Area and Volume Revisited — Question 8

Question 8 – Solution

Strategy. First express the volume as a function of the unknown cutting height $A$, then solve the resulting equation.

Step 1: Geometry and volume

The plane and paraboloid meet when $r^2=A$, so $0\le r\le\sqrt A$. Thus $$\begin{align*} V(A)&=\int_0^{2\pi}\int_0^{\sqrt A}(A-r^2)r\,dr\,d\theta\\ &=2\pi\left[\frac{Ar^2}{2}-\frac{r^4}{4}\right]_0^{\sqrt A} =\frac{\pi A^2}{2}. \end{align*}$$

Step 2: Recover the dimensions The condition $V(A)=18\pi$ yields $$\frac{\pi A^2}{2}=18\pi,\qquad A^2=36.$$ Since $A>0$, $$\boxed{A=6},\qquad \boxed{\text{base radius}=\sqrt 6}, \qquad \boxed{\text{height}=6}.$$

Verification Substitution gives $V(6)=\pi(36)/2=18\pi$. The positive root is required because $A$ is a height, and $r=\sqrt A$ follows from the intersection equation.