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Integration question againby lionely
Tags: integration 
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#1
Feb2313, 11:02 AM

P: 523

1. The problem statement, all variables and given/known data
A goldfish bowl is a glass sphere of inside diameter 20cm. Calculate the volume of water it contains when the maximum depth is 18cm. The attempt at a solution I don't really have an idea of how to attempt this, all I did so far was a draw a little sketch of the bowl and put in the dimensions Hmm.. should I just find the volume of the sphere, sketch the crosssection of it, and then try to use the principles of solids of revolution to find the volume? 


#2
Feb2313, 11:26 AM

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P: 2,943

Now consider the horizontal circular cross section of a disc of water taken at a certain ycoordinate. Find its radius via Pythagoras theorem, and hence its area. Hence figure out the volume of an infinitesimally small cylinder having that crosssection and a vertical height dy. Now do the integration, imposing the correct bounds for y. 


#3
Feb2313, 11:39 AM

P: 523

Umm wouldn't the radius be 10cm, because the diameter of the bowl is 20cm?
Cause maybe I drew my diagram badly I'm not seeing how I can use Pythagoras ' theorem. :S 


#4
Feb2413, 12:00 AM

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P: 2,943

Integration question again
But what about other y values. Hint: think of a right triangle, the hypotenuse being the constant radius of the bowl (10cm), the vertical height being y and the horizontal base being the radius of the crosssection. 


#5
Feb2413, 11:23 AM

P: 523

Is the radius 6cm?



#6
Feb2513, 12:40 AM

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P: 2,943

Did you make a proper sketch? 


#7
Feb2513, 06:22 PM

P: 523

Hm... Maybe I didn't on my diagram 10 is the hypotenuse and 8 is the perpendicular height.



#8
Feb2613, 03:48 AM

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P: 2,943

y varies from 10 (bottom) to +8 (top of water level). Do you understand this? 8cm is only the height when you're considering the area of the top surface of the water (y = 8). At that point, the radius is ##\sqrt{10^2  8^2} = 6cm##. Agree? At the bottom of the bowl (y = 10), the radius is zero, because the bottom is just a point, not a circle. Agree? The radius of the crosssection right through the level of y = 0 (center of the sphere) is 10cm (simply the radius of the sphere). Agree? Now you can take the cross section of water at *any* water level between the bottom and the top, not just those "special" levels. Your job is to find an expression, in terms of y, for the radius of this crosssection. Can you do this? Remember, what you get will be in terms of y  it'll have a y in the expression, not just a number. 


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