Region bounded by the parabola $\, y=x^2\;$ and the line $\, y=16\;$ What is the volume of the solid generated when R is rotated about $y = 17$?

Question

Let R be the region bounded by the parabola $ \, y=x^2\;$ and the line $\, y=16\;. $

What is the volume of the solid generated when R is rotated about the line $\, y=17 \; ? $

So there's outer radius, and inner radius. What's my outer radius? $17-16 = 1$?

And inner radius is just $x^2 - 16$?

So the formula for integration is is $\pi \cdot 1^2 + \pi \cdot (x^2 - 16)^2$

Right? And then the intervals are between $- \sqrt{17}$ and $\sqrt{17}$

But the answer I get: $\frac{4004 \sqrt(17) \pi}{15}$ seems to be wrong...

p.s. Okay, it has a hole in it, outer radius (actual thing I'm finding) - inner radius (the hole).

Answer 1

Outer radius $R$ should be the distance from the curve furthest from the axis of rotation and the axis. In your case, $R=17-x^2.$ Inner radius, $r$ is the distance between the curve closest to the axis of rotation and the axis. In your case, $r=17-16=1.$ The integral for your volume then is $$\pi\int_{-4}^4(R^2-r^2)dx.$$

Can you take it from here?

Answer 2

Notice:

$$\text{Outer radius }=r_o=17-x^2\qquad\text{and}\qquad\text{Inner radius }=r_i=17-16=1$$

Then, since $y=x^2$ meets the line $y=16$ at $(-4,16)$ and $(4,16)$, we have \begin{align*} V&=\int_{-4}^{4}\pi (r_o^2-r_i^2)\,\mathrm d x\\ &=\pi\int_{-4}^{4} \left[(17-x^2)^2-1^2\right]\,\mathrm d x\\ \end{align*}