Question

Calculate the flux of the vector field f(x, y, z) = 3hx + y, x − y, x2 + y 2 − 2zi through the surface s parametrized by φ(u, v) = u + 2v, u − 2v, u2 + 2v 2 with 0 ≤ u, v ≤ 1, and oriented by φu × φv

Answer 1

To calculate the flux of the given vector field through the surface S, you need to calculate the partial derivatives of the parametrization, then find the cross product of these partial derivatives, and finally evaluate the double integral over the surface using the flux formula.

Step-by-step explanation:

To calculate the flux of the vector field f(x, y, z) = 3hx + y, x − y, x2 + y2 − 2zi through the surface S parametrized by φ(u, v) = u + 2v, u − 2v, u2 + 2v2 with 0 ≤ u, v ≤ 1 and oriented by φu × φv:

1. Calculate the partial derivatives of φ(u, v) with respect to u and v.
2. Calculate the cross product φu × φv using the partial derivatives obtained in the previous step.
3. Plug the vector field f(x, y, z), the parametrization φ(u, v), and the cross product φu × φv into the flux formula Φ = ∫∫S f • (φu × φv) dA.
4. Evaluate the double integral over the surface S using the limits of integration 0 ≤ u, v ≤ 1.

Answer 2

I'm not sure what to make of the "h" and "i" in your question, so I'll just ignore them (and the 3). Looks like we have

`\vec f(x,y,z)=(x+y,x-y,x^2+y^2-2z)`

and a surface parameterzied by

`\varphi(u,v)=(u+2v,u-2v,u^2+2v^2)`

with
`0\le u\le1` and
`0\le v\le1`. Then

`\varphi_u*\varphi_v=(4u+4v,4u-4v,-4)`

so that the flux is given by the integral

`\displaystyle\iint_S\vec f\cdot\mathrm d\vec S=\int_0^1\int_0^1(2u,4v,4v^2)\cdot(4u+4v,4u-4v,-4)\,\mathrm du\,\mathrm dv`

`=\displaystyle8\int_0^1\int_0^1(u^2+3uv-4v^2)\,\mathrm du\,\mathrm dv=\boxed{-2}`