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JH
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Problem 53 Hard Difficulty

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?

Answer

a) $\frac{\pi h^{2}}{3}(3 a-h)$
b) $=\frac{1}{120 \pi} m / \sec$

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Video Transcript

So for part a of this problem, we're finding the volume of the water in the bowl to best do this, I always start out with drawing a picture. So using the plane of X and Y, we can make a that is they have a fait circle for a hemisphere global. So in this bowl, we're just drawing it upside down. So it's easier to Workways. We have water up to the level of H. So this level that distance is h and given in the problem, we know that the radius is a You can say this is equal to a This means that this little section in kier is equal to the difference in these two links. So that distance is a minus h. So using these and an additional equation of the volume which is what we're looking for, is equal to pi times the integral from A to B of r squared TV. Why? And we're using d Y because we're revolving around the y axis. So the next step this is to see what is art are, is what limits are volume. So we're trying to find the volume of this water in here And what is limiting that thing is this This line right there? So that line we can express in an equation which that equation because it's a circle will be ax squared. Plus why squared is equal to the radius squared, which in our problem, is a so that will be a square. Now, in order to fit it into this equation over here, we must isolate X and put it in terms of why. So in order to do that, we have X squared is equal to a squared minus y squared. And normally you would want to get rid of this squared by square rooting date function. But because we have the handy dandy square root or square over here, the square would cancel a square root. And thus we don't need to do this step so we can just plug it right back, right in. So this makes it pie times integral of a squared minus y squared times d. Why now? What do we need to do with the bounds here? Those bounds we look at the graph. So the point right here that point is equal to this distance here, so we can say that point is a minus H, which is where we're starting. And our goal is to get all the way up here in this upper point is equal to a because that is the link or the depth of the whole entire bowl. So now, knowing that we can continue by taking the integral So to do this, we say, Why is times because pies a concept, Um, and is also a constant in this case. So we can say that this is a squared times Why minus why cubed over three. And this is still bounded by a minus h to a continuing to simplify this, we can get that Hi is plugging in the upper bound A. We get a cubed minus a cubed over three. This is just plugging in where why is equal to a and then on the lower bound we say minus pie and plug in the lower bound. So this will be multiplying a squared times this value. Here we get a cubed minus a squared times h and then minus. We have up on top a minus H cube all over Now this could be a bit crazy, but we just keep simplifying it, and it turns out very nicely. So in this first section over here recon simplify it into, um where if we combined these terms So we can say that this is 3/3. Just making it easier on the bottom and matching these bottom numbers. We can say three a cubed minus. A cube is equal to two a cube, and this is all over three. Then moving on to the next section. Simplifying this over here we will break apart this fraction her net and the will distribute que so in order to do so, this is equal to a cube minus a squared times H minus. When you remember to keep that minus throughout the entire thing, we have a que minus three a squared H plus three a h squared minus h Q. This is all over three. Now, remember, because these air individual characters and there's one number in the base here, we can split them up into individuals fractions. So we have pie times to a cubed over three minus hi times a que minus So squared times h minus a que over three. And remembering this negative as well as the three we have plus three cancel out here. So it's just a squared H and then minus a h squared again, the three cancels out and then we have plus h cubed over again. We can keep on simplifying and eventually we will get. This is equal to pie times a h where'd minus h cubed over. In this, we can take out the like terms in order to get final volume which is equal to pie. Don't let me write it a little bit higher so we can get it on to the field. But volume is equal. Teoh pie h squared because both terms have church squared in them, divided by three times three day minus age. And that will get you the volume of the water at hi of H in a radius of a now moving on to part B. It says the radius of this bowl is five and the water is flowing at a rate of point to we must find how fast the water level is, when, how fast the water level is rising when the water level is four at four. So using the function that we just we just found, we can fine this next face. So let's write out what we already know. So we know this function here. What volume is equal to? We additionally know how fast the volume is increasing. So we have TV over. DT is increasing by 0.2. This is given to us now. What are we looking for? We're looking for how fast the water level is rising. So this we're looking for the change in hi or changed in which. So knowing this we can we want to find d of h over Dft in terms of using this equation here. So right now, we have three variables in there. So we want to get rid of the A because we're we're No, we're gonna use the V. So in order to do this, we must book bath and see what is given to us in this problem. In the problem, it says that a or the radius is equal to fight. This is a great way to get rid of the eight. So we can say that V is equal to you. Pi times h squared over three. It's equal to three times five minus h this. Now we can say is equal to pi Times H squared over three times 15 minus. We can then distributed this and make it equal to he has times the 15 we have pie and the 15 divided by three is equal to five. So pi times five h squared, mindless pie h cute of the three now moving on from here, we can say that we have three is equal to this function over here. So what we're looking for is h over t so we can take this whole function by Do you teach you and it is equal to using the h here. This would be to times pi times five times age to the Cube. Now, because we take the derivative it so it's d h over d t minus. We have again take the derivative because there's an H in that value. Hi! Over three and carry Multiply the three over. So we this three cancels out the other three so we can just get rid of that three. And it is just the pie out in front. And it's hh sward. And previously I messed up on here. This should not be a three here, so it's just a single h there, so we can Now we have this equation and this last term is also flight by this. So now we can take out thes guys because both of these terms or multiply by d h of d. T. And this makes the function Devi over de t is equal to d h over t t won't supplied by these two terms. Here we see, we can also simplify this guy by multiplying the to and the five to get just to make it a little bit simpler here. And you have minus pi h. Now, looking back at the quit the problem again, we see that we're looking for the rate of the height at when it is at the level of four meters deep. So in this case, H equals four. So plugging four in for H. We see that as well as this value is our left last variable because this value is the one we're looking for and given in the problem as well, we have that the change of rate of the volume 0.2 So plugging in both this and the H, we have this resulting equation. So we plug in four for H here minus pi and four for H. That's well, this can then equal. If we divide this section over, we have zero point to divided by 10 pie times for minus pi four squared. That is all in the bottom. And that is equal to what is left on this side of the equation, which is our goal plugging this into the calculator. We can get a result of one over 120 times on this.