Find the Volume Using Cylindrical Shells About the x-Axis

The bounded region between $x=16y^2-4y^3$ and $x=0$ lies above the $x$-axis. To use cylindrical shells when rotating it about the $x$-axis, take horizontal strips: each strip is parallel to the axis of rotation and sweeps out a cylindrical shell.

The shell method avoids solving the cubic relation for $y$ as a function of $x$. The resulting volume is $\frac{2048\pi}{5}$ cubic units.

Find the region’s limits

The curve meets the line $x=0$ where

$$16y^2-4y^3=0$$

Factor the expression:

$$4y^2(4-y)=0.$$

Thus the intersection heights are $y=0$ and $y=4$. For $0<y<4$, the curve has positive $x$-coordinates, so it is the right-hand boundary of the region and $x=0$ is the left-hand boundary.

Set up the shell integral

A horizontal strip at height $y$ has thickness $dy$. When revolved about the $x$-axis, it forms a thin cylindrical shell. Its dimensions are:

• Radius: $y$, the distance from the strip to the $x$-axis.
• Height: the horizontal distance across the region, or $x_{\text{right}}-x_{\text{left}}=16y^2-4y^3$.
• Thickness: $dy$.

The lateral area of a thin shell is approximately its circumference times its height, $2\pi y(16y^2-4y^3)$. Multiplying by the thickness and integrating over all heights gives

$$V=2\pi\int_0^4 y(16y^2-4y^3)\,dy.$$

The factor $2\pi y$ accounts for the circumference at radius $y$; the curve’s $x$-value supplies the shell height.

Evaluate the integral

First expand the integrand and integrate term by term:

$$\begin{aligned} V&=2\pi\int_0^4(16y^3-4y^4)\,dy\\ &=2\pi\left[4y^4-\frac45y^5\right]_0^4\\ &=2\pi\left(4(4^4)-\frac45(4^5)\right)\\ &=2\pi\left(1024-\frac{4096}{5}\right)\\ &=2\pi\left(\frac{1024}{5}\right)\\ &=\boxed{\frac{2048\pi}{5}}. \end{aligned}$$

The volume is approximately $1286.8$ cubic units.

Why the curve’s turning point does not change the setup

The curve’s $x$-coordinate reaches a maximum inside the interval: differentiating gives $dx/dy=32y-12y^2$, which is zero at $y=8/3$ (as well as at $y=0$). This does not require splitting the shell integral. At each fixed height $y$, the region still stretches from $x=0$ to the single point $x=16y^2-4y^3$. The shell height is that horizontal distance, whether the curve is increasing or decreasing as $y$ changes.

Common mistakes

• Using vertical strips: Vertical strips are perpendicular to the $x$-axis and suggest a disk-or-washer setup, not cylindrical shells. For shells around the $x$-axis, use horizontal strips and integrate with respect to $y$.
• Using the wrong radius: The radius is the distance to the axis of rotation, so it is $y$, not the curve’s $x$-coordinate.
• Using the wrong shell height: The height is right boundary minus left boundary: $(16y^2-4y^3)-0$.
• Choosing only part of the interval: The bounded region runs from $y=0$ to $y=4$, including both portions on either side of the curve’s maximum $x$-value.

Here all the shells have nonnegative radii because the region lies above the $x$-axis. If a region crosses the axis of rotation, shell radii must be treated as distances (absolute values), and the geometry may require splitting the region to avoid counting overlapping shells.