Question

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.

In other words, find the flux of F across S.

For closed surfaces, use the positive (outward) orientation.

F(x, y, z) = xy i + yz j + zx k S is the part of the paraboloid z = 3 − x2 − y2 that lies above the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and has upward orientation

Answer 1

2.794

Explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then

`\bf \displaystyle\iint_(S)FdS=\displaystyle\iint_(R)F(G(x,y))\cdot(\displaystyle(\partial G)/(\partial x)*\displaystyle(\partial G)/(\partial y))dxdy`

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

`\bf G(x,y) = (x, y, 3-x^2-y^2)`

with 0≤ x≤1 and 0≤ y≤1

So

`\bf \displaystyle(\partial G)/(\partial x)= (1,0,-2x)\\\\\displaystyle(\partial G)/(\partial y)= (0,1,-2y)\\\\\displaystyle(\partial G)/(\partial x)*\displaystyle(\partial G)/(\partial y)=(2x,2y,1)`

we also have

`\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)`

and so

`\bf F(G(x,y))\cdot(\displaystyle(\partial G)/(\partial x)*\displaystyle(\partial G)/(\partial y))=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2`

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and 0≤ y≤1

`\bf \displaystyle\int_(0)^(1)\displaystyle\int_(0)^(1)(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_(0)^(1)x^2dx\displaystyle\int_(0)^(1)ydy+6\displaystyle\int_(0)^(1)dx\displaystyle\int_(0)^(1)y^2dy-2\displaystyle\int_(0)^(1)x^2dx\displaystyle\int_(0)^(1)y^2dy-2\displaystyle\int_(0)^(1)dx\displaystyle\int_(0)^(1)y^4dy+\\\\+3\displaystyle\int_(0)^(1)xdx\displaystyle\int_(0)^(1)dy-\displaystyle\int_(0)^(1)x^3dx\displaystyle\int_(0)^(1)dy-\displaystyle\int_(0)^(1)xdx\displaystyle\int_(0)^(1)y^2dy`

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794