Evaluate the given integral by changing to polar coordinates

[insanerp]

double integral of 4(x+y)dA ,where R is the region that lies to the left of the y-axis between the circles x^2 + y^2 = 1 and x^2 + y^2 = 4. The value of integral is?
I found the value of ? to be ? / 2 to 3 ?/ 2 and r to be from 1 to 2 so the double integral of r dr d? to be 6pie but it seems to be wrong, please help, thank you.

 

[Deleted member 4993]

double integral of 4(x+y)dA ,where R is the region that lies to the left of the y-axis between the circles x^2 + y^2 = 1 and x^2 + y^2 = 4. The value of integral is?
I found the value of ? to be ? / 2 to 3 ?/ 2 and r to be from 1 to 2 so the double integral of r dr d? to be 6pie but it seems to be wrong, please help, thank you.

Where did (x+y) go?

 

[insanerp]

so I sub. r cos? in for x and r sin? in for y and got 56/3 as a final answer is this correct?

 

[BigGlenntheHeavy]

\(\displaystyle 4\int_{1}^{2}\int_{\pi/2}^{3 \pi/2}r^{2}[cos(\theta)+sin(\theta)]d\theta dr \ = \ \frac{56}{3}.\)

\(\displaystyle Looks \ good \ to \ me.\)