# Thread: Help me with double integrals

1. ## Help me with double integrals

$\int_D \int (2x-y) dA$
D is bounded by the circle with center the origin and radius 2.
help me to get started.

2. Originally Posted by zpwnchen
help me to get started.
Well, as that circle's equation is $x^2+y^2=4 \Longrightarrow y=\pm\sqrt{4-x^2}$, we can see that

$\iint\limits_D (2x-y)dA=\int\limits_{-2}^2\int \limits_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}(2x-y)dydx$

Tonio

3. Originally Posted by zpwnchen
help me to get started.
When dealing with circles, I would suggest converting to polar coordinates.

Do you know the following formulae:

$x = r \cos (\theta)$

$y = r \sin (\theta)$

Also, do you know the Jacobian matrix for a change of variables in double integration?

Once you have converted to polar coordinates, your variables of integration will be r, the radius, and $\theta$, the angle. Clearly, the limits for these are as follows:

$0 \leq r \leq 2$

$0 \leq \theta \leq 2\pi$