Evaluate the integral $\int_R\int \frac{x+y}{x^2+y^2+a^2}dA$ over the portion of the first quadrant lying inside the circle $x^2+y^2 = a^2$.

Question

Evaluate the integral $\int_R\int \frac{x+y}{x^2+y^2+a^2}dA$ over the portion of the first quadrant lying inside the circle $x^2+y^2 = a^2$.

Hi I treid finding a question on this but I couldn't find one with the added confusion of the $a$.

Now I am well aware of variable change and so I get that we need:

$x = acos(\theta)$ and $y = acos(\theta)$

However I am unsure what the integrand becomes. In a solution I have seen of this problem they do the following:

$\frac{x+y}{x^2+y^2+a^2} = [cos(\theta)+sin(\theta](\frac{r^2}{r^2+a^2})$

and then the inner integral can be evaluated easily.

However I do not understand why they don't make use of the fact that a is always equal to r and so it simply becomes:

$\frac{x+y}{x^2+y^2+a^2} = [cos(\theta)+sin(\theta](\frac{r^2}{r^2+r^2})$

which can then be simplified.

Anyone know why I cannot do the last step? Why does there need to be an a?

Any help much appreciated!

Answer 1

The domain in the first quadrant inside the circle $x^2+y^2=a^2$ means that you actually consider a quarter of a disk.

Thus, you want $x\geq 0$, $y\geq 0$ and $x^2+y^2\leq a^2$ (OBS!, not $x^2+y^2=a^2$). Since $r^2=x^2+y^2$, this means that you need to consider $r\leq a$ and not only $r=a$.

Answer 2

With a correct change of variables the given integral equals

$$ \int_{0}^{a}\int_{0}^{\pi/2}(\sin\theta+\cos\theta)\frac{\rho^2}{\rho^2+a^2}\,d\theta\,d\rho = 2\int_{0}^{a}\frac{\rho^2}{\rho^2+a^2}\,d\rho\stackrel{\rho=au}{=}2a\int_{0}^{1}\frac{u^2}{u^2+1}$$ that is $\color{red}{2a\left(1-\frac{\pi}{4}\right)}.$