# Integration question again

1. Feb 23, 2013

### lionely

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 cross-section of it, and then try to use the principles of solids of revolution to find the volume?

2. Feb 23, 2013

### Curious3141

Consider the vertical cross-section of the bowl. Let the centre of the bowl be the origin (0,0). y-coordinates range from -10 to +8.

Now consider the horizontal circular cross section of a disc of water taken at a certain y-coordinate. Find its radius via Pythagoras theorem, and hence its area.

Hence figure out the volume of an infinitesimally small cylinder having that cross-section and a vertical height dy.

Now do the integration, imposing the correct bounds for y.

3. Feb 23, 2013

### lionely

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. Feb 24, 2013

### Curious3141

Right through the centre (at y = 0), yes the radius of the cross-section would be 10cm.

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 cross-section.

5. Feb 24, 2013

### lionely

6. Feb 25, 2013

### Curious3141

Huh? The radius of the horizontal cross-section varies continuously depending on the level you're taking it at.

Did you make a proper sketch?

7. Feb 25, 2013

### lionely

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

8. Feb 26, 2013

### Curious3141

10cm is always the hypotenuse.

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 cross-section 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 cross-section. 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.

Last edited: Feb 26, 2013