Find $\int \int_X f(x, y)dxdy$

Question

Given $f(x, y) = x$ and $X$ is defined as $2rx \leq x^2 + y^2 \leq R^2$ and $0 < 2r < R$.

Calculate $\int \int_X f(x, y)dxdy$.

So $x^2 + y^2 \leq R^2$ is an area inside a circle with radius $R$.

$2rx \leq x^2 + y^2$ and $0 < 2r < R$ give us a second circle, which area depends on value of $x$ with center on somewhere on $(0, R)$, the leftest coordinate on $(0, 0)$ and the rightest on $(x_1, 0)$, where $x_1 < R$.

So the area of the integral is area of a big circle minus the area of a small circle.

Now let's denote the area of a big circle as $X_1: (x, y)| - R \leq x \leq R; -\sqrt{x} \leq y \leq \sqrt{x}$.

Small circle $X_2: (x, y) | 0 \leq x \leq 2r; -\sqrt{x(x - r)} \leq y \leq \sqrt{x(x - 2r)}$

I'm not quite sure about the bounds, especially of the small circle, so it may be some mistake here. Unfortunately I'm unable to find it.

Then $\int \int_{X_1} x = \int \limits_{-R}^{R} dx \int \limits_{-\sqrt{x}}^{\sqrt{x}} x dy = \int \limits^R_{-R}xdx\int \limits_{-\sqrt{x}}^{\sqrt{x}}dy = 2 \int \limits_{-R}^{R} x^{\frac{3}{2}} dx = \frac{4}{5}\left(R^{\frac{5}{2}} - \left(-R \right)^{\frac{5}{2}}\right)$.

And now I'm stuck on calculate the area of a small circle.

Also can somebody check the bounds which I found for $X_1$ and $X_2$, since it might be the reason I'm unable evaluate second integral.

Answer 1

For the big circle:

$x^2 + y^2 \leq R^2$.

Let $x=\rho\cos\phi$ and $y=\rho\sin\phi$

$$I_1=\int \int_{X_1} x dx dy =\int \limits_{0}^{2\pi}\int \limits_{0}^{R} \rho^2\cos\phi d\rho d\phi = \int \limits_{0}^{R} \rho^2 d\rho\int \limits_{0}^{2\pi}\cos\phi d\phi =0$$

For the small circle:

$2rx \geq x^2 + y^2 \rightarrow r^2 \geq (x-r)^2 + y^2$.

Let $x=\rho\cos\phi+r$ and $y=\rho\sin\phi$

$$I_2=\int \int_{X_2} x dx dy =\int \limits_{0}^{2\pi}\int \limits_{0}^{r} \rho(\rho\cos\phi+r) d\rho d\phi =\int \limits_{0}^{r} \rho^2 d\rho\int \limits_{0}^{2\pi}\cos\phi d\phi+ r\int \limits_{0}^{r} \rho d\rho\int \limits_{0}^{2\pi}1 d\phi =$$

$$=0+\frac{r^3}{2}2\pi=r^3\pi$$

Finally:

$$I=I_1-I_2=-r^3\pi$$

Answer 2

Well, firstly, I'd like you to realize that it would be best to deal with this in terms of polar coordinates, as we are dealing with circles.

Now let's analyse this.

$2rx\leq x^2+y^2\leq R^2$, $0\leq 2r\leq R$, where $R$ is the radius of the circle.

Now remember that in polar coordinates, $x=R\cos\theta,y=R\sin\theta$, therefore we have that $x^2+y^2=R^2$

$2rx\leq R^2\leq R^2$, which simplifies to $2rx\leq R^2$, or $2r(R\cos\theta)\leq R^2$, which means that $2r\cos\theta\leq R$.

Now we also have that $0\lt 2r\lt R$. Since these circles are not shifted vertically, the range for $\theta$ is $0\leq\theta\leq 2\pi$

What about the radius $R$? Well the radius is clearly between $0\leq R\leq 2r\cos\theta$

Now our original function $f(x,y)=x$ can be written as $R\cos\theta$.

So the integral is of the form:

$$\int\int\text{ f(x,y) R dr d$\theta$}=\int^{2\pi}_{0}\int^{2r\cos\theta}_{0}(R\cos\theta)\text{ dR d$\theta$}$$

Your integral will depend on $r$