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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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Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Hi there now for this problem we need to calculate the volume coming to these two spheres. No one convenient way to set these two spheres is to make that the center of each of spheres is as the distance art over to from zero so that we obtain them this picture And with that said we can notice that from symmetry the volume of a about the assets that we are going to call b minus and equals to the body of being rotated about the X assets. We call these the assets and we are going to call it B plus this one that is in red and so the total volume that we are going to obtain is just the sum of these two and this and since they are the same, we obtain this. So we just simply need to obtain um B plus so that we can determine and the net volume. So to calculate B plus, we used these methods and the end points in this case is very easily because deployment Stars in here in zero and ends at a distance R um Over two. So dad that does and points are A equal to zero and be equal to our over to right And from the equation of the red circle we know that the equation for a circle in this case um sorry, the blue one, it should be as the following. It is atZ plus are over two A squirt glass, y a square and this should be equal to the radius of the circle that it is simply are judas square from this um we can solve for why. So we are going to obtain why square is equal to r squared minus ads Glass are over two square. And with this we can simply obtain the volume B plus because it is defined as the integral from zero to our over to those are the endpoints that we're sitting here. Yeah. And the and this should be the integral of the area in this case is p times why it's work and we are integrated by delta X. Because we have um set the end points in that it's that confident so with that we substitute what we have obtaining here from the white part. So we will have zero for our or too high are square minus. It's loss are over two. Remember that art is just a constant. Is the radius of the circle And and it is not changing. The only party will in here is that so we can expand this. So we will have the following series over over time. This will be our square minus at this work blast. Um well minus minus as times are when we do this, cora's terms minus Our square over four. Okay, so this is what we have 10 and this is integral is in EDS. So from here we can see that the first term is just when we integrate it is pine. Uh huh ads-. The integral of the square is as Cube over three. The integral of ad is at square over two times. Are because it is a constant and the final one is just a cancer. So it should be are Our ads over four. So this Evaluated from 0 to our over to So when we have done this we can simply substitute dot. So in here we will have X. So we know that for zero. We know that when we evaluate an integral is the upper limit minus the upper leg. The upper function evaluated in the the functionality in the upper limit minus dysfunction. Have already in the lower limit and lower limit is zero. And we can see here that all these terms carry an X. So that term it will be zero. So we only evaluate for our over two. So that will be are overturned. This case minus what over three Or over three Cube are over to squirt Are over two Our square over four arts work. So with all of this we can just simply name obtained that so many. All of this we are going to obtain that the bottom is five high our cube over 24. And since we know the radios for so we are given that the radius of the circle is just art. So when it spread it like that. Mhm And we know that Finally the total volume is two times the volume that we have obtained. So that will be two times y. Prior Q Over 24. So this is five point our cube over 12 and this as that the total bully um of the common or the common volume between these two circles. So for this problem, um this is the result. Thank you.

Central University of Venezuela

Applications of Integration