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leonid [27]
2 years ago
9

Help me pleaseeeeeeeeeeeeeeeee

Mathematics
2 answers:
Julli [10]2 years ago
4 0

Answer:

7

Step-by-step explanation:

My name is Ann [436]2 years ago
4 0
7f=49
Divide both sides by 7
F=7
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) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle
Fynjy0 [20]

The surface (call it S) is a triangle with vertices at the points

x=0,y=0\implies z=2\implies(0,0,2)

x=0,y=2\implies z=-2\implies(0,2,-2)

x=2,y=0\implies z=-8\implies(2,0,-8)

Parameterize S by

\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)

with 0\le u\le1 and 0\le v\le1. Take the normal vector to S to be

\vec s_v\times\vec s_u=(20v,8v,4v)

Then the flux of \vec F across S is

\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv

=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}

8 0
3 years ago
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