid ornament on a lathe. Seen from the side, the radius of the ornament at distance $x$ centimetres from its base is $r(x) = 2 + \dfrac{x(6 - x)}{9}$ centimetres, for $0 \le x \le 6$ . So the ornament is $6$ cm tall, with a slim base and top and a fatter middle. It is cut from light pine of density $0.7$ g/cm $^3$ . The turner first cuts a plain cylinder, as tall as the ornament and as wide as its widest part, and then turns the ornament out of it. Find the widest radius of the ornament, the volume of the ornament, the mass of the finished piece, and the fraction of the plain cylinder that is left after turning. Give the fraction as a decimal. The cylinder is the least wood the turner could start with, so the fraction tells how much of that starting wood the finished piece keeps.

Need a reminder? The key ideaShow

For a region turned about the $x$ -axis, a slice is a disc of volume $\pi y^2\,dx$ , so $V = \pi\displaystyle\int_a^b y^2\,dx$ . With a hole, use washers: $V = \pi\displaystyle\int_a^b (R^2 - r^2)\,dx$ . About the $y$ -axis, use $\pi x^2\,dy$ and write $x^2$ in terms of $y$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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