- •Донецьк 2006
- •Integral calculus lecture no. 19. Primitive and indefinite integral
- •Point 1. Primitive
- •Properties of primitives
- •Point 2. Indefinite integral and its properties
- •Point 3. Integration by substitution (change of a variable)
- •Point 4. Integration by parts
- •Lecture no.20. Classes of integrable functions
- •Point 1. Rational functions (rational fractions)
- •Point 2. Trigonometric functions
- •Universal trigonometrical substitution
- •Other substitutions
- •Point 3. Irrational functions
- •Quadratic irrationalities. Trigonometric substitutions
- •Quadratic irrationalities (general case)
- •Indefinite integral: Basic Terminology
- •Lecture no. 21. Definite integral
- •Point 1. Problems leading to the concept ofa definite integral
- •Point 2. Definite integral
- •Point 3. Properties of a definite integral
- •I ntegration of inequalities
- •Point 4. Definite integral as a function of its upper variable limit
- •Point 5. Newton-leibniz formula
- •Point 6. Main methods of evaluation a definite integral Change of a variable (substitution method)
- •Integration by parts
- •Lecture no.22. Applications of definite integral
- •Point 1. Problem – solving schemes. Areas
- •Additional remarks about the areas of plane figures
- •Point 2. Arс length
- •Point 3. Volumes
- •Volume of a body with known areas of its parallel cross-sections
- •Volume of a body of rotation
- •Point 4. Economic applications
- •Lecture no. 23. Definite integral: additional questions
- •Point 1. Approximate integration
- •Rectangular Formulas
- •Trapezium Formula
- •Simpson10 formula (parabolic formula)
- •Point 2. Improper integrals
- •Improper integrals of the first kind
- •Improper integrals of the second kind
- •Convergence tests
- •Point 3. Euler г- function
- •Definite integral: Basic Terminology
- •Lecture no. 24. Double integral
- •Point 1. Double integral
- •Point 2. Evaluation of a double integral in cartesian coordinates
- •Point 3. Improper double integrals. Poisson formula
- •Point 4. Double integral in polar coordinates
- •Double integral: Basic Terminology
- •Differential equations lecture no.25. First and second order differential equations
- •Point 1. General notions
- •Point 2. Integrable types of the first order differential equations (of de - 1)
- •1. Separated de-1 (de-1 with separated variables)
- •2. Separable de-1 (de-1 with separable variables)
- •3. Homogeneous de-1
- •4. Linear de-1
- •5. Bernoulli de-1
- •Point 3. Order reducing second order differential equations
- •Lecture no.26. Second order linear differential equations
- •Point 1. General notions
- •Point 2. Linear dependence and independence
- •Point 3. Homogeneous equations Structure of the general solution of so lhde
- •So lhde with constant coefficients
- •Point 4. Nonhomogeneous equations Structure of the general solution of so lnde
- •Method of variation of arbitrary constants
- •Method of undetermined coefficients for so lnde with constant coefficients
- •Lecture no. 27. Systems of differential equations. Approximate integration of differential equations
- •Point 1. Normal systems of differential equations
- •Point 2. Approximate integration of differential equations Successive approximations method
- •Euler method
- •Differential equations: Basic Terminology
- •Bibliography textbooks
- •Problem books
- •Contents
Point 3. Volumes
Volume of a body with known areas of its parallel cross-sections
Let some body is situated between the planes , , and for any the area of its cross-section by a plane perpendicular to Ox-axis is known (see fig. 16). The volume of the body equals the Fig. 16 integral . ( 22 )
■ An element of the volume is the volu-me of a right circular cylinder with the base and the altitude , . Adding all the- Fig. 17 se elements we get the required volume represent-ted by the formula (22).■
Ex. 15. Find the volume of the triaxial ellipsoid (fig. 17)
It’s evident that . For any the cross-section of the body perpendicular to the Ox-axis is the ellipse with the semi-axes
and the area
.
Therefore the volume of the ellipsoid by the formula (22) equals
.
Volume of a body of rotation
A curvilinear trapezium (see fig. 1) rotates about -axis. Prove that the volume of the corres-ponding body (body of rotation, fig 18) equals the definite integral
. ( 23 )
Fig. 18 ■For any a cross-section of the body of ro- tation by a plane perpendicular to Ox-axis is a circle of radius (fig. 18). Therefore its area , and by the formula (22) the volume of the body in question is given by the formula (23).■
Let the same curvilinear trapezium (fig. 1) rotates about -axis which doesn’t pass through the interior of the trapezium9. Prove that the volume of the body of its rotation is represented by the next integral:
. ( 24 )
Instructions. As an element of the volume one can take the volume of a part of
the body generated by rotation of the rectangle with the sides about the Oy-axis. Then the element of the volume is whence it follows the formula (24).
Ex. 16. Let the arc of a sinusoid rotates about the Ox-, Oy-axes. Calculate the volumes of corresponding bodies of rotation.
With the help of the formulas (23), (24) we get
.
.
A curvilinear trapezium (fig. 6) rotates about -axis. Prove that the volume of the corresponding body of rotation is represented by the integral
. ( 25 )
Ex. 17. An ellipse rotates about the Ox-, Oy-axes. Calculate the volumes of corresponding bodies of rotation.
From the equation of the ellipse
and by the formulas (24), (25) we have
; .
Point 4. Economic applications
Problem 1 (produced quantity). Let the labour productivity of a some factory at a time moment t equals . It is known (see the formula (11) of the point 2 of prce-ding lecture) that its produced quantity U during the time interval [0, T] equals
.
Ex. 18. Let the labour productivity of a factory is . Then its produced quantity
.
Problem 2 (costs of conservation of goods). Let is the quantity of goods in the storage at a time moment t, and a constant quantity h represents the costs of conservation of unit of goods per unit of time. Then the costs of conservation of goods during a time interval (or an element of the costs of conservation) equals
.
Adding all these elements from to we get the costs of conservation of goods during the time interval , that is
.
Let, for example, we study the case of uniform consumption of all goods from at a time to at a time . In this case the quantity of goods at a time mo-ment t is
,
and the costs of conservation of goods equals
.