Finding the Volume of a Whisky Glass: Which Method Results in More Whisky?

[lionely]

Homework Statement

The region R shown in Fig. 1 is bounded by the line y=8(x-2), the axes, and the line y=h. Find by integration the volume formed when R is rotated through 360° about the y axis(see Fig. 2)

A whisky glass has the shape indicated in Fig 2 where the units are centimetres. A whisky taster
holds the glass upright and pours in whisky to a depth of 2cm. He then adds water to a further depth of 2cm. Show that if he had poured in the water first and then the whisky, each to a depth of 2cm the glass would have contained approximately 25% more whisky than by the first method.





The attempt at a solution

For the volume generated I got 55∏/6 cubic units.

It's the 2nd part of the question that is puzzling me.
Well I know realize that the cup is not uniform so.. there would be more whiskey in the 2nd method. But how do I show this mathematically?

HELP MUCH APPRECIATED!

 

[Dick]

Homework Statement

The region R shown in Fig. 1 is bounded by the line y=8(x-2), the axes, and the line y=h. Find by integration the volume formed when R is rotated through 360° about the y axis(see Fig. 2)

A whisky glass has the shape indicated in Fig 2 where the units are centimetres. A whisky taster
holds the glass upright and pours in whisky to a depth of 2cm. He then adds water to a further depth of 2cm. Show that if he had poured in the water first and then the whisky, each to a depth of 2cm the glass would have contained approximately 25% more whisky than by the first method. The attempt at a solution

For the volume generated I got 55∏/6 cubic units.

It's the 2nd part of the question that is puzzling me.
Well I know realize that the cup is not uniform so.. there would be more whiskey in the 2nd method. But how do I show this mathematically?

HELP MUCH APPRECIATED!

I'm confused as to how the volume you got for the first part doesn't depend on h. What did you use for h? For the second part you just need to compute the volumes of the two halves.

 

[lionely]

to get the volume I did this

∏∫x²δy (from 0 to 2)

= ∏[ y³/48 + y²/4 + 4y]

 

[lionely]

Also how would I calculate the volumes of the halves?

Would I call the top of the liquids y=2 and y=4 for each liquid?

since 4cm of liquid was added? hmm and integrate x^2 = (y^2/8 + 2)

for the certain limits?

 

[Infrared]

Check your volume. Firstly, you are integrating with respect to y, but you are using x=0 and x=2 as bounds. You want to convert these into y-values (Hint: The fact that your answer did not include an h when it obviously should affect your answer tells you that something is wrong). Also check your algebra. If y=8(x-2), then x= y/8+2 and x^2=y^2/64+y/2+4. You integrated the last two terms correctly, but ∫y^2/64 dy=y^3/(64*3) ≠ y^3/48

 

[lionely]

Wow sorry, I get (h^3 + 48h^2 + 768h)/192 Pi cubic units

 

[Dick]

Wow sorry, I get (h^3 + 48h^2 + 768h)/192 Pi cubic units

That's right. If you evaluate that at h=2 you get the volume of the first half. h=4 gives you the sum of the first half and the second half. So subtract to get the volume of the second half.

 

[lionely]

OHHHHHHH I see now because h wasn't in my volume , it made it SEEM SO CONFUSING ! When this question is somewhat easy... thank you guys! I feel soo dumb for using the x's as bounds why it was with respect to y :( .