21-32 Use spherical coordinates. (a) Find the centroid of a solid homogeneous hemisphere of radi

21-32 Use spherical coordinates. (a) Find the centroid of a...

Key Concepts

Spherical Coordinates

Spherical coordinates provide a system for representing points in three-dimensional space using a radius and two angles. This coordinate system is particularly useful when dealing with spherical or hemispherical regions, as it simplifies the integration process by aligning with the geometry of the region. It replaces the Cartesian triple integral with one that includes the Jacobian r² sin(?), which accounts for the volume element in spherical coordinates.

Centroid of a Solid

The centroid of a solid is the point where the entire mass of a uniformly dense object can be considered to be concentrated. For a homogeneous body, this point also represents its center of mass. Finding the centroid involves integrating the coordinates over the volume of the object, dividing by the total volume or mass, and in symmetric cases, leveraging the symmetry to simplify the computations.

Moment of Inertia

The moment of inertia measures how mass is distributed with respect to a given axis and quantifies the resistance of an object to rotational acceleration about that axis. It is computed by integrating the product of mass density and the square of the distance of each mass element from the axis of rotation. For a solid with a particular symmetry, such as a hemisphere, using the appropriate coordinate system and exploiting symmetry properties can simplify the calculation.

Jacobian in Coordinate Transformations

In any coordinate transformation, such as from Cartesian to spherical coordinates, the Jacobian determinant is a critical factor that modifies the differential volume element to accurately reflect the change in variables. In spherical coordinates, the differential volume element becomes r² sin(?) dr d? d?, which is essential for correctly calculating integrals over volumes in this coordinate system.

Transcript

00:01 Right, so you want to find a centroid of a solid homogeneous hemisphere.

00:04 So we have this hemisphere, radius a, and we want to find its centroid.

00:12 So the first thing we can do is, since it's homogeneous, we already know that it has a volume equal to his mass.

00:22 So at this point, we don't really need to do anything with that yet.

00:24 However, we can go ahead and start with our volume integral.

00:29 So we'll have our volume integral, just a triple integral over h .dv.

00:38 And a little trick we can use is that since we know it's just a hemisphere, we know that we already know the formula of a sphere.

00:45 So we can just do 1 .5 times this formula of a sphere to give us 2 thirds pi a cube as a volume.

00:56 So now we want to find our centroid.

00:58 And since it's a sphere, we know it has x and y of zero.

01:06 And we just want to find our z.

01:07 So our z, we just want over volume.

01:11 So our average z is just one over volume, triple integral of z dv.

01:17 And we can go ahead and start solving this.

01:19 We know that our volume, the bottom, is a hemisphere, so our bounds are the following.

01:28 And z is just row cosine phi, and dv is row squared sine phi with all the differential.

01:35 So we have a row cubed sine phi, cosine phi, d row, d -fi, d -fi, d -d -d -data.

01:42 And solving this, we can go ahead and start separating our intervals.

01:46 So we have 1 over a v, integral from 0 to 2 pi, d data, integral from 0 to pi over 2, sine -fi, cosine, phi, integral from 0 to a of row -cube, d row.

02:02 And now we can start solving everything.

02:04 We know our volume, the inverse volume, is just 3 over 2 pi a cubed.

02:10 We have 2 pi from this first integral.

02:14 Our second integral gives us sine squared phi over 2 when we use u sub.

02:19 And at pi over 2 and 0, it gives us 1 half.

02:23 And finally, our third integral gives us a to the 4th over 4.

02:27 So we can go ahead and start canceling a lot of stuff.

02:30 We have 2 pi's canceling a to the 4th in a cube, giving just 1a.

02:38 And finally, we have our final answer for our centroid z coordinate 3a over 8.

02:44 So our centroid has the coordinates 0 .0 .3 eighths a.

02:52 And now we want to find our moment of inertia about a diameter.

03:00 Now, there's a many days you can do that.