DHS 15.1 - Option 28. Decisions Ryabushko AP

Product description

• Dana function u (M) = u (x, y, z) and the point M1, M2. Calculate: 1) The derivative of this function in the direction of the point M1 M1M2 vector; 2) grad u (M1)

1.28. u (M) = x3 + xy2 - 6xyz, M1 (1, 3, -5), M2 (4, 2, -2)

• Calculate the surface integral of the first kind on the surface S, where S - part of the plane (p), cut off by the coordinate planes.

(P): 2x + 2y + z = 2

• Calculate the surface integral of the second kind.

where S - the surface of the cone x2 + y2 = z2 (normal vector n which forms an obtuse angle with the unit vector k), cut off by the plane z = 0 and z = 4.

• Calculate the flow vector field a (M) through the outer surface of the pyramid formed by plane (p) and the coordinate planes in two ways: a) determination using flow; b) using the formula Ostrogradskii - Gauss.

4.28. and (M) = (y + z) i + xj + (y - 2z) k, (p): 2x + 2y + z = 2

Additional information

Detailed solution. Designed in PDF format for easy viewing of IDZ solutions on smartphones and PCs. In MS Word (doc format) sent additionally.