Volume of a solid generated by rotating the region between a parabola and a line

Expert intro: This one is a classic washer problem. The axis of rotation sits above the region, so the key is to measure every radius from $y=16$.

Detailed walkthrough

We rotate the region bounded by $y=4x^2$ and $y=8-4x$ about the line $y=16$.

1) Find the intersection points

Set the curves equal:

$$4x^2 = 8 - 4x$$ $$x^2 + x - 2 = 0$$ $$(x+2)(x-1)=0$$

So the region runs from $x=-2$ to $x=1$.

2) Identify outer and inner radii

Since the axis $y=16$ is above both curves:

• outer radius = distance from $y=16$ to the lower curve $y=4x^2$:

$$R(x)=16-4x^2$$

• inner radius = distance from $y=16$ to the upper curve $y=8-4x$:

$$r(x)=16-(8-4x)=8+4x$$

3) Use the washer formula

$$V=\pi\int_{-2}^{1}\big(R(x)^2-r(x)^2\big)\,dx$$ $$V=\pi\int_{-2}^{1}\Big((16-4x^2)^2-(8+4x)^2\Big)\,dx$$

Expanding gives

$$(16-4x^2)^2-(8+4x)^2=16x^4-144x^2-64x+192$$

So

$$V=\pi\int_{-2}^{1}(16x^4-144x^2-64x+192)\,dx$$

4) Integrate

$$V=\pi\left[\frac{16}{5}x^5-48x^3-32x^2+192x\right]_{-2}^{1}$$

Evaluating gives

$$V=\frac{1728\pi}{5}$$

Correct choice: D.

💡 Pitfall guide

Two mistakes show up a lot here: mixing up which curve gives the outer radius, and forgetting that the axis is horizontal, so radii must be measured vertically. If the smaller radius is subtracted from the larger one in the wrong order, the volume expression breaks quickly.

🔄 Real-world variant

If the line of rotation were $y=0$ instead of $y=16$, the same region would still use washers, but the radii would become the curve heights above the x-axis. If the axis moved inside the region, the outer and inner radii would need to be rechecked carefully.

🔍 Related terms

washer method, intersection points, axis of rotation