Question

Question: To find: The rate of flow outward through the cylinder \({x^2} + {y^2} = 4,0 \le z \le 1\).

Short answer

The rate of flow outward through the cylinder \({x^2} + {y^2} = 4,0 \le z \le 1\) is \(0{\rm{Kg}}/{\rm{s}}\)..

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:

The density of the fluid is\(870\frac{{{\rm{kg}}}}{{{{\rm{m}}^3}}}\)and the velocity is\({\rm{v}} = z{\rm{i}} + {y^2}{\rm{j}} + {x^2}{\rm{k}}\frac{{\rm{m}}}{{\rm{s}}}\).

The equation of the cylinder is given by\({x^2} + {y^2} = 4,0 \le z \le 1\).

Formula used:

Calculate the Surface Integral.

By using the parametric representation of the cylinder, consider\(x = 2\cos u,y = 2\sin u,z = v\)and limits for\(u\)and\(v\)are 0 to\(2\pi \)and 0 to 1 .

\({\rm{r}}(u,v) = 2\cos u{\rm{i}} + 2\sin u{\rm{j}} + v{\rm{k}}\)

Substitute\(2\cos u\)for\(x,2\sin u\)for\(y\)and\(v\)for\(z\)in equation (2) and find\({r_u}\)as,

\(\begin{array}{l}{{\rm{r}}_u} = \frac{\partial }{{\partial u}}(2\cos u){\rm{i}} + \frac{\partial }{{\partial u}}(2\sin u){\rm{j}} + \frac{\partial }{{\partial u}}(v){\rm{k}}\\ = - 2\sin u{\rm{i}} + 2\cos u{\rm{j}} + 0{\rm{k}}\\ = - 2\sin u{\rm{i}} + 2\cos u{\rm{j}}\end{array}\)

Substitute\(2\cos u\)for\(x,2\sin u\)for\(y\)and\(v\)for\(z\)in equation (3) and find\({r_v}\).

To find , apply limits and substitute \(z{\rm{i}} + {y^2}{\rm{j}} + {x^2}{\rm{k}}\) for \(v,2\cos u{\rm{i}} + 2\sin u{\rm{j}}\) for \({{\rm{r}}_u} \times {{\rm{r}}_v}\),

Substitute \(2\cos u\) for \(x,2\sin u\) for \(y\) and \(v\) for \(z\),

On further simplification,

\(\begin{array}{l} = \rho \left( {\int_0^{2\pi } {(\cos u)} du + \int_0^{2\pi } {\left( {8{{\sin }^3}u} \right)} du} \right)\\ = \rho \left( {(\sin u)_0^{2\pi } + \frac{8}{{12}}(\cos (3u) - 9\cos (u))_0^{2\pi }} \right)\\ = \rho \left( {[\sin (2\pi ) - \sin (0)) + \frac{8}{{12}}(\cos (6\pi ) - 9\cos (2\pi ) - \cos (0) + 9\cos (0))} \right)\end{array}\)

Modify the equation.

Thus, the rate of flow outward through the cylinder \({x^2} + {y^2} = 4,0 \le z \le 1\) is \(0{\rm{Kg}}/{\rm{s}}\)..