Area and Volume Revisited — Question 3

Question 3 – Solution

Strategy. Subtract the lower paraboloid from the upper one over their circular projection.

Step 1: Intersection

Setting the surfaces equal gives $$r^2=6-r^2,\qquad r=\sqrt 3,\qquad z=3.$$ Hence the projection is the disk $0\le r\le\sqrt 3$.

Step 2: Volume The vertical height is $(6-r^2)-r^2=6-2r^2$. Therefore $$\begin{align*} V&=\int_0^{2\pi}\int_0^{\sqrt 3}(6-2r^2)r\,dr\,d\theta\\ &=2\pi\left[3r^2-\frac{r^4}{2}\right]_0^{\sqrt 3} =2\pi\left(9-\frac 92\right)=\boxed{9\pi}. \end{align*}$$

Verification On $0\le r\le\sqrt 3$, the height is $2(3-r^2)\ge 0$; it decreases from $6$ at the axis to $0$ at the intersection circle, matching the geometry.