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Numerade Educator



Problem 8 Easy Difficulty

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

$ y = 6 - x^2 $ , $ y = 2 $ ; about the x-axis


$V=\frac{384 \pi}{5}$


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Video Transcript

in this problem. We're finding the volumes of solid obtained by rotating the region bounded by why you go six minutes. That's where in y equals two only rotating about the X axis. We make this more balanced. So this is the wine ex axes the locals to nine is very simple. We call that point why it was, too. And that's the line. The Y equals six minus X squared is a little more complicated. So typically, we remember the y equals X squared line. As something like this, However, that's not what we need to do. What we first do is that we get it. So when we get it, we get this. Y equals negative X work. That's what it looks like. Then we're shifting it up by six because we're adding six to whatever this problem gives us. So when we do that, you get something like this where at this point, that six and that's two right there. So that's the region that were given. They're bounded between these two curves. Next part is to find the points of intersection so that we know what intervals to integrate over so we can do that by setting two equations equal to each others. In this case, we get two equal six minutes X squared. If you subtract both sides, buy six, you get negative. Four equals negative X square negate. So you get four equals X squared. And then you get ax equals negative too. And also to as well. So at this point, yeah, we can call this point negative too. Two, This point right here is two two. So these two points satisfied, but with equations. So now we know the general that we should integrate over, which is from X equals negative to two X equals two. You know this because we're rotating about the X axis. They also need to draw a typical washer. So in this case, you know some When we rotate, we don't wash it. That looks like this. And now we need to find the cross sectional area now washers composed. That's attracting the smallest circle out of a larger circle. So to find the cross sectional area in terms of X, you find the area of the larger circle said trapped area of the smaller shovel. You know, if you look back, the y equals x minus y equals six minutes. X Square is creating the larger circle because is constantly greater than two or the Y values constantly greater than two. So we know they are. The big R is equal to this. Why? So we can plug in six nine x squared for big R and then for the little R. It has to be the other line, which is why close to so we just plug into should you pie here and when we do this, simplify the south multiple head out. We get first blood pies. Sixth graders thirty six six times two is twelve X Squared square is up to the fourth. So that's what happens when you foil out to this part. And now that is just a minus for at the end, because Two Square is just for and so you can just simplify this as pie thirty two minus twelve ax squared. Plus, that's the fourth. So that is our final cross sectional area function. Now we can go ahead and proceed with the interval, then roll, as we showed before, was going from negative two to two, and we put in the cross sectional area function that's thirty two nice twelve x squared Plus for the fourth D X, and I'm gonna pull out the pie from before because the pie has no relation to the accidental that is equal to pi times. The anti dread of the thirty two is just very to pie or thirty two x The anti dribble of negative twelve X square is negative twelve times X to the third over three. So that's just minus four Excuse. And if you aren't sure of yourself, unis, take the derivative of this. So three times four is twelve. Three minus one is two. So that's what you get. So we did that incendiary correctly and the anti derivative of excellent Fourth is just extra fifty or five from negative to to to Now, why I do is Pugin and solved. So get hi four to three. Teo. So essentially I did was I first plugged in X equals two to this entire equation and they're plugged in. X equals mated to to this entire equation again attracted the first one from the second one. And when you do all of this out, you have to answer three, eight, four fly over five. That is our final answer