G. Set up a triple integral whose value gives the moment of inertia about the z-axis for a plastic conical 'top' bound by z = the Wolfram Alpha widget. (Ans: 50.2655) Vr + y? and the plane z = 2. The density of the plastic is p = 5. Evaluate using H. 1 the rooion mmon to the he his

We have shown in class that the acceleration of a particle can be decomposed into components that are tangential and normal to its trajector as a = kv²N+ÏT where T and N are the unit tangent vector and principal (unit) normal vector, respectively, is the length along the trajectory, v = is the speed, and is the curvature of the trajectory. The curvature is related to the radius of curvature p by k = 1/p. The plane formed by T and N is called the osculating plane, and can be thought of as the instantaneous plane of the trajectory. The unit vector B = TX N is called the binormal vector, and by definition is always perpendicular to the osculating plane. We have shown in class that both T and B are parallel to N. The torsion r(t) of the trajectory is defined by or, equivalently, by B = -TIN dB de = -T, N. (i) The concept of curvature and radius of curvature are defined by extending those con- cepts from circles to curves in general. The curvature & is defined as the rate (with respect to…

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…

I need complete solution here. And answer must be in 4 decimal places and also the graph.