Question

This glass had a circular base, radius 2cm and height was 10cm. The opposite base, the opened one where you drink from, was an ellipse having half axes 3cm and 4cm. The intermediate section were ellipse changing smoothly from the circular elliptic shape: any section is an ellipse.

Answer 1

x = (2 + u/5) cos t, y = (2 + u/10) sin t, z = u
with 0 ? t ? 2?; 0 ? u ? 10 seem reasonable enough.

Since this only gives the surface of the container, we need one more parameter for the 'thickness' (the inside of the glass) at any cross-section.

This is easy to remedy:
x = (r + u/5) cos t, y = (r + u/10) sin t, and z = u
with 0 ? r ? 2, 0 ? t ? 2?, and 0 ? u ? 10.

The Jacobian ?(x,y,z)/?(r,t,u) equals
|cos t...-(r + u/5) sin t... (1/5) cos t|
|sin t...(r + u/10) cos t... (1/10) sin t|
|....0.................0...........1......

= (r + u/10) cos^2(t) + (r + u/5) sin^2(t)
= r + (u/10) (cos^2(t) + 2 sin^2(t))
= r + (u/10) (1 + sin^2(t))
= r + (u/10) (1 + (1/2)(1 - cos(2t)))
= r + (u/20) (3 - cos(2t)).
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So, the volume ??? 1 dV equals
?(u = 0 to 10) ?(r = 0 to 2) ?(t = 0 to 2?) 1 * |(u/20) (3 - cos(2t))| dt dr du
= ?(u = 0 to 10) (u/20) du * ?(r = 0 to 2) dr * ?(t = 0 to 2?) (3 - cos(2t)) dt

= (1/40)u^2 {for u = 0 to 10} * 2 * (3t - sin(2t)/2) {for t = 0 to 2?}
= 30?.
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