It can be proved that if a bounded plane region slides along a helix in such a way that the region is always orthogonal to the helix (i.e., orthogonal to the unit tangent vector to the helix), then the volume swept out by the region is equal to the area of the region times the distance traveled by its centroid. Use this result to find the volume of the “tubeâ€� in the accompanying figure that is swept out by sliding a circle of radius 1 2 along the helix x = cos t , y = sin t , z = t 4 0 ≤ t ≤ 4 π in such a way that the circle is always centered on the helix and lies in the plane perpendicular to the helix.
It can be proved that if a bounded plane region slides along a helix in such a way that the region is always orthogonal to the helix (i.e., orthogonal to the unit tangent vector to the helix), then the volume swept out by the region is equal to the area of the region times the distance traveled by its centroid. Use this result to find the volume of the “tubeâ€� in the accompanying figure that is swept out by sliding a circle of radius 1 2 along the helix x = cos t , y = sin t , z = t 4 0 ≤ t ≤ 4 π in such a way that the circle is always centered on the helix and lies in the plane perpendicular to the helix.
It can be proved that if a bounded plane region slides along a helix in such a way that the region is always orthogonal to the helix (i.e., orthogonal to the unit tangent vector to the helix), then the volume swept out by the region is equal to the area of the region times the distance traveled by its centroid. Use this result to find the volume of the “tube� in the accompanying figure that is swept out by sliding a circle of radius
1
2
along the helix
x
=
cos
t
,
y
=
sin
t
,
z
=
t
4
0
≤
t
≤
4
π
in such a way that the circle is always centered on the helix and lies in the plane perpendicular to the helix.
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Area Between The Curve Problem No 1 - Applications Of Definite Integration - Diploma Maths II; Author: Ekeeda;https://www.youtube.com/watch?v=q3ZU0GnGaxA;License: Standard YouTube License, CC-BY