Question

Question: (a) To determine

The integral expression for the moment of inertia \({I_z}\)about the \({\bf{z}} - \)axis.

(b) To determine

The moment of inertia about \({\bf{z}} - \)axis of the funnel.

Short answer

(a) The integral expression for the moment of inertia \({I_z}\)about the \(z - \)axis is;

(b) The moment of inertia about \(z - \)axis of the funnel is \(\frac{{4329}}{5}\sqrt 2 \pi \).

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.

The equation of moment of inertia is

The given functions are \(\rho (x,y,z) = 10 - z,z = {x^2} + {y^2},1 \le z \le 4.\)

Formula used:

The greens theorem for the function \(F\)is

Calculate integral expression for the moment of inertia.

a. Calculate the moment of inertia about the \(z - \)axis as follows.

Note that for \(z - \)axis we have \(z = {x^2} + {y^2}\).

Substitute \({x^2} + {y^2}\) for \(xz\) in equation (1) as follows.

Thus, the integral expression for the moment of inertia \({I_z}\) about the \(z\)-axis is .

Calculate the moment of inertia .

(b)

Substitute \(10 - z\)for \(\rho (x,y,z)\) and \(\sqrt {{x^2} + {y^2}} \)for \(z\)in equation (2),

Modify equation (3) as follows.

Obtain\({{\rm{r}}_x}\)as shown below.

Modify equation (3) as follows.

\({{\rm{r}}_x} = \frac{{\partial x}}{{\partial x}}{\rm{i}} + \frac{{\partial y}}{{dx}}{\rm{j}} + \frac{{\partial z}}{{\partial x}}{\rm{k}}\)

Substitute\(x\)for\(x,y\)for\(y\)and

\(\begin{array}{l}\sqrt {{x^2} + {y^2}} \forall z\\{{\bf{r}}_x} = \frac{{\partial x}}{{\partial x}}{\rm{i}} + \frac{{\partial y}}{{\partial x}}{\rm{j}} + \frac{{\partial z}}{{\partial x}}{\rm{k}}\$ ,\end{array}\)

\(\begin{array}{l}{{\rm{r}}_x} = \frac{\partial }{{\partial x}}(x){\rm{i}} + \frac{\partial }{{\partial x}}(y){\rm{j}} + \frac{\partial }{{\partial x}}\left( {\sqrt {{x^2} + y} } \right.\\ = {\rm{i}} + (0){\rm{j}} + \left( {\frac{{2x}}{{2\sqrt {{x^2} + {y^2}} }}} \right){\rm{k}}\\ = {\rm{i}} + \left( {\frac{x}{{\sqrt {{x^2} + {y^2}} }}} \right){\rm{k}}\end{array}\)

Obtain\({r_y}\)as follows.

Modify equation (3) as shown below.

\({{\rm{r}}_y} = \frac{{\partial x}}{{\partial y}}{\rm{i}} + \frac{{\partial y}}{{\partial y}}{\rm{j}} + \frac{{\partial z}}{{\partial y}}{\rm{k}}\)