In Problems 17 to 30, for the curve y = x , between x = 0 and x = 2 , find 24, 25, 26, 27. The moments of inertia about the x axis of a lamina in the shape of the plane area under the curve; of a wire bent along the arc of the curve; of the solid of revolution; and of a thin shell whose shape is the curved surface of the solid (assuming constant density for all these problems).
In Problems 17 to 30, for the curve y = x , between x = 0 and x = 2 , find 24, 25, 26, 27. The moments of inertia about the x axis of a lamina in the shape of the plane area under the curve; of a wire bent along the arc of the curve; of the solid of revolution; and of a thin shell whose shape is the curved surface of the solid (assuming constant density for all these problems).
In Problems 17 to 30, for the curve
y
=
x
, between
x
=
0
and
x
=
2
, find 24, 25, 26, 27. The moments of inertia about the x axis of a lamina in the shape of the plane area under the curve; of a wire bent along the arc of the curve; of the solid of revolution; and of a thin shell whose shape is the curved surface of the solid (assuming constant density for all these problems).
If you compute the surface area of a sphere as the volume of revolution of the parametric curve c(t) = (cos t, sin t), as t ranges from 0
to 27. you get 0. If you use the same parametrization, but restrict t to the range from () to T, you get 47 (the correct surface area for the
sphere of radius 1).
Why did you get 0 when t ranged from () to 27?
O Having closed the circle parametrized by c, the displacement is zero, so the surface area is zero.
O Sine has the same (unsigned) area of its positive part on 0, 27 as its negative part, but on 0, 7|, it is all positive. So the sine contribution cancels out
on the longer interval and not on the shorter interval.
O Correct.
The contribution from t e [0, 7| is exactly cancelled by t e T, 27| because the positive contributions for y(t)
= cos tin the first interval are
exactly cancelled by the negative contributions in the second interval.
O The arc length contribution from the upper semicircle is negative because x' (t) < 0 on that semicircle…
- Another bowl has the shape of a paraboloid. The equation of the cross-
section with the (x, z)-plane is x² = 2RAZ, where RA is the curvature
radius at point A. One releases at a point with height z = h a point
mass with initial velocity equal to 0. What is in this case the magnitude
of the force exerted by the particle on the bowl when it passes point
A?
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Fundamental Theorem of Calculus 1 | Geometric Idea + Chain Rule Example; Author: Dr. Trefor Bazett;https://www.youtube.com/watch?v=hAfpl8jLFOs;License: Standard YouTube License, CC-BY