12. Reduction of double integral to repeated one in a case of curvilinear area.

Let’s consider area (P) bounded by two curves y=y₁(x) and y=y₂(x) and by lines x=a and x=b.

Th 2. If for function f(x,y) defined in area (P) double integral exists and for any permanent value x in [a,b] simple define integralexists then repeated integralalso exists and(2.5)

Proof. Proof is based on reduction of this case to considered one above in Th 1. Let’s draw rectangle R=[a,b,c,d] containing area (P), where and define function f*(x,y) such thatLet’s demonstrate that this function satisfies to conditions of Th.1. At first it is integrated in area (P) because here it coincide with integrated function f(x,y), therefore. On other hand f*(x,y)=0 out of (P) and because that is integrated in remaining part of rectangle (R) and. Then according to property 2 function f* is integrated in all rectangle (R) and

(2.6) . For permanent value x in [a,b] there integral exists because any of three integrals on right exist. Really, first and third integrals exist being equal to zero and second integral coincide with integral of function f(x,y). Finally we have got(2.7) . And according to Th.1 repeated integral exists too and is equal to double one. Taking into account (2.6) and (2.7) we can see that this formula is analogical to formula (2.5) .Q.e.d. In a case when area is bounded by curves x=x₁(y), x=x₂(y) and by lines y=c, y=d we can got formula (2.5*). Such areas are called first type area and second type area accordingly. Besides that some area may be considered both as first and second type (example). And, finally, any complex closed area may be divided into parts representing itself areas of first or second type.

13. Change of variables in double integrals.

Let’s consider double integral (1.1) where area (D) bounded by continuous curve (L) and function f is continuous in this area. Suppose now that area (D) is connected with another area () by formulas(1.2) so that between points belonged to (D) and () one-one mapping exists. It’s necessary to present integral (1.1) spread in area () by changing variables in it. Let’s divide area () into parts () (i=1,2,…,n); at the same time area (D) also will be divided into parts (D_i). In each part (D_i) let’s choose arbitrary point (x_i,y_i); and finally let’s compose the integral sum for integral (1.1) .It’s known that squares D_i and Δ_i connected with formula where J is jacobian -andis some point belonged area (). This point is defined by mean value theorem and we can’t choose it arbitrarily, but point (x_i,y_i) in area (D_i) we can take arbitrarily. Using this fact we suppose that . Then integral sum takes form. In this form it represent integral sum for integral(1.3) . Existence of the last integral follows from continuity of function f. Finally, we can write

(1.4).

So, rule for changing variables in double integral: it’s necessary to change variables x and y by formulas (1.2) in integrating function and multiply integrand by jacobian.

The most usable is polar coordinates system: . Jacobian for this system is

.

Then formula (1.4) takes form (1.5)