Question

Evaluate the double integral ∬R(2x−y)dA, where R is the region in the first quadrant enclosed by the circle x^2+y^ 2= 4 and the lines x = 0 and y = x, by changing to polar coordinates.

Answer 1

Answer:

Step-by-step explanation:

We are to integrate the function

$f(x,y) = 2x-y$

over the area enclosed by the circle

$x^2+y^2 =4 \\x=0\\y=x$

Convert into polar coordinates

$dA=rdr dt$

x=rcost and y = rsint

Since the lines are x=0 and y=x we find that t varies from 0 to pi/4

$2x-y =2rsint-rcost$

Double integral will become now

$\int\limits^\frac{\pi}{4} _0 \int\limits^2_0 2rcost -rsint dt\\=\int\limits^\frac{\pi}{4} _0 r^2 cost -r^2sint /2 dt\\= r^2 sint - r^2 cost/2$

(since variables are independent here)

Substitute for r and t

4 sin pi/4 -2 cospi/4

= $\sqrt{2}$