14. Mechanical applications of double integral.

Double integral represents itself volume of some cylindrical body bounded by plane area at bottom and by surface f(x,y) at top. Suppose that f(x,y)=1, then, obviously, the volume will coincide numerically with square of area (D):(2.1) Next, suppose that f(x,y)=p(x,y) represents itself some density spread in area (D), thenmass of this area may be calculated by formula

(2.2) – we must take density in each point and multiply it by elementary square dxdy and sum up all products passing to limit; and by definition we have got double integral (2.2). Other applications are based on this result. For example, static moments and moments of inertia in relation to axes x and y may be calculated by formulas (2.3)

Next, coordinates center of gravity may be calculated by formulas

(2.4)

15. Triple integral. Task about mass calculation. Definition and conditions of existence.

Let some body (V) is given with density in each point. It is necessary to find the mass m of body. We divide body (V) into parts (V₁),(V₂),…,(V_n) . Then choose arbitrary point in each part. Approximately we can suppose that density is constant and equals to density of chosen pointwithin the limits of each part (Vi). Then mass of this part is

. And the mass of whole body will be . If diameters of all parts approaches to zero, then if we pass to limit this equality will become exact(1) and problem is solved.

Limits of this kind are called triple integrals. The last result may be rewrote as

.(2) Now let’s define triple integral in general sense. Let function f(x,y,z) is defined in some space area (V). Let’s divide this space into parts (V₁),(V₂),…,(V_n) with the help of surface network. After this we take arbitrary point in each part, multiply function valuein this point by the volume V_i of this part and, finally, compose the integral sum .

Def. Finite limit I of integral sum when the most of diameters Vi approaches to zero is called triple integral of function f(x,y,z) in area (V). It denotes by symbol

Finite limit of this kind may exist only for bounded function; for such functions Darboux sums are considered besides the integral sum where. By usual way it may be got that for triple integral existence conditionoris necessary and sufficient, where- oscillation of function in area (V_i). From this the next statement follows: every continuous function is integrated.

16. Properties of integrated function and triple integrals.

Most of properties is analogical to properties of double integral.

1. 1) If (V)=(V’)+(V’’) then and from existence of the left integral existence of the right integrals follows and vice versa.

2. 2) If k=const then and from existence of the right integral existence of the left integral follows.

3. 3 ) If function f and g are integrated in area (V) then function is integrated too and

4. 4) If for integrated in area (V) functions f and g inequality takes place then.

5. 5) If function f is integrated then function is integrated too and inequality takes place.

6) If integrated in area (V) function f satisfies to inequality then. By other words, we have mean value theorem. In a case of continuous function this formula may be rewrote as(3).

17. Calculation of triple integral in a case of parallelepiped.

Let’s suppose that area (V) is parallelepiped (T)=[a,b,c,d,e,f] projecting to plane yz as rectangle (R)=[c,d,e,f].

Th. If for function f(x,y,z) triple integral (4) exists and for any constant x in [a,b] double integral(5) exists then repeated integral(6) also exists and equality(7) takes place.

Proof. Proof is analogical to proof of theorem for double integral. Here we only remember the main steps. Segments [a,b],[c,d],[e,f] divide into parts by inserting points, at the same time area (T) will be divided into elementary parallelepipeds and area (R) into elementary rectangles. Then we consider mean value theorem for double integral in elementary rectangle. After that we fix arbitrary value of x in segment [a,b] and sum up inequalities for j and k; then multiply inequalities byand sum up once more in relation to i. Finally we have got. Extremes are Darboux sums for integral (4) and approaches to this integral, so middle integral sum also approaches to it. This fact proves both existence of integral (6) and equality (7).

If we suppose that integral (8) exists for any valuesthen double integral in equality (7) may be changed by repeated one and finally we have got(9) . The variables x,y,z in the last expression may be ordered arbitrary.