   Chapter 15.3, Problem 39E

Chapter
Section
Textbook Problem

Use polar coordinates to combine the sum ∫ 1 / 2 1 ∫ 1 − x 2 x x y   d y   d x   +   ∫ 1 2 ∫ 0 x x y   d y   d x   +   ∫ 2 2 ∫ 0 4 − x 2 x y   d y   d x into one double integral. Then evaluate the double integral.

To determine

To evaluate: The given double integral by changing into the polar coordinates.

Explanation

Given:

The function, 1211x2xxydydx+120xxydydx+2204x2xydydx.

Formula used:

If f is a polar rectangle R given by 0arb,αθβ, where 0βα2π, then, Rf(x,y)dA=αβabf(rcosθ,rsinθ)rdrdθ (1)

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (2)

Calculation:

From the limit points of the given three integrals, draw the figures as given below.

From the three figures (Figure1, Figure2, and Figure3), it is observed that the union is the annular region of the two disks whose radius is 1 and 2, respectively.

Therefore, r varies from 1 to 2 and θ varies from 0 to π4

Substitute x=rcosθ and y=rsinθ and convert the given function completely into polar coordinates inclusive of the limit points,

f(x,y)=(rcosθ)(rsinθ)=r2cosθsinθ

Therefore, by the equation (1) the given integral becomes,

1211x2xxydydx+120xxydydx+2204x2xydydx=0π412r2cosθsinθ(r)drdθ=0π412r3cosθsinθdrdθ

Integrate the function with respect to r and θ by using the equation (2)

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