How to chose the bounds of an integral?

Question

How to evaluate the integral $$\int_{-1}^1 \int_0 ^{\sqrt{1-x^2}}x^2(x^2+y^2)^2dydx$$ via polar coordinates?

I substituted $x^2+y^2$ by $r^2(\cos^2(\phi)+\sin^2(\phi))$ and so on but how to chose the bounds of integration? Thanks in advance.

Answer 1

The domain of integration is the upper semi-circle.

\begin{align*} \int_{-1}^{1} \int_{0}^{\sqrt{1-x^2}} x^2(x^2+y^2)^2 \,dy \,dx &= \int_{0}^{\pi} \int_{0}^{1} (r^2 \cos^2 \theta)(r^4)\, r\, dr \, d\theta \\ &= \left( \int_{0}^{\pi} \cos^2 \theta \, d\theta \right) \left(\int_{0}^{1} r^7 dr \right) \\ &= \frac{\pi}{2} \times \frac{1}{8} \\ &= \frac{\pi}{16} \end{align*}

Answer 2

The set described by the aboves is $\{x^2 + y^2 \le 1, y\ge 0 \}$

Obviously this is a filled half unit circle above the x-axis… is that clear to you?

So: $0 \le r \le 1, \phi \in [0,\pi]$.