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) = x i + 2y j + 5z k S is the cube with vertices (±1, ±1, ±1)

Step-by-step explanation:

To solve this problem, we will use the following two theorems/definitions:

- Given a vector field F of the form (P(x,y,z),Q(x,y,z),W(x,y,z)) then the divergence of F denoted by

- (Gauss' theorem)Given a closed surface S, the following applies

where n is the normal vector pointing outward of the surface and V is the volume bounded by the surface S.

Let us, in our case, calculate the divergence of the given field. We have that

Hence, by the Gauss theorem we have that

So, we must calculate the volume V bounded by the cube S.

We know that the vertices are located on the given points. We must determine the lenght of the side of the cube. To do so, we will take two vertices that are on the some side and whose coordinates differ in only one coordinate. Then, we will calculate the distance between the vertices and that is the lenght of the side.

Take the vertices (1,1,1) and (1,1-1). The distance between them is given by

.

Hence, the volume of V is . Then, the final answer is

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}