A bowl is shaped like a hemisphere with diameter 20 cm. A...

A bowl is shaped like a hemisphere with diameter 20 cm. A heavy ball with diameter 12 cm is placed in the bowl and water is poured into the bowl to a depth of h centimeters. Find the volume $V$ of water in the bowl. (Consider two cases: one in which the ball is not completely submerged and the other in which it is.) $V = 10pi h^2 - (pi h^3)/3$ cm$^3$ (the ball is not completely submerged) $V = 6pi h^2 - (pi h^3)/3$ cm$^3$ (the ball is completely submerged)

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00:01 Hi, in this question we are given with a bowl having a diameter that is equal to 40 cm and hence the radius will be equal to 20 centimetre of a bowl and a heavy ball with a diameter d -dash let it be and this is equal to 8 cm so we get r -dash to be equals to 4 centimeter this is of a ball which is poured into the bowl to a depth of each centimeter we need to find the volume of water in the bowl when we will consider the two cases the first one when the ball is not completely submerged as in this case we have when h is less than smaller than this eight diameter of the ball and hence expressing bowel and ball as a function of x as revolving around y x and we will use washers method and and outside radius or the diameter we can say first we will find the diameter is x square plus y minus 20 whole square is equal to 20 square or solving this further we get x is equals to under root of 400 minus y minus 20 whole square.

02:02 Now, inner dia will be x square plus y minus 4 whole square is equal to 4 square.

02:16 Or further, we get x to b equals to under root of 16 minus y minus 4 whole square.

02:26 Now we get the area a as a function of y that will be equals to pi times 5.

02:34 Outer dia that is 400 minus y minus 20 whole square and whole square minus minus under root of 16 minus y minus 4 whole square and whole square and whole square and solving this further we get this to be equals to 5 times solving this we get this to be equals to 5 times solving this we get 400 minus y minus 20 whole square minus solving this we get 16 minus y minus 4 whole square and solving this further we get a as a function of y to be equals to 5 times 32 y and now we can find the value for volume as a function of h that will be equals to pi times integration from 0 to h 32 times y times d y and this will be equals to pi times 32 times y square by two limits from zero to h now here we can see this two will cancel this 32 and we get 16 and solving further we get our final result for v h when the ball is not completely submerged and this will be equals to 16 pi h square now moving further to solve for the second part b when the ball is completely submerged...

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