Evaluate the integral from x = 0 to 1 of the integral from y = x to root(2 - x^2) of x/root(x^2 + y^2) dy dx by changing the order of integration.

Step-by-step solution

1. 1

   Draw the region. The bounds say $0\le x\le 1$ and $x\le y\le\sqrt{2-x^2}$: the region sits above the line $y=x$, inside the circle $x^2+y^2=2$, and left of $x=1$. The line meets the circle at $(1,1)$, so the region is a curved wedge with corners $(0,0)$, $(0,\sqrt2)$ and $(1,1)$.

2. 2

   Reverse the order carefully — the region needs two slices. For a fixed $y$ the left edge is always $x=0$. For $0\le y\le 1$ the right edge is the line $x=y$; for $1\le y\le\sqrt2$ the right edge is the circle $x=\sqrt{2-y^2}$. So $I=\int_0^1\!\!\int_0^{y}\frac{x\,dx}{\sqrt{x^2+y^2}}\,dy+\int_1^{\sqrt2}\!\!\int_0^{\sqrt{2-y^2}}\frac{x\,dx}{\sqrt{x^2+y^2}}\,dy.$ Writing a single slice $x$ from $y$ to $\sqrt{2-y^2}$ would describe the region below $y=x$ instead, and is the standard trap here.

3. 3

   Do the inner integral once. With $u=x^2+y^2$, $du=2x\,dx$, $\int\frac{x\,dx}{\sqrt{x^2+y^2}}=\sqrt{x^2+y^2}.$

4. 4

   Evaluate the two slices. First slice: $\left[\sqrt{x^2+y^2}\right]_0^{y}=\sqrt2\,y-y$, so $\int_0^1(\sqrt2-1)y\,dy=\dfrac{\sqrt2-1}{2}$. Second slice: $\left[\sqrt{x^2+y^2}\right]_0^{\sqrt{2-y^2}}=\sqrt2-y$, so $\int_1^{\sqrt2}(\sqrt2-y)\,dy=2-\sqrt2-\dfrac12$.

5. 5

   Add them. $I=\frac{\sqrt2-1}{2}+\left(\frac32-\sqrt2\right)=1-\frac{\sqrt2}{2}\approx 0.292893.$

6. 6

   Confirm in polar coordinates. With $x=\rho\cos\theta$ the integrand is just $\cos\theta$ and $dA=\rho\,d\rho\,d\theta$. On $\tfrac\pi4\le\theta\le\tfrac\pi2$ the constraint $x\le 1$ is slack (since $\sec\theta\ge\sqrt2$), so $\rho$ runs from $0$ to $\sqrt2$: $\int_{\pi/4}^{\pi/2}\!\!\int_0^{\sqrt2}\cos\theta\,\rho\,d\rho\,d\theta=\bigl[\sin\theta\bigr]_{\pi/4}^{\pi/2}=1-\frac{\sqrt2}{2}.$