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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 hemi- spherical 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) $V(h)=\frac{\pi h^{2}(3 a-h)}{3}$

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

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Numerade Educator

Campbell University

Harvey Mudd College

University of Nottingham

in this problem. We have a hemispherical bowl and the first part of the problem. We're asked to determine the volume of the bowl that is, when it is filled to a height H that is less than the radius of the bowl. And the second part where asked to figure out how fast the water level in the bowl is rising. Given the volume flow rate that is being released into the bowl eso What our strategy will be for the first part is will find an equation for our shape in, uh, two D and then using adding up discs, we confined a volume for this level up to height from why, from zero to h and then for our second part, will take that volume flow rate and used the chain role with related rates to turn it into a rate of change for height. All right, so for our part, a eso what is our equation for this circle look like that we've drawn on our Cartesian coordinate plain eso It's just gonna be a circle Ah, with radius A and we've just moved it in our Y coordinate by distance A to have X squared plus why minus a squared equal to a squared will be equation for our circle. And we want that in terms of acts, eso will say X equals to square it of a squared minus Why menace a squared wow, and that will give us square it to a Y minus y squared for expression for X in terms of why And so now we want Thio add up discs. So we're going to take disks with Radius X and we're going to add those up to get our volume. So our volume integral So our devi equal to these disks, ah, times are why element and so that our is going to be X so that t v equals two pi x squared ey And so we just found X here. So that's gonna be equal to pi turns a to a y minus to a y minus y squared de y. And so we're finding that volume we're integrating. And so we're taking up all these discs and we're going from, uh, mhm. Why equals zero to y equals h. So have rpai to a Y minus y squared inte Grint d y and so that's going to give us a pie. Mm hmm. A Y squared minus y cubed over three. And so if we work that out, we get and let's get a pie. Uh, we got a H squared minus h squared over three. And we can write that as a factor on H squared square Pi h squared and a minus h over three. And that will be our volume for part a. So our ball of radius A filled to height h and part B. So we're given the were given some data here, so we're given our radius of our bowl is 5 m. The volume Florida DVD T is 0.2 cubic meters per second. We're filling the bowl to a height of 4 m and we want to know what the height flow rate with the rate that the height is rising at or what is D h d t and soldiers used the expression for the volume that we found in part a So where we found volume is equal to in a row of pie to a Y minus y squared de y And so we can say our TV d h be equal thio pie on to H when this h squared and that will be equal to by times 10 inch minus h squared for our A equals to five. All right, now So we've got this I d v d t and we can relate that to d h d t with the general So we'll say DVD t equals DV th times d h d t on. We want d h d t so d h d t equals to TV GT divided by d v d h And so were given a d v d. T equal to 0.2. And we found our DVD h was equal to pi times 10 h minus X squared. So our h is for here, So plugging a four for H 10 times four on this four squared And so that's gonna work out to be 0.2 over 24 pi and weaken right that as one over 120 pie meters per second, that will be our height flow rate for part