Verified answer

They want a flux, but the wording of the question is bizarre. The "boundaries" given are themselves surfaces; probably what was meant was that you are to consider the volume bounded by the four given surfaces, and find the flux outward through the boundary of that volume.

To get a picture of the volume, note that y=0, z=0, and y+z = 2 are the boundaries of an infinitely long shaft (in the x direction) with a triangular cross-section. The fourth boundary, z = 1 - x^2, cuts it down into a rather short shaft (in the y direction) with a curved top and two flat ends.

To find the outward flux, first calculate div(F) and then integrate it over the volume. The divergence of F is 2 + x + xz, not too hard to deal with. I would integrate in the y direction first: the integrated value along each little matchstick parallel to the y axis will be (2 + x + xz)(2 - z), because the length of the matchstick is y = 0 to y = 2-z. Now you want to integrate this function (4 - 2z + 2x - xz + 2xz - xz^2) in the z-direction. Simplify the function a bit, it's (4 - 2z + 2x + xz - xz^2). The limits in the z direction are from z = 0 to z = 1 - x^2. Integrating with respect to z gives

(4z - z^2 + 2zx + (1/2)xz^2 - (1/3)xz^3), and plug in z = (1 - x^2) to get

(4(1-x^2) - (1-x^2)^2 + 2x(1-x^2) + (1/2)x(1-x^2)^2 - (1/3)x(1-x^2)^3),

and probably you should expand and simplify this before doing the final integration in the x-direction.

When the above expression has been simplified, integrate it from x = -1 to x = 1, and the answer will be an actual number (!) which will give you the outward flux through the boundary of the volume described.

BTW I think your teacher is a moron for saying S is a surface bounded by four surfaces. An INCORRECT way of posing the question.

Later … I like Indica's shortcut, it seems legitimate. AND Indica brings out the fact that I made a booboo in saying div(F) = 2 + x + xz, of course it's not; instead it's 2 + x + xy.

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Volume is a tunnel along the y-axis with ceiling z=1−x², floor z=0 and plane ends y=0, y+z=2

The best integration order for this is y,z,x where 0≤y≤2−z, 0≤z≤1−x², −1≤x≤1

div(𝑭)=2+x+xy so by DT flux out = ∭ (2+x+xy) dydzdx, [y=0,2−z], [z=0,1−x²], [x=−1,1]

Notice that the x+xy part of integrand is an odd function of x so over [−1,1] there is no contribution.

Hence integral can be simplified to ∭ (2) dydzdx, [y=0,2−z], [z=0,1−x²], [x=−1,1]

= ∬ (4−2z) dzdx, [z=0,1−x²], [x=−1,1]

= ∫ (4(1−x²)−(1−x²)²) dx, [x=−1,1] = ∫ (3−2x²−x⁴)dx, [x=−1,1]

= 3(2) – (2/3)(2) – (1/5)(2) = 64/15

A double integral is defined as a limit of a Riemann sum.