Area and Volume Revisited — Question 4

Question 4 – Solution

Strategy. At each fixed $z$, the two cylinder inequalities give independent bounds on $x$ and $y$, producing a square.

Step 1: Cross-section

For $-a\le z\le a$, $$|x|\le\sqrt{a^2-z^2},\qquad |y|\le\sqrt{a^2-z^2}.$$ Thus the slice is a square with area $$A(z)=\left(2\sqrt{a^2-z^2}\right)^2=4(a^2-z^2).$$

Step 2: Integrate $$\begin{align*} V&=\int_{-a}^{a}4(a^2-z^2)\,dz =8\int_0^a(a^2-z^2)\,dz\\ &=8\left[a^2z-\frac{z^3}{3}\right]_0^a =\boxed{\frac{16a^3}{3}}. \end{align*}$$

Verification The cross-sectional area is even in $z$, justifying the factor $2$. The answer is proportional to $a^3$, the correct scaling for a volume.