Calculate $\iint_D x dxdy$ using polar coordinates

Question

Using polar coordinates, I want to calculate $\iint_D x dxdy$, where $D$ is the disk with center $(2,3)$ and radius $2$.

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I have done the following:

We have $D=\{(x,y)\mid (x-2)^2+(y-3)^2\leq 4\}$.

We use $(x,y)=(r\cos \theta, r\sin \theta)$.

From the inequality $$(x-2)^2+(y-3)^2\leq 4\Rightarrow x^2-4x+4+y^2-6y+9\leq 4 \Rightarrow x^2+y^2-4x-6y\leq -9$$ we get $$r^2\cos^2\theta+r^2\sin^2\theta-4r\cos\theta-6r\sin\theta\leq -9 \Rightarrow r^2-r(4\cos\theta-6\sin\theta)+9\leq 0$$ To find for which values of $r$ that inequality is true, we have to find first the roots of $r^2-r(4\cos\theta-6\sin\theta)+9=0$.

The roots are $$2\cos \theta+3\sin\theta\pm \sqrt{12\cos\theta\sin\theta-5\cos^2\theta}$$ Therefore, we get the inequality $r^2-r(4\cos\theta-6\sin\theta)+9\leq 0$ for $$2\cos \theta+3\sin\theta-\sqrt{12\cos\theta\sin\theta-5\cos^2\theta}\leq r\\ \leq 2\cos \theta+3\sin\theta+\sqrt{12\cos\theta\sin\theta-5\cos^2\theta}$$ or not?

So, at the integral do we use these limits for $r$ ? And what about $\theta$ ? Does it hold that $0\leq \theta\leq 2\pi$ ?

Answer 1

OFC your area of $\theta$ has to be restricted to the area where it "hits" the circle and as you already figured out your ansatz leads you to a very complicated calculation.

The reason is that your circle is not centered so first shift your integration area that you have a centered circle by substituting $\overline{x} = x-2, \overline{y} = y-3$ so you get $$\iint_D x dxdy = \iint_\overline{D} (\overline{x} + 2)\;d\overline{x}d\overline{y}$$ with $$\overline{D} =\{(\overline{x},\overline{y})\mid \overline{x}^2+\overline{y}^2\leq 4\}$$

What can easily be solved by using polar coordinates.

Answer 2

Hint: Use substitution \begin{cases} x=2+r\cos\theta,\\ y=3+r\sin\theta,\\ \end{cases} then $$\iint_D xdxdy=\int_0^{2\pi}\int_0^2r(2+r\cos\theta)\,dr\,d\theta$$