Question

In Exercises 25 and 26, use spherical coordinates to find the center of mass of the solid of uniform density. Hemispherical solid of radius \(r\)

Short answer

The center of mass of the hemispherical solid is \((r,\frac{3r}{8},0)\) in spherical coordinates

Step-by-step solution

Understanding Spherical coordinates

Firstly, it's important to understand that spherical coordinates use three parameters to specify a point in three-dimensional space: the radial distance (r), the inclination angle (θ), and the azimuthal angle (ϕ). Here, the radial distance \(r\) is from the origin to the point, whereas both the azimuthal angle and inclination angle are relative to the origin.

Setting up the Hemisphere

A hemisphere is a solid half-sphere of radius \(r\). The spherical coordinates of any point in the hemisphere are \((r,\theta , \phi)\) where \(0 \leq r \leq R\), \( 0 \leq \theta \leq 2\pi\) and \(0 \leq \phi \leq \pi/2 \). The center of mass of the hemisphere can be found by integrating over these ranges.

Locating the Center of Mass

Since the hemisphere has uniform density, the center of mass is also the geometric center. Considering the hemisphere's symmetry, the center of mass in the x and y directions are simply at the axis of symmetry. In the z-direction, it can be shown that center of mass is at a height of \(3r/8\) from the base of the hemisphere. Therefore, in spherical coordinates, the center of mass is located at \((r,\frac{3r}{8},0)\)