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).
Best Answer
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?