To calculate the quantities m, and you need to use formulas (4), (5) and (7). As a result, we get formulas for the coordinates of the center of mass of a thin plate :
Example 4 (calculating the coordinates of the center of mass of a homogeneous plate)
Find the coordinates of the center of mass of a homogeneous figure bounded by lines and.
Having constructed the figure, we notice that it is geometrically symmetrical with respect to a straight line.Since the figure is made of a homogeneous material, it has not only geometric, but also physical symmetry, that is, the mass of its part, which is located to the left of the axis of symmetry, is equal to the mass of the part, which located on the right. Then according to the well-known physical properties center of mass, we conclude that it is on the axis of symmetry, that is
To calculate, we compose the static moment and use formulas (4) and (5):
 | ;
 |
Answer: C.
Triple integral applications
Applications of triple integrals are similar to applications of double integrals, but only for three-dimensional bodies.
If we use one of the properties of the triple integral (about its value of a function, identically equal to one), it turns out formula for calculating the volume of any spatial body :
We write the formula for the volume in terms of the triple integral and calculate the triple integral in cylindrical coordinates:
Answer: (units of volume).
Formula for calculating the mass of a three-dimensional object occupying volume V, has the form:
(13)
Here is the bulk density of the mass distribution.
Example 6 (calculating the mass of a three-dimensional body)
Find the mass of a ball of radius Rif the density is proportional to the cube of the distance from the center and per unit distance is k.
V: elementary volume and.
Note that here, when calculating the threefold integral, we got the product of integrals, since the inner integrals turned out to be independent of the variable outer integrals.
Answer: (units of mass).
Mechanical characteristics for volume V (static moments, moments of inertia, coordinates of the center of mass) are calculated by the formulas that
are compiled by analogy with formulas for two-dimensional bodies.
Elementary static moments and moments of inertia about the coordinate axes:
elementary moments of inertia about the coordinate planes and the point of origin:
Further, to calculate the mechanical characteristic of the entire volume V, it is necessary to sum the elementary terms of this characteristic over all parts of the partition (since the computed characteristic possesses the additivity property), and then go to the limit in the resulting sum, provided that all elementary parts of the partition decrease (contract into points) indefinitely. These actions are described as the integration of the elementary term of the calculated mechanical characteristic over the volume V.
The result is the following formulas for calculating static moments M and moments of inertia I of three-dimensional bodies :
In practice, it is useful not only to use these formulas as ready-made, but also to derive them in the problem being solved.
Examples 7 (calculation of mechanical characteristics of three-dimensional bodies)
Find the moment of inertia of a homogeneous cylinder whose height is h and base radius R, about the axis coinciding with the base diameter.
Find the distance dfor an arbitrary point of the cylinder:
distance from point with coordinates to axis – this is the length of the perpendicular drawn from this point to the axis . Let's construct a plane perpendicular to the axis so that the point belongs to this plane. Then any straight line intersecting the axis and belonging to this plane will be perpendicular . In particular, the straight line connecting point and point will be perpendicular to the axis, and the distance between these points will be the desired distance d... We calculate it according to the well-known formula for the distance between two points.
3 Applications of double integrals
3.1 Theoretical introduction
Consider applications double integral to the solution of a number of geometric problems and problems of mechanics.
3.1.1 Calculating the area and mass of a flat plate
Consider a thin material plate Dlocated in the plane Ooh. Area S this plate can be found using a double integral by the formula:
3.1.2 Static moments. Center of mass of a flat plate
Static moment M x about the axis Ox material point P(x;y) lying in the plane Oxy and having mass m, is called the product of the mass of a point by its ordinate, i.e. M x \u003d my... The static moment is determined similarly M y about the axis Oy: M y \u003d mx. Static moments flat plate with surface density γ = γ (x, y) are calculated by the formulas:
As is known from mechanics, the coordinates x c , y c the center of mass of a flat material system are determined by the equalities:
where m Is the mass of the system, and M x and M y - static moments of the system. Flat plate weight m is determined by formula (1), the static moments of a flat plate can be calculated by formulas (3) and (4). Then, according to formulas (5), we obtain an expression for the coordinates of the center of mass of a flat plate:
A typical calculation contains two tasks. Each problem has a flat plate D, limited by the lines indicated in the problem statement. D(x, y) Is the surface density of the plate D... Find for this plate: 1. S - area; 2. m - mass; 3. M y , M x - static moments about the axes Oy and Oh respectively; 4., - coordinates of the center of mass.
3.3 Procedure for performing a typical calculation
When solving each problem, it is necessary to: 1. Draw a drawing of the specified area. Select the coordinate system in which the double integrals will be calculated. 2. Write the area as a system of inequalities in the selected coordinate system. 3. Calculate area S and mass m plates according to formulas (1) and (2). 4. Calculate static moments M y , M x by formulas (3) and (4). 5. Calculate the coordinates of the center of mass using formulas (6). Mark the center of mass on the drawing. In this case, there is a visual (qualitative) control of the results obtained. Numerical responses must be received with three significant digits.
3.4 Examples of performing a typical calculation
Objective 1. Plate D limited by lines: y
= 4 – x 2 ;
x =
0; y
= 0 (x
≥ 0; y ≥ 0) Surface density γ
0
= 3.
Decision. The domain specified in the problem is bounded by a parabola y
= 4 – x 2, coordinate axes and lies in the first quarter (Fig. 1). We will solve the problem in a Cartesian coordinate system. This area can be described by a system of inequalities: 
Figure: one
Area S plate is equal to (1): Since the plate is homogeneous, its mass m
= γ
0 S \u003d 3 · \u003d 16. By formulas (3), (4) we find the static moments of the plate: 
The coordinates of the center of mass are found by the formula (6): 
Answer:
S ≈
5,33; m
= 16; M x
= 25,6; M y
= 12;
=
0,75;
=
1,6.
Objective 2. Plate D limited by lines: x 2 + at 2 = 4; x = 0, at = x (x ≥ 0, at ≥ 0). Surface density γ (x, y) = at. Decision. The plate is bounded by a circle and straight lines passing through the origin (Fig. 2). Therefore, to solve the problem, it is convenient to use a polar coordinate system. Polar angle φ varies from π / 4 to π / 2. A ray drawn from a pole through a plate “enters” it at ρ \u003d 0 and “leaves” on a circle, the equation of which is: x 2 + at 2 = 4 <=> ρ \u003d 2.
Figure: 2
Therefore, the given area can be written as a system of inequalities: 
We find the area of \u200b\u200bthe plate by the formula (1): 
We find the mass of the plate by formula (2), substituting γ
(x, y)
= y \u003d ρ sin φ
:
To calculate the static moments of the plate, we use formulas (3) and (4): 
We obtain the coordinates of the center of mass by the formulas (6): Answer:
S ≈
1,57; m
≈ 1,886; M x
= 2,57; M y
= 1;
=
0,53;
=
1,36.
3.5 Reporting
The report should contain all the calculations performed, neatly executed drawings. Numerical answers should be received with three significant digits.
– calculating the center of gravity of a flat bounded figure... Many readers intuitively understand what the center of gravity is, but, nevertheless, I recommend repeating the material of one of the lessons analytic geometrywhere I parsed the problem of the center of gravity of a triangle and deciphered the physical meaning of this term in an accessible form.
In independent and test assignments for the solution, as a rule, the simplest case is proposed - a flat bounded homogeneous a figure, that is, a figure of constant physical density - glass, wood, tin, cast-iron toys, difficult childhood, etc. Further, by default, we will only talk about such figures \u003d)
First rule and simplest example: if flat figure there is center of symmetry, then it is the center of gravity of this figure... For example, the center of a round uniform plate. It is logical and understandable in everyday life - the mass of such a figure is "fairly distributed in all directions" relative to the center. Verti - I don't want to.
However, in harsh realities, you are unlikely to be thrown sweet elliptical chocolate, so you have to arm yourself with a serious kitchen tool:
The coordinates of the center of gravity of a flat homogeneous bounded figure are calculated by the following formulas:
, or:
, where is the area of \u200b\u200bthe area (figure); or very shortly:
where
The integral will be conventionally called the “x” integral, and the integral will be called the “game” integral.
Note-help
: for flat limited heterogeneous figures whose density is given by a function, the formulas are more complex:
where 
- the mass of the figure;in the case of uniform density, they are simplified to the above formulas.
On the formulas, in fact, all the novelty ends, the rest is your skill solve double integralsBy the way, now there is a great opportunity to practice and improve your technique. And as you know, there is no limit to perfection \u003d)
Let's throw in an invigorating portion of parabolas:
Example 1
Find the coordinates of the center of gravity of a uniform flat figure bounded by lines.
Decision: the lines are elementary here: it sets the abscissa axis, and the equation is a parabola, which is easily and quickly constructed using geometric transformations of graphs:
– parabola shifted 2 units to the left and 1 unit down.
I will complete the entire drawing at once with the finished point of the center of gravity of the figure: 
Second rule: if the figure has axis of symmetry, then the center of gravity of this figure necessarily lies on this axis.
In our case, the figure is symmetrical about straight , that is, in fact, we already know the "x" coordinate of the "um" point.
Also note that the vertical center of gravity is shifted closer to the abscissa axis, since the figure is more massive there.
Yes, perhaps not everyone has yet fully understood what the center of gravity is: please, lift your index finger up and mentally place the shaded “sole” on it with a dot. In theory, the piece should not fall.
We find the coordinates of the center of gravity of the figure by the formulas 
where.
The order of traversing the area (figure) is obvious here: 
Attention! Determine the most advantageous traversal order once - and use it for all integrals!
1) First, we calculate the area of \u200b\u200bthe figure. In view of the relative simplicity of the integral, the solution can be formulated compactly, the main thing is not to get confused in the calculations:
We look at the drawing and estimate the area by cells. It turned out about the case.
2) The x coordinate of the center of gravity has already been found by the "graphical method", so you can refer to the symmetry and go to the next point. However, I still do not advise you to do this - there is a high probability that the solution will be rejected with the wording "use the formula".
Note that here you can do only with oral calculations - sometimes it is not at all necessary to bring fractions to a common denominator or torment the calculator.
Thus: 
, which was required to be obtained.
3) Find the ordinate of the center of gravity. Let us calculate the "game" integral:
But here it would be hard without a calculator. Just in case, I will comment that as a result of multiplying polynomials, 9 terms are obtained, and some of them are similar. Similar terms I brought orally (as is usually done in similar cases) and immediately wrote down the total amount.
As a result: 
, which is very, very similar to the truth.
On final stage mark a point on the drawing. According to the condition, it was not required to draw anything, but in most tasks, we willy-nilly have to draw a figure. But there is a definite plus - a visual and rather effective check of the result.
Answer: 
The next two examples are for independent decision.
Example 2
Find the coordinates of the center of gravity of a uniform flat figure bounded by lines 
By the way, if you have an idea of \u200b\u200bhow the parabola is located and saw the points at which it crosses the axis, then here you can actually do without a drawing.
And more complicated:
Example 3
Find the center of gravity of a uniform flat figure bounded by lines
If you have trouble plotting, study (repeat) parabola lesson and / or Example No. 11 of the article Double Integrals for Dummies.
Sample solutions at the end of the lesson.
In addition, a dozen or two similar examples can be found in the corresponding archive on the page Ready-made solutions for higher mathematics.
Well, I can't help but please lovers higher mathematics, which often ask me to sort out difficult problems:
Example 4
Find the center of gravity of a uniform flat figure bounded by lines. Draw the figure and its center of gravity in the drawing.
Decision: the condition of this task already categorically requires the execution of the drawing. But the requirement is not so formal! - even a person with an average level of training is able to imagine this figure in his mind: 
A straight line cuts the circle into 2 parts, and an additional disclaimer (cm. linear inequalities)
indicates that we are talking about a small shaded piece.
The figure is symmetrical about a straight line (shown by a dotted line), so the center of gravity should lie on this line. And, obviously, its coordinates are equal modulo... An excellent guideline, almost eliminating the erroneous answer!
Now the bad news \u003d) On the horizon looms an unpleasant integral from the root, which we analyzed in detail in Example # 4 of the lesson Effective methods for solving integrals... And who knows what else will be drawn there. It would seem that due to the presence circles profitable, but not so simple. The straight line equation is converted to the form 
and the integrals are also not sugar ones (although fans trigonometric integralswill appreciate). In this regard, it is more prudent to stop at Cartesian coordinates.
The order of the figure traversal: 
1) Calculate the area of \u200b\u200bthe figure:
It is more rational to take the first integral under the differential sign:
And in the second integral we carry out the standard replacement:
Let us calculate the new limits of integration: 
2) Find.
Here, in the 2nd integral, we again used method of bringing a function under the differential sign... Practice and adopt these optimal (in my opinion) methods for solving typical integrals.
After difficult and lengthy calculations, we again turn our attention to the drawing (remember that the points we don't know yet! ) and we get deep moral satisfaction from the found value.
3) Based on the previous analysis, it remains to be sure that.
Excellent: 
Let's draw a point 
in the drawing. In accordance with the wording of the condition, we write it down as the final answer: 
A similar task for an independent solution:
Example 5
Find the center of gravity of a uniform flat figure bounded by lines. Execute a drawing.
This problem is of interest in that a figure of sufficiently small size is given in it, and if you make a mistake somewhere, then there is a high probability of not getting into the region at all. Which is definitely good from the point of view of decision control.
A sample design at the end of the lesson.
Sometimes it is advisable transition to polar coordinates in double integrals... It depends on the shape. I was looking for a good example, but I could not find it, so I will demonstrate the solution on the 1st demo problem of the above lesson: 
Let me remind you that in that example we went to polar coordinates, figured out the order of traversing the area 
and calculated its area
Let's find the center of gravity of this shape. The scheme is the same: 
... The value is viewed directly from the drawing, and the "x" coordinate should be shifted slightly closer to the ordinate axis, since the more massive part of the semicircle is located there.
In the integrals, we use the standard transition formulas:
Plausibly, most likely not wrong.