Find the volume of the solid bounded by the surfaces x =

Find the volume of the solid bounded by the surfaces $x = \frac{1}{2}(y^2 + z^2)$ and $x = a$.

We slice the solid into circular disks perpendicular to the $x$-axis.

Step 1: Cross-section at position $x$

The paraboloid gives $y^2+z^2=2x$, a circle of radius $r(x)=\sqrt{2x}$, for $0\le x\le a$.

Step 2: Area of the cross-section

\[A(x) = \pi r^2 = \pi(2x) = 2\pi x\]

Step 3: Integrate

\[V = \int_0^a 2\pi x\,dx = 2\pi\left[\frac{x^2}{2}\right]_0^a = \pi a^2\]

\[\boxed{V = \pi a^2 \text{ cubic units}}\]

Step 4: Comparison with a cone

A cone with the same base radius $\sqrt{2a}$ and height $a$ would have volume $\frac13\pi(2a)(a)=\frac{2\pi a^2}{3}$. Our paraboloid's volume $\pi a^2$ is exactly $\frac32$ times that, reflecting the paraboloid's "fuller" profile compared to a cone's straight sides.

Summary

\[\boxed{V = \pi a^2 \text{ cubic units}}\]

Key insight: Because the cross-sectional area of a paraboloid grows linearly with $x$ (rather than quadratically, as for a cone), the volume integral is trivial, giving the remarkably clean result $\pi a^2$ — with no factor of $\frac13$ or higher powers of $a$.