IDZ 9.2 – Option 8. Solutions Ryabushko A.P.

1. Calculating the area of ​​a figure bounded by lines

The equation is given: 1.8 ρ2 = 2sin2φ

Let's find the limits of integration:

1.8 ρ2 = 2sin2φ

ρ2 = 2/(1.8sin2φ)

ρ = sqrt(2/(1.8sin2φ))

The constraint is given: 0 ≤ φ ≤ π/4

Then the limits of integration will be: 0 ≤ ρ ≤ sqrt(2/(1.8sin2φ))

Thus, the area of ​​the figure will be equal to:

S = ∫∫D ρ dφ dρ

S = ∫0^(π/4) ∫0^sqrt(2/(1.8sin2φ)) ρ drρ dφ

S = 1.8/2 ∫0^(π/4) (2/(1.8sin2φ)) dφ

S = 1.8/2 [1/2 ln(tan(π/8)) - 1/2 ln(tan(0))] ≈ 0.32

Answer: The area of ​​the figure bounded by the indicated lines is approximately 0.32.

1. Calculating the arc length of a line

Dano's equation: 2.8 y = 1− lncosx, (0 ≤ x ≤ π/6)

Let's find the first derivative:

y' = -(2.8/cos(x)) * (-sin(x))

y' = 2.8 * tan(x)

Then the arc length will be equal to:

L = ∫0^(π/6) sqrt(1 + (y')^2) dx

L = ∫0^(π/6) sqrt(1 + (2.8tan(x))^2) dx

L = ∫0^(π/6) sqrt(1 + 7.84tan^2(x)) dx

Let's make a replacement: t = tan(x)

dx = dt / (1 + t^2)

L = ∫0^tan(π/6) sqrt(1 + 7.84t^2) dt / (1 + t^2)

L = ∫0^tan(π/6) sqrt((1 + 0.84t^2) / (1 + t^2)) dt

Let's make a replacement: u = 1 + 0.84t^2

du = 1.68t dt

L = 1.68 ∫1.84^(1.84tan(π/6)^2) sqrt(u / (u - 1.84)) du / (1.68u - 1.4352)

L ≈ 1.05

Answer: The arc length of this line is approximately 1.05.

1. Calculation of the volume of a body obtained by rotating the figure Ф around the coordinate axis

Given equation: 3.8 Ф: y2 = (x – 1)3, x = 2, Ox

Let's find a function that describes the figure:

y = (x – 1)^(3/2)

Let us find the volume of the body obtained by rotating the figure around the Ox axis:

V = ∫2^3 πy^2 dx

V = π ∫2^3 (x – 1)^3 dx

V = π [(x – 1)^4 / 4]│2^3

V = π (81 / 4)

Answer: the volume of the body obtained by rotating the figure Ф around the coordinate axis is equal to 20.09.

1. Calculation of the surface area formed by rotating the arc of a curve L around a specified axis

Given equation: 4.8 L: x = cost, y = 3 + sint, Ox

Let us find the function describing the arc of the curve L:

x^2 + (y – 3)^2 = 1

From here we get:

y = 3 + sqrt(1 – x^2)

Let us find the surface area formed by the rotation of this arc around the Ox axis:

S = 2π ∫0^1 y √(1 + (y')^2) dx

S = 2π ∫0^1 (3 + sqrt(1 – x^2)) √(1 + x^2 / (1 – x^2)) dx

Let's make a replacement: t = √(1 – x^2)

x = √(1 – t^2)

dx = (-t / √(1 – t^2)) dt

S = 2π ∫0^1 (3 + t) √(1 + 1 / t^2) (-t / √(1 – t^2)) dt

S = -2π ∫0^1 (3t + t^2) / √(1 – t^2) dt

Let's make a replacement: u = 1 – t^2

you = -2t dt

S = π ∫0^1 (u + 1) / √u you

S = π [2/3u^(3/2) + 2u^(1/2)] │0^1

S = 4π/3

Answer: The surface area formed by the rotation of the arc of the curve L around the indicated axis is equal to 4π/3.

Product description:

IDZ 9.2 – Option 8. Solutions Ryabushko A.P.

IDZ 9.2 – Option 8. Solutions Ryabushko A.P. is a unique digital product designed for students and students who want to receive detailed and understandable solutions to problems in mathematics. The product was developed by an experienced mathematics teacher - A.P. Ryabushko. and contains solutions to problems in various branches of mathematics.

Beautiful design in HTML format gives the product an attractive appearance and ease of use. Solutions are presented in a clear and easy-to-understand form, allowing you to quickly and efficiently learn the material.

IDZ 9.2 – Option 8. Solutions Ryabushko A.P. is an excellent choice for students who want to improve their knowledge and skills in mathematics, as well as for teachers who are looking for quality materials for preparing school assignments.

Product IDZ 9.2 – Option 8. Solutions Ryabushko A.P. is a digital material that contains detailed solutions to problems in various branches of mathematics. It is intended for students and students who wish to improve their knowledge and skills in mathematics, as well as for teachers who are looking for quality materials for preparing study assignments.

The product was developed by an experienced mathematics teacher - A.P. Ryabushko. and contains solutions to problems in various branches of mathematics. Beautiful design in HTML format gives the product an attractive appearance and ease of use. Solutions are presented in a clear and easy-to-understand form, allowing you to quickly and efficiently learn the material.

Product IDZ 9.2 – Option 8. Solutions Ryabushko A.P. is an excellent choice for students and students who want detailed and understandable solutions to problems in mathematics. In addition, it is a useful tool for teachers who are looking for quality materials for preparing teaching assignments.

IDZ 9.2 – Option 8. Solutions Ryabushko A.P. is a digital product containing detailed solutions to mathematics problems. The product was developed by an experienced mathematics teacher and is intended for students who want to improve their knowledge and skills in mathematics, as well as for teachers looking for quality materials for preparing educational assignments. The product is designed in HTML format, which gives it an attractive appearance and makes it easy to use. Solutions are presented in an understandable form, which allows you to quickly and efficiently absorb the material. General information about the product: IDZ 9.2 – Option 8. Solutions Ryabushko A.P. contains solutions to problems in various branches of mathematics and is an excellent choice for anyone who wants to improve their knowledge and skills in this science.

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IDZ 9.2 – Option 8 is a set of mathematical problems and solutions to them, prepared by the author Ryabushko A.P. The product description indicates that the solutions to the problems are formatted in Microsoft Word 2003 and use a formula editor for more convenient presentation of mathematical expressions.

The first task is to calculate the area of ​​the figure bounded by the indicated lines, namely: 1.8 ρ2 = 2sin2φ. The second problem requires calculating the arc length of the line given by Equation 2.8 y = 1− lncosx, for 0 ≤ x ≤ π/6. The third task is related to calculating the volume of a body obtained by rotating the figure Ф around the specified coordinate axis. The figure Ф is given by the equation y2 = (x – 1)3, x = 2, Ox. Finally, the fourth problem requires calculating the surface area formed by rotating the arc of the curve L around a specified axis. The L curve is defined by the equations x = cost, y = 3 + sint, Ox.

Solutions to these problems are contained in the document that comes with the product. All problems are solved with an accuracy of two decimal places.

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