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Suppose you make napkin rings by drilling holes with different diameters through two wooden balls (which also have different diameters). You discover that both napkin rings have the same height $ h $, as shown in the figure.

(a) Guess which ring has more wood in it.

(b) Check your guess: Use cylindrical shells to compute the volume of a napkin ring created by drilling a hole with radius $ r $ through the center of a sphere of radius $ R $ and express the answer in terms of $ h $.

A. the rings have the same amount of wood. $\\$

B. $\frac{\pi h^{3}}{6}$

Applications of Integration

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Missouri State University

Campbell University

Oregon State University

Idaho State University

first problem, we would guess that the smaller ring has more wood. Um, since it is all closer together, uh, the bigger one is more spread out with more would that was drilled away. So on the other hand, if we were able to squeeze the bigger ring together to make it closer to the size of the smaller one, it would appear to be about the same. Um, so it does appear to be about the same. And then for part B, what we want to do is check our guests. So we're gonna use the cylindrical shell method. We have our circle right here, and then we have our This is the origin and we have a radius right here of big are and then a smaller radius of little are. Then we want to get the height of each cylindrical shell and and multiplied by two. So we get that X squared, plus y squared equals R squared solving for wine. We get that y equals the square root. And this is the height right here at the height. Mhm, um y equals the square root of R squared minus X squared so that our volume is going to equal to pipe of the integral from our to our of the shell radius times the shell height, DX So what that's gonna look like is the shell radius of X and the show height, which is gonna be two times the y value. So two times the square root of R squared minus x squared dx. When we integrate this, we end up getting that are volume is going to be four thirds pi Our big R squared minus little r squared to the three house power. Um Then what we see is that this r squared minus r squared we've seen before because this is the same thing as a church squared, Um, so we can replace that with h squared. Thus, in doing so, we would, um, looking at each squared we have r squared plus h squared over four equals big R squared so R squared minus big R squared minus r squared. If we bring this over, is gonna end up giving us I'm eight spread over four and then we multiply by four to get each squared. So looking at that, we see that we can rewrite this as V equals four thirds pi of H squared over four to the three house. And this simplifies to be 16 Hi, h cute.

California Baptist University

Applications of Integration