A 15cm long toothpaste tube starts off as a circle of radius 2cm and then finishes as

[student001]

A 15cm long toothpaste tube starts off as a circle of radius 2cm and then finishes as a straight line of length 4cm. Cross sections taken perpendicular to the circle are isosceles triangles. Find the volume of the toothpaste in the tube when full. Answer is 30pi cubic centimetres.

I can get 30pi by taking cross sections parallel to the circular end as ellipses with a (semi major axis) 2cm and b (semi minor axis) as 2h/15 where h is the length from the base and runs from 0 to 15. Using pi.a.b as the area of an ellipse and integrating, I get integral from 0 to 15 of pi(2)(2h/15)dh which comes out as 30pi. The question suggests using isosceles triangles as slices however. Using that approach, I can't get that to work out as 30pi.

Can anyone help? Thanks.

 

[Deleted member 4993]

A 15cm long toothpaste tube starts off as a circle of radius 2cm and then finishes as a straight line of length 4cm. Cross sections taken perpendicular to the circle are isosceles triangles. Find the volume of the toothpaste in the tube when full. Answer is 30pi cubic centimetres.

I can get 30pi by taking cross sections parallel to the circular end as ellipses with a (semi major axis) 2cm and b (semi minor axis) as 2h/15 where h is the length from the base and runs from 0 to 15. Using pi.a.b as the area of an ellipse and integrating, I get integral from 0 to 15 of pi(2)(2h/15)dh which comes out as 30pi. The question suggests using isosceles triangles as slices however. Using that approach, I can't get that to work out as 30pi.

Can anyone help? Thanks.

If you take a section at a distance x from the origin (center of the circle at the bottom), you will get a triangle of base 2y and height 15 cm.

The volume of the slice = 1/2 * 2y * 15 dx = 15 * y dx = 15 * √(4 - x^2) dx

Now integrate..

 

[student001]

thanks

Thanks for the help with this problem. I was making the two equal sides of the isosceles triangle both equal to 15 and then solving for the height of the triangle using Pythagoras. This gave an integral of a product of two square roots which was not able to be done. By letting the height of the isosceles triangle be 15, as you have done (and now I read it carefully is what is implied in the question), we then get a nice integral, which, yay, ends up as 30pi.