Asked by Sean
What would the radius and height be of a 3-dimensional cylinder that was formed by rotating a square, with 3 inch sides, around the y-axis?
Answers
Answered by
Reiny
Where was the y-axis located as the square rotated around it?
length of a diagonal in the square is
√(3^2 + 3^2) = √18 = 3√2
case 1, the y-axis runs down the centre of the square, so the
radius is (3/2)√2
volume = π r^2 h = π(9/2)(3) inches^3 = 13.5π inches^3 or appr 42.4 inches^3
case 2. the square is rotated with the y-axis along one of the edges,
then the radius would be 3√2
What would be the volume of the cylinder in that case?
length of a diagonal in the square is
√(3^2 + 3^2) = √18 = 3√2
case 1, the y-axis runs down the centre of the square, so the
radius is (3/2)√2
volume = π r^2 h = π(9/2)(3) inches^3 = 13.5π inches^3 or appr 42.4 inches^3
case 2. the square is rotated with the y-axis along one of the edges,
then the radius would be 3√2
What would be the volume of the cylinder in that case?
Answered by
oobleck
surely for case 1 the radius is just 3/2 if you want a cylinder and not two cones joined at their base.
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