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Find the volume common to two spheres, each with radius $r$

if the center of each sphere lies on the surface of the other

sphere.

$$

\frac{5}{12} \pi r^{3}

$$

Applications of Integration

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Missouri State University

Harvey Mudd College

Idaho State University

Boston College

{'transcript': "of course, to find this all you common to these two spheres. Let's just just go ahead and try to trawl what we're doing here. So here is going to be our first fear. And then they say that the center lies along the surface of the other one. So it looks something kind of like this for our second one like that. So what we're interested is like this region in between these two. Well, why don't we try to project this into the X Y plane here? I kind of get an idea. Well, this would actually just be two circles, actually. Put this red first and I'm going to center the black one here at the origin. And then over here, our blue circle local something like this a little bit better. It was something kind of like that there. And we know that the distance between these two is going to be our well, that area in green right here from the volume. The volume over there in green, I should say, is gonna be that same kind of area here. If we were to rotate this around the X axis so well, we can try to do is let's get rid of this bottom half of our function and let's try to define what these functions are well over here. Since this black line is supposed to be a circle of radius R, that's going to be X squared. Plus y squared is able to r squared. And now this blue one Well, our center isn't at 00 is going to be here at our zero. So that means this blue equation is going to be x minus are squared, plus why square is equal to our script. So we have that. And now what our goal is. Well, we would want to integrate this so we would be integrating well, first, whatever this area here is, But in order for us to do that, it looks like we have to break this up into two different intervals because notice right here the function that is on top in the function that is on bottom changes. So we first want to do this first area. I'll call one here and then this other one I'll call to so that first area is where the blue line is on top. So we want to find so I'll call this A to B to C So we want to find the area where the volume from A to B and this volume formula we're going to be using remember is pi r squared two pi r squared. And this is for the blue one there. And then we're going to be adding, when we integrate from the to see dogs, that black ones then pi r squared. Well, in this case, we just need to figure out Well, what is that blue equation going to give us? Well, we would need to solve for why? Well, let's go ahead and do that really quickly over here, So we would subtract the X minus R squared over. So I guess that's why squared is equal to r squared minus X minus are squared. And then we would take the square root so he should get plus or minus for this. But you might recall that, um, we won't need to do the minus because notice that our blue line here is completely above the X axis, so that's gonna be all positives. That's why we're just have that there. So go ahead and plug that in over here. So this would be the integral. Maybe I'll keep that on. So the integral of pie and then r squared minus x minus R squared, squared, almost square. We're here and then our aid to be Well, we know A is supposed to be zero because we're starting from the origin here. So that zero we don't quite know what b is. So let's just leave it for Beaver right now that will come back to figure that out and then over here So again, we don't know what B is, but see Should just be our radius r So we have that. And then we would have high times. Well, we could just go ahead and solve for r here, just like we did before. Why? I mean, so it was tracked X squared over Take the square root. So let's just be why is into the square root of R squared minus X squared. So we have our square minus X squared, square rooted and then squared DS. Right now we need to figure out Well, what is this value for B going to be, well, notice. That's where these two functions intersect each other. So what we can do is, since this is supposed to be just why Square is equal to for both of these, we could just set these functions equal to each other. So let's just set X squared plus y squared equal to X minus R squared plus y squared. I'll do that over here in this corner here, so we have X squared. Plus y squared is equal to X minus R squared plus y squared, and we'll notice the's Y squares cancel out with each other, and then we could just take the square. Roots that gives us X is equal to x minus are once we square root each side and then we can subtract the are over. And so that just does this r zero. So that doesn't seem like it works. So what we need to remember, though, for this is when we're going to rue. If we take the positive route of something like this, we should also get plus or minus. So write this up. So when we take the square root here, since this is supposed to be even route, we'd have plus or minus X is equal to X minus R, and then we can go ahead and subtract the are over. And so then that gives us plus or minus X minus X is equal to negative are. And if we distribute that are that tells us our is equal to so we'll end up with two different one. So if we subtract them, that is going to be just two X and then if we add them, this is just going to be zero. So that's what we got the first time. So again, this radius of zero didn't really make sense for us. Since if we had a race of zero, we just would be right here, like we wouldn't move it all. So we would need to use this to X so we can divide over by two. So that tells us exes even toe are over two. So this here is going to be our value for be. So let's go ahead and plug that into both of these. Now we get rid of those plug in B, plug in beef, and now we can simplify down our algebra a little bit. So I'm going to just pull that pie out. So it be zero. Why did I rewrite Beat. This is our over to are over to and that will just be r squared minus x minus are squared. Uh, dx are over two. And then over here, this would be Plus, I are over to tow our oh, just r squared. Minus expert. Yes, on that. I'm gonna go ahead and expand this over here to make it a little bit easier. So this should be expert minus two our X our X plus X squared, But we're minus ing it. So for subtracting those knows they are squares. Cancel out with each other. So this is just going to be pie is 20 are to, And then we would have to our x minus X squared The X plus high integrated from our over to tow r r squared my expert DX. Um and it doesn't really look like we can do much more to simplify this down. So let's just go ahead and start integrating each of these. So here to integrate. So remember, the R is a constant. So we're just looking at X, so this would be to our x squared. And then we divide by two by the power rule and then X square. That would be execute over three. I'm evaluating this from zero toe are over to And then over here we have the pie on the outside. So remember, R squared is just some constant. So this would just be r squared X minus and then for X squared here, this would be one third execute. And we're evaluating this from our over to to our. So now we can plug all these values in. So when we plug in our over to eso, actually first notice the's twos, cancel out with each other and then we would have it looks like our cubed over four minus are cute over eight, divided by three. So they are cute over 24. And then when we plug in zero, that's just going to be zero. And now over here, Well, when we plug in, our that gives us are huge, minus one third are cute. And then when we plug in are half that's going to give us. So our cubed over to minus and then again we get are cute over because it would be our cubed over eight and then times one third so 24 All right, So now we can go ahead and simplify these here. So when we do 1/4 minus 1/24 That gives us 5/24. So this is gonna be pi times five R, cubed over 24 and then plus well over here argued. Minus one third are cubes of the two thirds are cute and then minus. So one half minus 1/24 is going to be 11/24. So I have 11/24 are cute on, then just rewriting this soap by our cute 24 pi plus. Well, then we need to just do two thirds minus 11/24. So that gives us 5/24 when we do that. So this is going to be plus five are cute over 24 pi, and we can just go ahead and add those together. So that should be 10 r cubed over 24 high. And then that would simplify. Down to five are acute over 12 pie. So then this here is going to be the volume of the area between those two spheres"}

University of North Texas

Applications of Integration