I am hoping to solve the following integral.
$$\int_0^2\int_y^{\sqrt{8-y^2}} \frac{1}{\sqrt{(1+x^2+y^2)}}dxdy$$
I know that this has to be a polar conversion followed by U substitution, but I am unsure of what the new bound will be. I am unable to think of this geometrically.
I had gotten to this point, but wasn't sure if what I did was correct.
$$\int_0^{\pi/2} 2\sqrt{(1+r^2)}d\theta$$ from $r=rsin(\theta)$ to $\sqrt8$
I think there must be a problem in my bounds, specifically $r=rsin(\theta)$
Not very correct, but you can write the integral this way:
$$I=\int_0^2\int_{x_1=y}^{x_2=\sqrt{8-y^2}} \frac{1}{\sqrt{1+x^2+y^2}}dxdy$$
The lower limit is the line $y=x$ and the upper $x^2+y^2=8$, corresponding to a circle centered in the origin. The conversion to polars simplifies the integral, indeed. The area of integration is a circular sector limited by the $x$ axis and the line $x=y$ and radius $\sqrt{8}$
In polars
$x^2+y^2=r$ and $\mathrm dx\mathrm dy=r\mathrm dr\mathrm d\theta$
$$I=\int_0^\sqrt{8}\int_0^{\pi/4} \frac{r}{\sqrt{1+r^2}}\mathrm d\theta\mathrm dr=$$
$$=\frac\pi4\int_0^\sqrt{8}\frac{r}{\sqrt{1+r^2}}\mathrm dr=$$
$$=\left.\frac\pi4\sqrt{1+r^2}\right\vert_0^\sqrt{8}=\frac\pi2$$