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Volume of a bowl

a. A hemispherical bowl of radius $a$ contains water to a depth $h .$ Find the volume of water in the bowl.

b. Related rates Water runs into a sunken concrete hemispherical bowl of radius 5 $\mathrm{m}$ at the rate of 0.2 $\mathrm{m}^{3} / \mathrm{sec} .$ How fast is the water level in the bowl rising when the water is 4 $\mathrm{m}$ deep?

a) $\frac{\pi h^{2}(3 a-h)}{3}$

b) $\frac{1}{120 \pi} \mathrm{m} / \mathrm{s}$

Applications of Integration

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University of Nottingham

so we have a hemispherical break. It's a that contains water to a depth of H, and we want to first find the volume of the water in the bowl. And then if we have water running into a something concrete hemispherical bowl radius five meters at the rate of 0.2 meters per second cube her meters cute per second. How fast is the water level in the bowl rising with the water is or eaters? So let's first go ahead and start solving for the volume. So we want to figure out some access that we can rotate this oh, bowl about, or the slice of the bull that we have here draw on the board so we can find the volume. And since we know this is a hemispherical, but we know that we can represent this red line here as the equation a squared minus y square square root is equal to expert. Actually, I should have where I should say that this represents the right hand side. So just this side of the bowl and not the left and the second will become clear why? I'm only talking about the right hand side of the bowl and that we take this line here and we rotated about the Y access we can see it tells us the volume of that portion. So what we're going to do is rotate about why access, Which tells us anytime we rotate around of why access, we can use the formula. So a B fines of our radius squared with respect. What do y And in this case, we would have no in a radius, so we don't have to subtract anything off from that. Now all we need to do is figure out what our pounds of integration are going to be because we already know that our outer radius should be just this curve here because you can see from the green, we go straight out from there and we hit that lot. So I'm just gonna go ahead and plug it in and right off to the side for so are why is going to equal to square root of a squared minus y squared. So squaring that would just be left little they have on the inside. So a squared minus y squared Bye, Interval Pi. Now we need to figure out what our bounds of integration with this should end up being and, well, we at least know our range for why value should be from the bottom to where the water is. A sore bottom bound should be negative, eh? And our top bound Well, that's a distance h away. So that would be H minus eight. So now that we have the balance of integration and everything, we can go ahead and start integrate. So this is going to equal to pi on Sunday, too. And we can go ahead and just she's powerful for each of these. So a is a constant with respect. Why? So this should be a squared. Why minus why now to the third power and then divide by the new power. And we're gonna evaluate this from negative A to H minus a. Hi. And now we could go ahead and plug in each mind, say, and negative a into this, and I'm just going to spare you all the algebra and go ahead and say what we should end up with in the So if you were to plug both of those men, we should get each squared three a minus h all over three. So this here should be the somebody actually say, this is the volume of the water. All right. And now that we have the volume, we can move on to the next part. And what we want to do is find how the height of this is changing. So we know that be of H is equal to what We found it. Heart, eh? So here, so is going to be, And I'm just gonna distribute that h squared. So be 1/3 three a eight squared minus h cute. And now we're going to take the derivative of this with respect to time. And so doing that this year will become D B D T is equal to 1/3. So we'll end up with having to do the power will for each of these on the inside. So it'll be so each springs derivative is too great. So it should be six a. H. And then we're going to have to multiply by D h duty because we're doing implicit differentiation. And then the derivative of H coud will be three h squared. And then once again, we multiplied by D h I. All right now let's go ahead and factor that D h by dtl and doing that would give DVD t is equal to 1/3 six a h minus three h squared e h by d too. And since we want to find what a huge by DT is, we're going to go ahead and start plugging everything in that work told for the problem. So I should probably broke this down. But let me just write this on the side right here. So we know that the rate of change Oh, our volume is going to be 0.2 meters cubed per second. We also know that our radius of the bull should be five. So that means a is equal to five. And lastly, we want to know when the height of the water is for and both of these Aaron meters. So let's go ahead and plug. All listen, so I'm just gonna drop the units, but all the unit should work out in the end to give us what we're gonna write. Down 0.2 is equal to 1/3 so six times five is 30 times four is 120 minus and then our height, ISS 16. They're our height. It's four story that gives us 16 and then 16 times Re is 48. Where do you eat when we still have a D H by DT there and then 1 20 minus 48 is 72 and then divide that by three is 24. So when the 0.2 is equal to 24 th by dt and now we just divide by 24 and we'll get so D H by E t. This is equal to 0.2 divided by 24 which is one over 20. And I actually dropped something really important. Ah, this pie right here. So all of this should be actually multiplied by pie going all the way down. And then this should actually be 120 over pie. And let me just look there. Look, let's make sure I didn't do anything else that may have just got dropped. That seems to be the only thing. What? The only other thing I maybe should have did was on this step here, since this was supposed to be at height is evil toe for sometimes people would write this evaluated at H is equal to four like this, but it just kind of depends on how you are working the problems. But this here would be the rate of change and the units that should accompany this should be meter seconds, so the rate of change of the height would be won over 120 pie meters per second.

University of North Texas

Applications of Integration