Lectures 8. Applications definite integral.
The application of the integral to physical problems is based on the property of the additivity of the integral over a set. Therefore, using the integral, quantities can be calculated that are themselves additive in the set. For example, the area of a figure is equal to the sum of the areas of its parts. The length of the arc, surface area, volume of the body, and mass of the body have the same property. Therefore, all these quantities can be calculated using a definite integral.
You can use two methods to solve problems: method of integral sums and method of differentials.
The method of integral sums repeats the construction of a definite integral: a partition is constructed, points are marked, the function is calculated at them, the integral sum is calculated, and passage to the limit is performed. In this method, the main difficulty is to prove that in the limit the result is exactly what is needed in the problem.
The differential method uses indefinite integral and the Newton–Leibniz formula. The differential of the quantity to be determined is calculated, and then, by integrating this differential, the required quantity is obtained using the Newton–Leibniz formula. In this method, the main difficulty is to prove that it is the differential of the required value that has been calculated, and not something else.
Calculation of areas of plane figures.
1. The figure is limited by the graph of a function defined in a Cartesian coordinate system.
We came to the concept of a definite integral from the problem of the area of a curved trapezoid (in fact, using the method of integral sums). If a function takes only non-negative values, then the area under the graph of the function on a segment can be calculated using a definite integral. Note that 
therefore, the method of differentials can also be seen here.
But a function can also take negative values on a certain segment, then the integral over this segment will give a negative area, which contradicts the definition of area.
You can calculate the area using the formulaS=. This is equivalent to changing the sign of the function in those areas in which it takes negative values.
If you need to calculate the area of a figure bounded above by the graph of the function and below by the graph of the function, then you can use the formulaS=
, because .
Example. Calculate the area of the figure bounded by straight lines x=0, x=2 and graphs of functions y=x 2, y=x 3.
Note that on the interval (0,1) the inequality x 2 > x 3 holds, and for x >1 the inequality x 3 > x 2 holds. That's why
2. The figure is limited by the graph of a function specified in a polar coordinate system.
Let the graph of a function be given in a polar coordinate system and we want to calculate the area of a curvilinear sector bounded by two rays and the graph of a function in a polar coordinate system.
Here you can use the method of integral sums, calculating the area of a curvilinear sector as the limit of the sum of the areas of elementary sectors in which the graph of the function is replaced by a circular arc 
.
You can also use the differential method: 
.
You can think like this. Replacing the elementary curvilinear sector corresponding to the central angle with a circular sector, we have the proportion . From here 
. Integrating and using the Newton–Leibniz formula, we obtain 
.
Example. Let's calculate the area of the circle (check the formula). We believe. The area of the circle is 
.
Example. Let's calculate the area bounded by the cardioid 
.
3 The figure is limited by the graph of a function defined parametrically.
The function can be specified parametrically in the form . We use the formula S=
, substituting into it the limits of integration over the new variable. 
. Usually, when calculating the integral, those areas where the integrand function has a certain sign are identified and the corresponding area with one or another sign is taken into account.
Example. Calculate the area enclosed by the ellipse.
Using the symmetry of the ellipse, we calculate the area of the quarter of the ellipse located in the first quadrant. In this quadrant. That's why .
Calculation of volumes of bodies.
1. Calculation of volumes of bodies from the areas of parallel sections.
Let it be required to calculate the volume of a certain body V from the known cross-sectional areas of this body by planes perpendicular to the line OX drawn through any point x of the line segment OX.
Let's apply the method of differentials. Considering the elementary volume above the segment as the volume of a right circular cylinder with base area and height, we obtain 
. Integrating and applying the Newton–Leibniz formula, we obtain
2. Calculation of volumes of bodies of rotation.
Let it be necessary to calculate OX.
Then 
.
Likewise, volume of a body of revolution around an axisOY, if the function is given in the form , can be calculated using the formula .
If the function is specified in the form and you need to determine the volume of a body of rotation around an axisOY, then the formula for calculating the volume can be obtained as follows.
Passing to the differential and neglecting the quadratic terms, we have 
. Integrating and applying the Newton–Leibniz formula, we have .
Example. Calculate the volume of the sphere.
Example. Calculate the volume of a right circular cone bounded by a surface and a plane.
Let's calculate the volume as the volume of a body of rotation, formed by rotation around the OZ axis right triangle in the OXZ plane, the legs of which lie on the OZ axis and the line z = H, and the hypotenuse lies on the line.
Expressing x in terms of z, we get 
.
Calculation of arc length.
In order to obtain formulas for calculating the length of an arc, recall the formulas derived in the 1st semester for the differential of the arc length.
If the arc is the graph of a continuously differentiable function, the arc length differential can be calculated using the formula
. That's why 
If a smooth arc is specified parametrically, That
. That's why 
.
If the arc is specified in a polar coordinate system, That
. That's why 
.
Example. Calculate the length of the arc of the graph of the function, . 
.
Let us present some applications of the definite integral.
Calculating the area of a flat figure
The area of a curved trapezoid bounded by a curve (where 
), straight 
,
and a segment 
axes 
, calculated by the formula
.
Area of a figure bounded by curves 
And 
(Where 
) straight 
And 
calculated by the formula
|
|
.If the curve is given by parametric equations 
, then the area of a curvilinear trapezoid bounded by this curve by straight lines 
,
and a segment 
axes 
, calculated by the formula
,
Where 
And 
are determined from the equations 
,
, A 
at 
.
The area of a curvilinear sector bounded by a curve given in polar coordinates by the equation 
and two polar radii 
,
(
), is found by the formula
.
Example 1.27. Calculate the area of a figure bounded by a parabola 
and straight 
(Figure 1.1).
|
|
Solution. Let's find the intersection points of a straight line and a parabola. To do this, we solve the equation
Where
|
,
.
,
. Then by formula (1.6) we have
.Calculating the arc length of a plane curve
If the curve 
on the segment 
- smooth (that is, derivative 
continuous), then the length of the corresponding arc of this curve is found by the formula
.
When specifying a curve parametrically 
(
- continuously differentiable functions) length of the arc of the curve corresponding to a monotonic change in the parameter 
from 
to 
, calculated by the formula
Example 1.28. Calculate the arc length of a curve 
,
,
.
Solution. Let's find derivatives with respect to the parameter 
:
,
. Then from formula (1.7) we obtain
.
2. Differential calculus of functions of several variables
Let each ordered pair of numbers 
from some area 
corresponds to a certain number 
. Then 
called function of two variables
And 
,
-independent variables
or arguments
,
-domain of definition
functions, and a set 
all function values - range of its values
and denote 
.
Geometrically, the domain of definition of a function usually represents some part of the plane 
, bounded by lines that may or may not belong to this area.
Example 2.1. Find the domain of definition 
functions 
.
|
|
Solution. This function is defined at those points of the plane |
, in which 
, or 
. Points of the plane for which 
, form the boundary of the region 
. Equation 
defines a parabola (Fig. 2.1; since the parabola does not belong to the region 
, then it is depicted by a dotted line). Further, it is easy to check directly that the points for which 
, located above the parabola. Region 
is open and can be specified using a system of inequalities:If the variable 
give some increment 
, A 
leave constant, then the function 
will receive an increment 
, called private increment of function 
by variable 
:
Likewise, if the variable 
gets increment 
, A
remains constant, then the function 
will receive an increment 
, called private increment of function 
by variable
:
If there are limits:
,
,
they are called partial derivatives of a function 
by variables 
And 
respectively.
Remark 2.1. Partial derivatives of functions of any number of independent variables are determined similarly.
Remark 2.2. Since the partial derivative with respect to any variable is the derivative with respect to this variable, provided that the other variables are constant, then all the rules for differentiating functions of one variable are applicable to finding partial derivatives of functions of any number of variables.
Example 2.2.
.
Solution. We find:
,
.
Example 2.3. Find partial derivatives of a function 
.
Solution. We find:
,
,
.
Full function increment
called difference
Main part of the full function increment
, linearly dependent on increments of independent variables 
And 
,is called the total differential of the function
and is designated 
. If a function has continuous partial derivatives, then the total differential exists and is equal to
,
Where 
,
- arbitrary increments of independent variables, called their differentials.
Similarly, for a function of three variables 
the total differential is given by
.
Let the function 
has at point 
first order partial derivatives with respect to all variables. Then the vector is called gradient
functions 
at the point 
and is designated 
or 
.
Remark 2.3.
Symbol 
is called the Hamilton operator and is pronounced “nambla”.
Example 2.4. Find the gradient of a function at a point 
.
Solution. Let's find the partial derivatives:
,
,
and calculate their values at the point 
:
,
,
.
Hence, 
.
Derivative
functions
at the point 
in the direction of the vector
called the limit of the ratio 
at 
:
, Where 
.
If the function
is differentiable, then the derivative in a given direction is calculated by the formula:
,
Where 
,
- angles, which is a vector 
forms with axes 
And 
respectively.
In the case of a function of three variables 
the directional derivative is defined similarly. The corresponding formula is
|
|
,
Where 
- direction cosines of the vector 
.
Example 2.5. Find the derivative of a function 
at the point 
in the direction of the vector 
, Where 
.
Solution. Let's find the vector 
and its direction cosines:
,
,
,
.
Let's calculate the values of partial derivatives at the point 
:
,
,
;
,
,
.
Substituting into (2.1), we get
.
Second order partial derivatives are called partial derivatives taken from partial derivatives of the first order:
,
,
,
Partial derivatives 
,
are called mixed
. The values of mixed derivatives are equal at those points at which these derivatives are continuous.
Example 2.6. Find second order partial derivatives of a function 
.
Solution. Let us first calculate the partial derivatives of the first order:
,
.
Differentiating them again, we get:
,
,
,
.
Comparing the last expressions, we see that 
.
Example 2.7. Prove that the function 
satisfies Laplace's equation
.
Solution. We find:
,
.
,
.
.
Dot 
called local maximum point
(minimum
) functions 
, if for all points 
, different from 
and belonging to a sufficiently small neighborhood of it, the inequality
(
).
The maximum or minimum of a function is called its extremum . The point at which the extremum of the function is reached is called extremum point of the function .
Theorem 2.1
(Necessary conditions for an extremum
).
If the point 
is the extremum point of the function 
, or at least one of these derivatives does not exist.
Points for which these conditions are met are called stationary or critical . Extremum points are always stationary, but a stationary point may not be an extremum point. For a stationary point to be an extremum point, sufficient conditions for the extremum must be satisfied.
Let us first introduce the following notation :
,
,
,
.
Theorem 2.2
(Sufficient conditions for an extremum
).
Let the function 
twice differentiable in a neighborhood of a point 
and period 
is stationary for the function 
. Then:
1.If 
, then point 
is an extremum of the function, and 
will be the maximum point at 
(
)and the minimum point at 
(
).
2.If 
, then at the point 
there is no extreme.
3.If 
, then the extremum may or may not exist.
Example 2.8. Examine the extremum function 
.
Solution. Since in in this case partial derivatives of the first order always exist, then to find stationary (critical) points we solve the system:
,
,
where 
,
,
,
. Thus, we got two stationary points: 
,
.
,
,
.
For a point 
we get:, that is, there is no extremum at this point. For a point 
we get: and 
, hence
at this point this function reaches a local minimum: .
Area of a curvilinear trapezoid bounded above by the graph of a function y=f(x), left and right - straight x=a And x=b accordingly, from below - the axis Ox, calculated by the formula
Area of a curvilinear trapezoid bounded on the right by the graph of a function x=φ(y), above and below - straight y=d And y=c accordingly, on the left - the axis Oy:
Area of a curvilinear figure bounded above by the graph of a function y 2 =f 2 (x), below - function graph y 1 =f 1 (x), left and right - straight x=a And x=b:
Area of a curvilinear figure bounded on the left and right by graphs of functions x 1 =φ 1 (y) And x 2 =φ 2 (y), above and below - straight y=d And y=c respectively:
Let us consider the case when the line limiting the curvilinear trapezoid from above is given by parametric equations x = φ 1 (t), y = φ 2 (t), Where α ≤ t ≤ β, φ 1 (α)=a, φ 1 (β)=b. These equations define some function y=f(x) on the segment [ a, b]. The area of a curved trapezoid is calculated by the formula
Let's move on to a new variable x = φ 1 (t), Then dx = φ" 1 (t) dt, A y=f(x)=f(φ 1 (t))=φ 2 (t), therefore \begin(displaymath)
Area in polar coordinates
Consider a curvilinear sector OAB, bounded by line, given by the equation ρ=ρ(φ) in polar coordinates, two rays O.A. And O.B., for which φ=α , φ=β .
We will divide the sector into elementary sectors OM k-1 M k ( k=1, …, n, M 0 =A, M n =B). Let us denote by Δφk angle between rays OM k-1 And OM k, forming angles with the polar axis φ k-1 And φk respectively. Each of the elementary sectors OM k-1 M k replace it with a circular sector with radius ρ k =ρ(φ" k), Where φ"k- angle value φ
from the interval [ φ k-1 , φ k], and the central angle Δφk. The area of the last sector is expressed by the formula 
.
expresses the area of a “stepped” sector that approximately replaces a given sector OAB.
Sector area OAB is called the limit of the area of the “stepped” sector at n → ∞ And λ=max Δφ k → 0:
Because 
, That
Curve arc length
Let on the segment [ a, b] a differentiable function is given y=f(x), the graph of which is the arc. Segment [ a,b] let's split it into n parts with dots x 1, x 2, …, xn-1. These points will correspond to points M 1, M 2, …, Mn-1 arcs, we connect them with a broken line, which is called a broken line inscribed in the arc. The perimeter of this broken line will be denoted by s n, that is
Definition. The length of the arc of a line is the limit of the perimeter of the broken line inscribed in it, when the number of links M k-1 M k increases unlimitedly, and the length of the largest of them tends to zero:
where λ is the length of the largest link.
We will count the length of the arc from some point, for example, A. Let at the point M(x,y) arc length is s, and at the point M"(x+Δ x,y+Δy) arc length is s+Δs, where,i>Δs is the length of the arc. From a triangle MNM" find the length of the chord: .
From geometric considerations it follows that
that is, an infinitesimal arc of a line and the chord subtending it are equivalent.
Let us transform the formula expressing the length of the chord:
Passing to the limit in this equality, we obtain a formula for the derivative of the function s=s(x):
from which we find
This formula expresses the differential of an arc of a plane curve and has a simple geometric meaning : expresses the Pythagorean theorem for an infinitesimal triangle MTN (ds=MT, ).
The differential of the arc of a spatial curve is determined by the formula
Consider the arc of a spatial line defined by parametric equations
Where α ≤ t ≤ β, φi(t) (i=1, 2, 3) - differentiable functions of the argument t, That
Integrating this equality over the interval [ α, β ], we get a formula for calculating the length of this line arc
If the line lies in the plane Oxy, That z=0 in front of everyone t∈[α, β], That's why
In the case where a flat line is given by the equation y=f(x) (a≤x≤b), Where f(x) is a differentiable function, the last formula takes the form
Let the plane line be given by the equation ρ=ρ(φ) (α≤φ≤β ) in polar coordinates. In this case we have parametric equations lines x=ρ(φ) cos φ, y=ρ(φ) sin φ, where the polar angle is taken as a parameter φ . Because
then the formula expressing the length of the arc of the line ρ=ρ(φ) (α≤φ≤β ) in polar coordinates, has the form
Body volume
Let's find the volume of a body if the area of any cross section of this body perpendicular to a certain direction is known.
Let us divide this body into elementary layers by planes perpendicular to the axis Ox and defined by equations x=const. For any fixed x∈ known area S=S(x) cross section given body.
Elementary layer cut off by planes x=x k-1, x=x k (k=1, …, n, x 0 =a, x n =b), replace it with a cylinder with height Δx k =x k -x k-1 and base area S(ξ k), ξ k ∈.
The volume of the indicated elementary cylinder is expressed by the formula Δv k =E(ξ k)Δx k. Let's sum up all such products
which is the integral sum for a given function S=S(x) on the segment [ a, b]. It expresses the volume of a stepped body consisting of elementary cylinders and approximately replacing this body.
The volume of a given body is the limit of the volume of the specified stepped body at λ→0 , Where λ - length of the largest of the elementary segments Δx k. Let us denote by V the volume of a given body, then by definition
On the other side,
Therefore, the volume of the body according to the given cross sections calculated by the formula
If a body is formed by rotation around an axis Ox a curved trapezoid bounded at the top by an arc of a continuous line y=f(x), Where a≤x≤b, That S(x)=πf 2 (x) and the last formula takes the form:
Comment. The volume of a body obtained by rotating a curved trapezoid bounded on the right by the graph of the function x=φ(y) (c ≤ x ≤ d), around the axis Oy calculated by the formula
Surface area of rotation
Consider the surface obtained by rotating the arc of the line y=f(x) (a≤x≤b) around the axis Ox(assume that the function y=f(x) has a continuous derivative). Fixing the value x∈, we will give an increment to the function argument dx, which corresponds to the “elementary ring” obtained by rotating the elementary arc Δl. Let us replace this “ring” with a cylindrical ring - the lateral surface of a body formed by the rotation of a rectangle with a base equal to the differential of the arc dl, and height h=f(x). By cutting the last ring and unfolding it, we get a strip with the width dl and length 2πy, Where y=f(x).
Therefore, the surface area differential is expressed by the formula
This formula expresses the surface area obtained by rotating the arc of a line y=f(x) (a≤x≤b) around the axis Ox.
