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A hole of radius $ r $ is bored through the center of a sphere of radius $ R > r $. Find the volume of the remaining portion of the sphere.

$V=\frac{4}{3} \pi\left(R^{2}-r^{2}\right)^{3 / 2}$

Applications of Integration

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Missouri State University

Harvey Mudd College

University of Nottingham

Idaho State University

So for this problem, we're going to be using the washer method. We have to integrate along the Y axis, focusing on the inner diameter and the outer diameter. So we're going to use symmetry and double the integral. So rather than doing it from a negative square root to a positive square root, we'll just go from y equals zero to y equals the square root of R squared minus R squared. Bigger and little are and then the upper limit is going to be found by using the Pythagorean theorem on the right triangle in the following picture. So what we have is this we have our circle. Um, there's this smaller radius right here. This smaller radius is the little r. And then the larger radius right here is the big are we also have a larger radius straight here, Obviously, which is going to be the same since its circle. Um, so we want to solve the circle equation for X. So first, what we have is r squared equals X squared plus y squared. So solving for X, we're going to get the X equals the square of R squared minus y squared. Then that means that our outer radius will be are one and that is R squared minus y squared and then our inner radius we already see as just are. So now we're going to integrate with the washer formula That's the equals pi times the bounds Where'd you decided that the bounds are zero to the square root of R squared minus y squared And then we plug in our our one squared minus r two squared. Once we do that, we end up being able to simplify a little bit further and it's going to be r squared minus y squared minus little r squared. Do you mind simplifying this further? Using integral zoned properties, we know we get four thirds our big R squared minus little r squared to the three halves. Um, and then we have the cylindrical shell method that we could also use. So what we have is r squared equals X squared plus y squared. And in this case, what we have is solving for why we'd get the square root of r squared my x squared, Um, and this is using this special method, but it would end up giving us the same answer and that is a volume of four thirds pie, um, four thirds pi. And that should be here to four thirds pi four thirds pi r squared minus little r squared to the three halves.

California Baptist University

Applications of Integration