Question

Question: To find: The center of mass of the hemisphere.

Short answer

The center of mass of the hemisphere \((\bar x,\bar y,\bar z)\) is \(\left( {0,0,\frac{1}{2}a} \right)\).

Step-by-step solution

Concept of Surface Integral and Formula Used.

A surface integral is a surface integration generalisation of multiple integrals. It can be thought of as the line integral's double integral equivalent. It is possible to integrate a scalar field or a vector field over a surface.

Given data:

\({x^2} + {y^2} + {z^2} = {a^2}........(1)\)

Formula used:

Calculate the Surface Integral.

Mass of the hemisphere \((m)\) is,

Rearrange the equation.

\(K = \frac{m}{{2\pi {a^2}}}\)

Due to symmetry\({M_{xz}} = {M_{yz}} = 0\)that is\(\bar x = \bar y = 0\).

Therefore,

By using the spherical coordinates to parameterize the sphere, consider

\({\rm{r}}(f,\theta ) = a\sin f\cos \theta {\rm{i}} + a\sin f\sin \theta {\rm{j}} + a\cos f{\rm{k }}\forall {\rm{ }}0 \le \theta \le 2\pi {\rm{ , }}0 \le f \le \frac{\pi }{2}{\rm{. }}\)

Find\({{\bf{r}}_f}\).

Substitute\(a\sin f\cos \theta \)for\(x,a\sin f\sin \theta \)for\(y\)and\(a\cos f\)for\(z\)in equation (3),

\(\begin{array}{l}{{\rm{r}}_\phi } = \frac{\partial }{{\partial \phi }}(a\sin \phi \cos \theta ){\rm{i}} + \frac{\partial }{{\partial \phi }}(a\sin \phi \sin \theta ){\rm{j}} + \frac{\partial }{{\partial \phi }}(a\cos \phi ){\rm{k}}\\ = (a\cos \theta )\frac{\partial }{{\partial \phi }}(\sin \phi ){\rm{i}} + (a\sin \theta )\frac{\partial }{{\partial \phi }}(\sin \phi ){\rm{j}} + (a)\frac{\partial }{{\partial \phi }}(\cos \phi ){\rm{k}}\\ = a\cos \theta \cos \phi {\rm{i}} + a\sin \theta \cos \phi {\rm{j}} - a\sin \phi {\rm{k}}\end{array}\)

Find\({{\rm{r}}_\theta }\).

Substitute\(a\sin f\cos \theta \)for\(x,a\sin f\sin \theta \)for\(y\)and\(a\cos f\)for\(z\)in equation (4),

\(\begin{array}{l}{{\rm{r}}_\theta } = \frac{\partial }{{\partial \theta }}(a\sin \phi \cos \theta ){\rm{i}} + \frac{\partial }{{\partial \theta }}(a\sin \phi \sin \theta ){\rm{j}} + \frac{\partial }{{\partial \theta }}(a\cos \phi ){\rm{k}}\\ = (a\sin \phi )\frac{\partial }{{\partial \theta }}(\cos \theta ){\rm{i}} + (a\sin \phi )\frac{\partial }{{\partial \theta }}(\sin \theta ){\rm{j}} + (a\cos \phi )\frac{\partial }{{\partial \theta }}(1){\rm{k}}\\ = (a\sin \phi )( - \sin \theta ){\rm{i}} + (a\sin \phi )(\cos \theta ){\rm{j}} + (a\cos \phi )(0){\rm{k}}\\ = - a\sin \phi \sin \theta {\rm{i}} + a\sin \phi \cos \theta {\rm{j}}\end{array}\)

Find \({{\rm{r}}_f} \times {{\rm{r}}_\theta }\).

Simplify the equation.

\(\begin{array}{l}\left| {{{\rm{r}}_f} \times {{\rm{r}}_\theta }} \right| = \sqrt {{a^4}{{\sin }^4}f + {a^4}{{\cos }^2}f{{\sin }^2}f} \\ = \sqrt {{a^4}{{\sin }^2}f} = {a^2}\sin f\end{array}\)

Centre of mass of Hemisphere.

The portion of the sphere corresponds to \(0 \le \phi \le \frac{\pi }{2}\)and \(0 \le \theta \le 2\pi \)

Find

Modify equation (1) as follows.

Apply limits and substitute \(a\cos f\) for \(z\) and \({a^2}\sin f\) for \(\left| {{{\rm{r}}_f} \times {{\rm{r}}_\theta }} \right|\),

Simplify the equation.

Substitute \(\frac{m}{{2\pi {a^2}}}\) for \(K\),

Thus, the center of mass of the hemisphere \((\bar x,\bar y,\bar z)\) is \(\left( {0,0,\frac{1}{2}a} \right)\).