Question

Question: To find: The rate of flow outward through the hemisphere \({x^2} + {y^2} + {z^2} = 9,z \ge 0\).

Short answer

The rate of flow outward through the hemisphere \({x^2} + {y^2} + {z^2} = 9,z \ge 0\) is \(0\frac{{{\rm{kg}}}}{{\rm{m}}}\).

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:

Density is\(1025\frac{{{\rm{kg}}}}{{{m^3}}},{\rm{v}} = y{\rm{i}} + x{\rm{j}}\frac{m}{s}\)and hemisphere with\({x^2} + {y^2} + {z^2} = 9,z \ge 0\).

Formula used:

Calculate the Surface Integral.

By using the spherical coordinates to parameterize the hemisphere, consider \({\rm{r}}(f,\theta ) = 3\sin f\cos \theta {\rm{i}} + 3\sin f\sin \theta {\rm{j}} + 3\cos f{\rm{k}}\) with limits are \(0 \le \theta \le 2\pi \) and \(0 \le f \le \frac{\pi }{2}.\)

Find\({r_f}\).

Substitute\(3\sin f\cos \theta \)for\(x,3\sin f\sin \theta \)for\(y\)and\(3\cos f\)for\(z\),in equation (2),

\(\begin{array}{l}{{\rm{r}}_\phi } = \frac{\partial }{{\partial \phi }}(3\sin \phi \cos \theta ){\rm{i}} + \frac{\partial }{{\partial \phi }}(3\sin \phi \sin \theta ){\rm{j}} + \frac{\partial }{{\partial \phi }}(3\cos \phi ){\rm{k}}\\ = (3\cos \theta )\frac{\partial }{{\partial \phi }}(\sin \phi ){\rm{i}} + (3\sin \theta )\frac{\partial }{{\partial \phi }}(\sin \phi ){\rm{j}} + (3)\frac{\partial }{{\partial \phi }}(\cos \phi ){\rm{k}}\\ = 3\cos \theta \cos \phi {\rm{i}} + 3\sin \theta \cos \phi {\rm{j}} - 3\sin \phi {\rm{k}}\end{array}\)

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

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

in equation (3),

\(\begin{array}{l}{{\rm{r}}_\theta } = \frac{\partial }{{\partial \theta }}(3\sin \phi \cos \theta ){\rm{i}} + \frac{\partial }{{\partial \theta }}(3\sin \phi \sin \theta ){\rm{j}} + \frac{\partial }{{\partial \theta }}(3\cos \phi ){\rm{k}}\\ = (3\sin \phi )\frac{\partial }{{\partial \theta }}(\cos \theta ){\rm{i}} + (3\sin \phi )\frac{\partial }{{\partial \theta }}(\sin \theta ){\rm{j}} + (3\cos \phi )\frac{\partial }{{\partial \theta }}(1){\rm{k}}\\ = (3\sin \phi )( - \sin \theta ){\rm{i}} + (3\sin \phi )(\cos \theta ){\rm{j}} + (3\cos \phi )(0){\rm{k}}\\ = - 3\sin \phi \sin \theta {\rm{i}} + 3\sin \phi \cos \theta {\rm{j}}\end{array}\)

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

Calculate the rate of flow outward through the hemisphere .

Apply limits and substitute \(y{\rm{i}} + x{\rm{j}}\) for \(v,9{\sin ^2}f\cos \theta {\rm{i}} + 9{\sin ^2}f\sin \theta {\rm{j}} + 9\cos f\sin f{\rm{k}}\) for \({{\rm{r}}_f} \times {{\rm{r}}_\theta }\),

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

Simplify the equation.

Modify the equation.

Thus, the rate of flow outward through the hemisphere \({x^2} + {y^2} + {z^2} = 9,z \ge 0\) is \(0\frac{{{\rm{kg}}}}{{\rm{m}}}\).