IDZ Ryabushko 4.2 Option 13

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

a) -16x2 + y2 + 4z2 - 32 = 0;

б) 6x2 + y2 - 3z2 = 0.

To solve the problem, it is necessary to bring the equations of surfaces to canonical form.

For surface a) we have:

-16x2 + y2 + 4z2 - 32 = 0

Let's move the free term to the right side of the equation:

-16x2 + y2 + 4z2 = 32

Divide both sides of the equation by 32:

-0.5x2 + 0.125y2 + 0.25z2 = 1

Thus, the surface equation has the canonical form:

x^2/(-2) + y^2/8 + z^2/4 = 1

The resulting surface is an ellipsoid.

For surface b) we have:

6x2 + y2 - 3z2 = 0

Let's move the free term to the right side of the equation:

6x2 + y2 = 3z2

Divide both sides of the equation by 3:

2x2 + y2/3 = z2

Thus, the surface equation has the canonical form:

z^2 = 2x^2 + (y^2/3)

The resulting surface is a hyperbolic paraboloid.

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 make a drawing:

а) z2 = 2y; Yes;

This line is a parabola bounded in the yz plane. When this parabola rotates around the Oy axis, we obtain a surface of rotation - a parabolic cylinder. The surface equation can be obtained by replacing the parabola y in the equation with √(z/2):

z^2/2 = 2y

z^2/2 = 2√(z/2)

z^2 = 8z

Thus, the surface equation has the canonical form:

z^2 - 8z = 0

or

z(z - 8) = 0

The resulting surface is a parabolic cylinder whose axis is the Oy axis.

б) 2x2 + 3z2 = 6; Oz.

This line is an ellipse bounded in the xz plane. When this ellipse rotates around the Oz axis, we obtain a surface of revolution - an elliptical paraboloid. The surface equation can be obtained by replacing z in the ellipse equation with √((6-2x^2)/3):

2x^2 + 3z^2 = 6

2x^2 + 3(6-2x^2)/3 = 6

2x^2 + 6 - 2x^2 = 6

Thus, the surface equation has the canonical form:

y = 6 - 2x^2

The resulting surface is a paraboloid whose axis is the Oz axis.

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

a) y = x; x = 2; y = 0; z = 0;

First, let's plot the surface y = x in three-dimensional space. To do this, note that this is a straight line passing through the origin and point (2, 2). Then we construct planes x = 2, y = 0 and z = 0, which intersect this line at given points. The resulting planes form a parallelepiped, which is the desired body.

б) x + y = 2; ... ; z = 2x; z = 0.

First, let's plot the surface x + y = 2 in three-dimensional space. To do this, note that this is a plane passing through the points (2, 0, 0), (0, 2, 0) and (0, 0, 2). Then we construct the planes z = 2x and z = 0, which intersect this plane at given points. The resulting surfaces form a pyramid with a triangular base, which is the desired body.

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IDZ Ryabushko 4.2 Option 13 is a set of tasks in mathematical geometry, including the construction of surfaces and bodies, as well as writing equations and determining their type. The first task requires you to construct surfaces and determine their appearance. In the second task, you need to write down an equation and determine the type of surface obtained by rotating a given line around the specified coordinate axis and draw it. The third task requires you to construct a body bounded by the specified surfaces and indicate their equations.

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