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A bowl is shaped like a hemisphere with diameter 30 cm. A heavy ball with diameter 10 cm is placed in the bowl and water is poured into the bowl to a depth of $ h $ centimeters. Find the volume of water in the bowl.

$1000\pi+\pi(15^2(h-10)$$.-\frac{1}{3}\left(5^{3}-(15-6)^{3}\right))$

Applications of Integration

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So in this problem we are given a bowl that is 30 cm in diameter And we drop a 10 cm diameter ball in this bowl. Heavy ball we begin to fill with water and the water level in here. Right? Is it some height? H. Okay. And the question is for us to figure out the formula on the volume of water in the bowl based on that depth. H. Okay. So first of all we know that for the bowl the diameter is 30 centimeters so therefore the radius is 15 centimeters. Okay let's assume that we do this right? We have the origin be at the very bottom of the bowl here and the X. And Y axis go through the middle of it. Just like that drawing right there. Okay then using the center radius form of a circle, this equation of this bulk is the hemisphere. It's a half a circle, right? Is X squared plus why minus 15 squared? I guess I should have put my X axis appear at the top 15 squared Is equal to 15 squared. Okay. Mhm. So as I've said we'll put our X axis up here. Okay now let's calculate this for values of X then. Right, so this is X squared plus Y squared. Actually we'll just say that X squared equals 225. That's 15 squared minus y minus 15 squared. And so then X is 225-. Why -15 squared square to that? And this is going to be for zero. It will be valid for zero is less than or equal to Y is less than or equal to 15. Okay now which matches we can't have a height Greater than 15 cm because our bowls only 15cm. Right? So then the volume it's going to be found by rotating these values rotating this with respect to the Y axis in it. Alright just rotate those those why values around those X. Failures around the y axis. Okay so that means I end up with pie I was integral from 0 to H. of the square root of 2 25 Why minus 15 squared. And you square this I. R. Squared do you? Y. Right okay so that means then that I get high Times interval from 0 to H. Uh huh. 2 25 minus Y squared minus C. two b plus 30 y minus 2 25. All of this dy Well to 25 -2. 25 cancels out doesn't it? Okay so that means I have pie times. We'll see the integral of 30 Y is 30 y over two minutes, y squared minus Why cubed over three. We're going to evaluate this from 0 to H. Okay well when I put in zero and therefore why every time that's all zeros that's all gonna can't drop out. And so that means I'm left with 15 H squared. Let me do it this way. Let me do it this way and make it clear Pie times 15 H squared minus H. Cube over three. Okay. And of course this is from zero is less cynical age is less than or equal to 15. So this is the volume in the bowl. All right. So similarly we do the same thing for the ball, right? For the ball we have diameter is 10 cm and so The radius is five cm. Okay, again, doing the same thing like we did earlier we have the ball here for X and Y this way. Okay, so we write the equation for that circle. Well that's X squared plus. Why? Minus five squared is five squared? And so that means X is 25 minus Y -5 Squared. Sorry, that's X squared. So the X. Is the square root of 25 -6 -5 squared. All right, So rotate this, rotate with respect two the y axis on this to do the volume and then we have the volume of the ball. Is pie Times The integral from 0 to H. of the square root of 25 minus y minus five squared squared power squared. Right do you want? Okay so this is Pie, I'm senator from 0 to age of 25 minus y minus five squared. Do you Why? And so this is pie I was integral from 0 to H. Of 25 minus Y squared plus 10 Y minus 25. Do you why? and 25 -25 zero. So those cancel out. And so I end up with pi times the integral of 10 Y. So that's 10 Y squared over two minus Y cubed over three, Evaluated from 0 to H. Okay So that means I have pie. We put zero in there for why? That's gone. So this is pi times five Y or not? Why it's H. Now? Right. Eight squared- H. Cubed Over three. Okay. And this is only valid from zero two H To 10. Right? With h between zero and 10 because the name of the ball is 10. All right. So volume the water in the bowl is the volume of the ball minus the environment. The ball takes up at any point at H. Right? So for zero is less than equal to age is less than equal 10. So for height of 10 cm or less than we have. The volume of the water is our two formulas together. So that's pi times 15 H squared- Age cubed over three minus pi times five H squared minus H cube over three. All right, well, so distribute this out in here. And what do we see? We'll see that this and this cancel out. And so we'll have yeah 15 pi eight square -5 x eight squared. That's when we left with 10 pi H squared is the volume of the water for zero is less than equal to H is less than equal to 10 For the 1st 10 cm of height. So then for 10 is less than H is less than or equal to 15. Right then the volume of the water is the volume of the ball minus the total volume of the ball. No, so that means I have pi times 15 H squared minus Age cubed over three minus The total volume of the ball which is 4/3 pi r cubed by definition of a sphere. Okay, so since the radius is five centimeters this is high times 15 H squared minus H. Cute. Over three minus four thirds. Hi times five cubed. Well five cubed is 125. Okay, so that means I'm left with 15 pi H squared minus pi H cubed over three minus four times 125. So that's 500 Over three time spy. And so there's the formula right? For 10 is less than H is less than April 15 for when the height is over 10 centimetres. And this is the formula When were from 0 to 10 cm. Hi

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Applications of Integration