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# POINT 4. DOUBLE INTEGRAL IN POLAR COORDINATES

Let we study a double integral

over a domain D of the xOy-plane, and we pass to polar coordinates

, ( 12 )

coinciding the pole O with the origin and the polar axis with positive se-miaxis of the Ox-axis of Cartesian coordinate system. The domain D transforms into some domain of the -plane and the double integral passes in that over the do-main .

To show up how the element dS of the area changes we generate an element dD of the domain D by two circles of radii centered at the origin and by two rays starting from the pole under angles to the Ox-axis (fig. 17 a). We can consider
Fig. 17 dD as curvilinear rectangle PQRT with the area

.

Therefore

,

and the formula of passing to polar coordinates in a double integral can be written as

. ( 13 )

In applications we often meet the case of a domain D bounded by two rays

( 14 )

and two lines with polar equations

( 15 )

(fig. 17 a). One can describe such the domain by two double inequalities

, ( 16 )

whence it follows that a domain Δ (fig. 17 b), in which D transforms after passage to polar coordinates, is that of the first type. Therefore, by the formula (5)

. ( 17 )

If a line degenerates in the pole O we get a curvilinear sector D bounded by two rays

( 18 )

and a line with a polar equation

Fig. 18 ( 19 )

(fig. 18 a). We describe it by inequalities

, ( 20 )

whence it follows that a domain Δ (fig. 18 b) is also that of the first type. Therefore, by the formula (5)

. ( 21 )

Let a domain D contains the pole O, and every ray intersects the boundary of the domain in unique point (fig. 19 a). If (19) is its polar equation,
then ( 22 )
Fig. 19 (fig. 19 b), and therefore

. ( 23 )

Ex. 7. Evaluate the mass of a plate D containing betweem two curves

for (fig. 20), if its surface density at any point is proportional to the polar radius
Fig. 20 of OP this point and equals 8 at the point .

The surface density of the plate D

and by virtue of the mechanic sense of a double integral (see (3)) we have to calculate a double integral

.

Compleating the squares we make sure that the curves are circles with radii 2 and 4 and centres correspondingly:

.

If we carry out the transition (12) to polare coordinates, we’ll get the polar equations of

and describe the domain D by two double inequalities

.

Therefore by the formula (17)

.

Ex. 8. Find the area of a figure bounded by a curve (Bernoulli lemniscate, fig. 21)

.
Fig. 21 We have studied this curve in the Lecture 8, Point 2. Its polar equation is

.

Making use of the formula (2) we can write

where a domain D is the shaded area on the fig. 21. It’s evident that

.

Hence in correspondence with the formula (20) the area in question equals

In generale case the substitution , when a domain D of the plane xOy changes in a domain of a plane , there is the next formula

.

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