Question

Let S be the part of the plane 5x + 4y +z= 4 which lies in the first octant, oriented upward. Find the flux of the vector field F=1i+2j + 3k across the surface S.

Answer 1

To find the flux of the vector field F across the surface S, we need to evaluate the surface integral ∬S F · dS. First, we rewrite the equation of the plane as z = 4 - 5x - 4y and write it in vector form as r(x, y) = (x, y, 4 - 5x - 4y). Next, we substitute the values of F and dS into the integral and simplify the expression (∂r/∂x) × (∂r/∂y). Finally, we evaluate the integral by multiplying the components of F and dS and integrating over the region in the xy-plane defined by the equation of the plane.

Step-by-step explanation:

To find the flux of the vector field across the surface S, we need to use the surface integral. The equation of the plane is given by 5x + 4y + z = 4. We can rewrite this equation as z = 4 - 5x - 4y to get a function that represents the plane in terms of x and y. The normal vector to the plane is (5, 4, 1), so we can write the equation of the plane in vector form as r(x, y) = (x, y, 4 - 5x - 4y).

The surface integral of a vector field F across a surface S is given by ∬S F · dS, where F is the vector field and dS is the differential of surface area. In this case, the vector field F is given by F = 1i + 2j + 3k. The differential of surface area dS is given by dS = (∂r/∂x) × (∂r/∂y) dA, where (∂r/∂x) and (∂r/∂y) are the partial derivatives of r with respect to x and y, and dA is the differential of area in the xy-plane.

To compute the flux, we need to evaluate the following surface integral: ∬S F · dS. Substituting the values of F and dS into the integral, we get ∬S (1i + 2j + 3k) · ((∂r/∂x) × (∂r/∂y)) dA. We can simplify this expression further by calculating (∂r/∂x) × (∂r/∂y), which is equal to the cross product of the vectors (1, 0, -5) and (0, 1, -4). The cross product is given by ((-5)(1) - (-4)(0))i - ((-5)(0) - (-4)(1))j + ((1)(1) - (0)(0))k = -5i + 4j + k.

Finally, we can evaluate the integral by multiplying the components of F and dS, and then integrating over the region in the xy-plane defined by the equation of the plane. The result will be the flux of the vector field across the surface S.

Answer 2

The flux of the vector field F = i + 2j + 3k across the surface S, which is the part of the plane 5x + 4y + z = 4 lying in the first octant and oriented upward, is √14/3 units^2.

Step-by-step explanation:

To find the flux of the vector field F across the surface S, we'll first determine the normal vector to the surface. The given plane equation, 5x + 4y + z = 4, can be rearranged in the form z = 4 - 5x - 4y. This represents a plane in the first octant.

The surface integral for flux involves the dot product of the vector field F and the normal vector to the surface. The normal vector to the plane is the gradient of the surface equation, which is (-5i - 4j + k). Normalize this vector to get the unit normal vector, N = (-5i - 4j + k) / √(5^2 + 4^2 + 1^2) = (-5i - 4j + k) / √42.

The dot product of F and the unit normal vector gives the flux. F dot N = (i + 2j + 3k) dot (-5i - 4j + k) / √42 = -5 + (-8) + 3 / √42 = -10 / √42.

As the surface is oriented upwards, we take the absolute value of the dot product to get the flux: |F dot N| = |-10 / √42| = 10 / √42 = √14 / 3 units^2. This value represents the flux of the vector field F across the surface S in the first octant, oriented upward.