Sketch and evaluate the integral region $x \ge 0, x^2+y^2 \le 2, x^2+y^2\ge 1$

Question

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!

Answer 1

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$$\int_{-\sqrt2}^{-1}\int_0^{\sqrt{2-x^2}}x^2\,\mathrm dx\,\mathrm dy+\int_{-1}^1\int_{\sqrt{1-x^2}}^{\sqrt{2-x^2}}x^2\,\mathrm dx\,\mathrm dy+\int_1^{\sqrt2}\int_0^{\sqrt{2-x^2}}x^2\,\mathrm dx\,\mathrm dy.$$

Answer 2

Alternative approach:

Note
In this note, the constant of integration will be ignored.

Let $I = \int \sin^2(u) ~du.$

Then, $~\cos(2u) = 1 - 2\sin^2(u)~$ which implies $~\int [1 - \cos(2u)] du = 2I.$

Therefore $~~~\displaystyle u - \frac{1}{2}\sin(2u) = 2I.$

Therefore $$\frac{1}{2}u - \frac{1}{4}\sin(2u) = I.\tag1$$

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Note
In both of the integrals below, trig substitutions will be employed where the pertinent angle $\theta$ (et al) is restricted to $0 \leq \theta \leq \pi/2$. Under this restriction, both $\cos(\theta)$ and $\sin(\theta)$ will be non-negative throughout the interval.

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For the smaller circle: $y = \sqrt{1 - x^2}$

So, you have $\int_0^1 2\sqrt{1 - x^2} ~dx.$

Because of the nature of the integral, integration by trig substitution suggests itself.

Let $\cos(u) = x \implies \sqrt{1 - x^2} = \sin(u)$.
Then $-\sin(u)du = dx.$
Further, as $x$ goes from $(0)$ to $(1)$, $u$ goes from $(\pi/2)$ to $(0).$

Therefore, the first integral becomes

$$\int_{\pi/2}^0 -2\sin^2(u) du.\tag2$$

Using (1) above, (2) is evaluated as

$\displaystyle (-2)\left[\frac{1}{2}u - \frac{1}{4}\sin(2u)\right]_{u = \pi/2}^{u = 0}.$

Since $\sin(2 \times 0) = 0 = \sin(2 \times \pi/2)$, (2) evaluates to $(\pi/2).$

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For the larger circle: $y = \sqrt{2 - x^2}$

So, you have $\int_0^{\sqrt{2}} 2\sqrt{2 - x^2} ~dx.$

The trig substitution here is a little trickier.

Let $\sqrt{2}\cos(u) = x \implies \sqrt{2 - x^2} = \sqrt{2}\sin(u)$.
Then $-\sqrt{2}\sin(u)du = dx.$
Further, as $x$ goes from $(0)$ to $(\sqrt{2})$, $u$ goes from $(\pi/2)$ to $(0).$

Therefore, the second integral becomes

$$\int_{\pi/2}^{0} -4\sin^2(u) du.\tag3$$

Using (1) above, (3) is evaluated as

$\displaystyle (-4)\left[\frac{1}{2}u - \frac{1}{4}\sin(2u)\right]_{u = \pi/2}^{u = 0}.$

Again, since $\sin(2 \times 0) = 0 = \sin(2 \times \pi/2)$, (3) evaluates to $(\pi).$

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Addendum
For what it's worth, the inspiration behind the trig substitutions was the assumption that you have a 1st quadrant circular arc [centered at $(0,0)$] of radius $r$. Here, as $u$ goes from $(0)$ to $(\pi/2)$, $(x,y)$ may be represented by $[r\cos(u), r\sin(u)].$