Volume of revolution?

Question

triangle whose sides are 2 ft, 2 ft and 3 ft is revolved about its longest side. find the volume generated

its not a cone
its a turos

Answer 1

You get two cones - back to back.

Radius of base = altitude of triangle to longest side = r
r^2 = 2^2 - (3/2)^2 = 1.75ft^2
height of each cone = 3/2 = 1.5ft

Volume of a cone = (1/3)pi*(r^2)*h
Total volume = (1/3)pi*1.75*1.5 + (1/3)pi*1.75*1.5 ft^2
= 1.75pi ft^2
= 5.4978 ft^3

It will be a pair of cones. Try cutting out a triangular piece of paper and rotating it about one of its sides. A torus is a completely different shape. "In geometry, a torus (pl. tori) is a surface of revolution generated by revolving a circle in three dimensional space about an axis coplanar with the circle, which does not touch the circle."
Como calculated the volume integration, and got the same result. He got a bit of a rounding error by plugging in the numbers early, but otherwise our answers are the same, using two different methods.

Answer 2

Slit the triangle down the middle to give two right triangles a hypotenuse of 2 and the other side 1.5 (where you've cut the 3 in half). Calculate the other side gives 1.32, then simply place the two triangles together to give a rectangle of 1. 50 x 1.32 ( the diagonal is 2). Now pivot at the 1.5 to create a cylinder with a radius of 1.32 and 1.5 high.
Answer: 8.214

Answer 3

V=(11/4)π

Answer 4

Consider isosceles triangle OAB where O is the origin.
OA = AB = 2 ft
OB = 3 ft
OB lies on x axis
C is mid point of OB
OC = 1.5 ft
AC² = 2² - 1.5²
AC² = 0.5 x 3.5
AC² = 1.75
AC = 1.32
Gradient of OA = 1.32 / 1.5 = 0.89
Equation of line OA is y = 0.89 x
Volume of revolution of triangle OAB when rotated about x axis is:-
V = 2.∫ π y² dx between limits of x = 0 to x = 1.5
V = 2π ∫ (0.89)² x² dx (lims 0 to 1.5)
V = 4.98 x³ / 3 (lims 0 to 1.5)
V = 5.6 units³

Answer 5

no