Sketch the region defined by $x \ge 0, x^2+y^2 \le 2, x^2+y^2\ge 1$. Write down the integral over the region in each of the two possible orders of $f(x,y) = x^2$. Evaluate both integrals.
In defining the regions of integration, I imagine it to look something like this:
By taking the integral verticallyt in respect to $y$ I then get: $x \le\sqrt{2-y^2}$, so $y$ has the bounds $-\sqrt{2-x^2}<y<\sqrt{2-x^2}$, though I am unsure of this as the inequality $x^2+y^2 \le 1$ throws me off on what I should do with $x$ and $y$, any ideas will be appreciated!
Best Answer
If you want to integrate first in order to $y$ and then in order to $x$, then start by noting that $0\leqslant x\leqslant\sqrt2$ (see the picture below). If $0\leqslant x\leqslant1$, then $y$ takes values from $-\sqrt{2-x^2}$ to $-\sqrt{1-x^2}$ and from $\sqrt{1-x^2}$ to $\sqrt{2-x^2}$, whereas if $1\leqslant x\leqslant\sqrt2$, then $y$ takes values from $-\sqrt{2-x^2}$ to $\sqrt{2-x^2}$. So, you have$$\int_0^1\left(\int_{-\sqrt{2-x^2}}^{-\sqrt{1-x^2}}x^2\,\mathrm dy+\int_{\sqrt{1-x^2}}^{\sqrt{2-x^2}}x^2\,\mathrm dy\right)\,\mathrm dx+\int_1^{\sqrt2}\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}x^2\,\mathrm dy\,\mathrm dx.$$
And if you integrate first in order to $x$ and then in order to $y$, by a similar reason you should get$$\int_{-\sqrt2}^{-1}\int_0^{\sqrt{2-x^2}}x^2\,\mathrm dx\,\mathrm dy+\int_{-1}^1\int_{\sqrt{1-x^2}}^{\sqrt{2-x^2}}x^2\,\mathrm dx\,\mathrm dy+\int_1^{\sqrt2}\int_0^{\sqrt{2-x^2}}x^2\,\mathrm dx\,\mathrm dy.$$