Answer :
Answer:
Step-by-step explanation:
We are to integrate the function
[tex]f(x,y) = 2x-y[/tex]
over the area enclosed by the circle
[tex]x^2+y^2 =4 \\x=0\\y=x[/tex]
Convert into polar coordinates
[tex]dA=rdr dt[/tex]
x=rcost and y = rsint
Since the lines are x=0 and y=x we find that t varies from 0 to pi/4
[tex]2x-y =2rsint-rcost[/tex]
Double integral will become now
[tex]\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[/tex]
(since variables are independent here)
Substitute for r and t
4 sin pi/4 -2 cospi/4
= [tex]\sqrt{2}[/tex]
The result of evaluating the double integral is [tex]\int\int R(2x-y) dA = \frac{4(4 - 3\sqrt 2)}{3}[/tex]
How to evaluate the double integral?
The given parameters are:
[tex]\int\int R(2x-y) dA[/tex]
x^2 + y^2 = 4
Lines x = 0 and y = x
By polar coordinates, we have:
x = rcost and y = rsint
dA = rdrdt
Substitute x = rcost and y = rsint in 2x - y
2x - y = 2rcost - rsint
So, the integral becomes
[tex]\int\int R(2x-y) dA = \int\limits^a_b \int\limits^a_b ( 2r\cos (t) - r \sin(t) )\ rdrdt[/tex]
The lines x = 0 and y = x imply that the integral varies from 0 to 2 and π/2 to π/4.
So, we have:
[tex]\int\int R(2x-y) dA = \int\limits^{\pi/4}_{\pi/2} \int\limits^2_0 ( 2r\cos (t) - r \sin(t) )\ rdrdt[/tex]
Rewrite as:
[tex]\int\int R(2x-y) dA = \int\limits^{\pi/4}_{\pi/2} \int\limits^2_0 ( 2\cos (t) - \sin(t) )\ r^2drdt[/tex]
Split the integral
[tex]\int\int R(2x-y) dA = \int\limits^{\pi/4}_{\pi/2} ( 2\cos (t) - \sin(t) ) dt \int\limits^2_0 r^2dr[/tex]
Integrate
[tex]\int\int R(2x-y) dA = [2\sin (t) - \cos(t)]\limits^{\pi/4}_{\pi/2} * [\frac{r^3}{3}]\limits^2_0[/tex]
Expand
[tex]\int\int R(2x-y) dA = [2\sin (\pi/2) + \cos(\pi/2) - 2\sin (\pi/4) - \cos(\pi/4)] * [\frac{2^3 - 0^3}{3}][/tex]
Simplify the above expression
[tex]\int\int R(2x-y) dA = [2*1 + 0 - \sqrt 2 - \frac{\sqrt 2}{2}] * [\frac{8}{3}][/tex]
[tex]\int\int R(2x-y) dA = [\frac{4 - 3\sqrt 2}{2}] * [\frac{8}{3}][/tex]
Evaluate the product
[tex]\int\int R(2x-y) dA = \frac{4(4 - 3\sqrt 2)}{3}[/tex]
Hence, the result of evaluating the double integral is [tex]\int\int R(2x-y) dA = \frac{4(4 - 3\sqrt 2)}{3}[/tex]
Read more about double integrals at:
https://brainly.com/question/19053586