IDZ Ryabushko 4.2 Option 7

No. 1. It is necessary to construct surfaces and determine their type for the equations:

а) 4x2 + 6y2 – 24z2 = 96; б) y2 + 8z2 = 20x2.

First, let's make a number of transformations. For equation (a), divide both sides by 96 and get:

(x^2) / 6 + (y^2) / 16 - (z^2) / 4 = 1

Thus, we obtain the equation of an ellipsoid whose center is at the origin.

For equation (b) it is also necessary to make a number of transformations. Divide both sides by 20 and get:

(y^2) / 20 + (z^2) / 2 = (x^2)

Thus, we obtain the equation of a parabolic cylinder, the base of which is a parabola directed along the x-axis, and the height is 2√5.

No. 2. It is necessary to write down the equation and determine the type of surface obtained by rotating this line around the specified coordinate axis, and draw the corresponding picture.

а) x^2 + 3z^2 = 9; Oz;

First, let's determine the type of line given by the equation. For z = 0 we get x^2 = 9, that is, it is a horizontal line passing through the points (-3, 0, 0) and (3, 0, 0). At x = 0 we get 3z^2 = 9, that is, this is a vertical line passing through the points (0, 0, -3) and (0, 0, 3). Thus, the line is an ellipse located in the xz plane.

To obtain a surface, it is necessary to rotate this ellipse around the Oz axis. The resulting surface is an ellipsoid, the center of which is at the origin, the axes coincide with the coordinate axes, and the semi-axes are equal to 3 and √3.

б) x = 4; z = 6; Oy.

First, let's determine the type of line given by the equation. The equation x = 4 defines a vertical plane parallel to the y-axis and passing through the point (4, 0, 0). The equation z = 6 defines a horizontal plane parallel to the xz plane and passing through the point (0, 0, 6). Thus, the line is a straight line segment passing through the points (4, 0, 6) and (4, 0, -6).

To obtain a surface, it is necessary to rotate this segment around the Oy axis. The resulting surface is a cylinder whose base is a circle of radius 6 and center at the point (4, 0, 0), and whose height is 12.

No. 3. It is necessary to construct a body bounded by the specified surfaces:

a) y = x; y = 0; x = 1; ... ; z = 0.

First, let's construct the planes defined by the equations y = x and y = 0. This intersection of these planes gives us a line passing through the points (0, 0, 0) and (1, 1, 0). We then construct the planes x = 1 and z = 0, which form a rectangle with vertices (1, 0, 0), (1, 1, 0), (1, 1, 1) and (1, 0, 1). Thus, we obtain a body limited by the indicated surfaces, which is a pyramid with a triangular base and height 1.

б) x^2 + y^2 + z^2 = 4; x^2 + y^2 = z^2; x ≥ 0; z ≥ 0.

First, let's construct the surfaces given by the equations x^2 + y^2 + z^2 = 4 and x^2 + y^2 = z^2. The equation x^2 + y^2 + z^2 = 4 describes a sphere of radius 2, and the equation x^2 + y^2 = z^2 describes a cone with its vertex at the origin and its axis coinciding with the z axis.

To construct a body bounded by these surfaces, we consider a region bounded by a sphere and a cone, as well as by the planes x = 0 and z = 0. Thus, we obtain a body bounded by these surfaces, which represents half a sphere and half a cone located in the first octant.

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IDZ Ryabushko 4.2 Option 7 is a math task that contains three different tasks:

1. It is necessary to construct surfaces and determine their appearance using the given equations: a) 4x2 + 6y2 – 24z2 = 96; b) y2 + 8z2 = 20x2.

2. It is necessary to write down the equation and determine the type of surface obtained by rotating this line around the specified coordinate axis, and also make a drawing: a) x2 + 3z2 = 9; rotation axis Oz; b) x = 4; z = 6; rotation axis Oy.

3. It is necessary to construct a body bounded by the specified surfaces: a) y = x; y = 0; x = 1; ... ; z = 0; b) x2 + y2 + z2 = 4; x2 + y2 = z2; x ≥ 0; z ≥ 0.

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