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The region consisting of all points between (but not on) the spheres of radius $ r $ and $ R $ centered at the origin, where $ r < R $.

$r^{2} < x^{2}+y^{2}+z^{2} < R^{2}$

Calculus 3

Chapter 12

Vectors and the Geometry of Space

Section 1

Three-Dimensional Coordinate Systems

Vectors

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in this problem. We're trying to find an inequality to describe the region of space between two different spheres centered at the origin. One with radius. Little are shown in blue and one with radius big are shown in red. You can imagine this region of space as a thick spirit Kal Shell, where the center is hollow. But the region between Radius little, R and big are is solid. So let's answer this question using ideas from the chapter. First, we know that the equation for a sphere centred at the origin is X squared, plus y squared plus C squared. Equals are square. Now I'm using a curse of our because the printers are already in use. Once you specify a value for cursive R, you get all of the points X y Z that lie on the surface of this sphere. What if we want to know all the points strictly inside this fear? Not on the surface, but actually inside? Well, it's almost exactly the same thing. It's X squared plus y squared plus C squared. But the equality changes to one inequality, a less fan in this case, since for inside. So that says X squared plus y squared plus C squared less than cursive r squared. And you kind of look at that that inequality and say, Well, that actually looks like all the points on every sphere that has radius less fan course of our and you would be exactly right. That's exactly the same thing as the whole volume inside anyway, Now we can use this knowledge. Thio, Thio, Compute the answer this question. It really just falls right out. We have two conditions are first condition is that our points have to be outside. Little are their second condition is that they have to be inside capital. Mathematically, this means we have X squared plus y squared plus C squared is greater than little r squared, and we also have X squared plus y squared. Plus C squared is less than a Capital R squared, and we're actually done. But we can combine these inequalities in a nice, happy way like so throw that little R squared over to the left. We have little R squared is less than X squared, plus y squared plus C squared. It's less than Capital R squared, and now it's clear that we have the set of all X y Z that is in between the two radio I little R and Capital R, and that is the answer.

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