Use polar coordinates to find the volume of the region above the cone $z=\sqrt{x^2+y^2}$ and below the sphere $x^2+y^2+z^2=1$
Drawing some sketch of the problem, I divided the region in two parts, the first one is covered from the origin to the intersection of the solids at $2(x^2+y^2)=1$, with $r=\frac{1}{\sqrt{2}}$, and the second one goes from the intersection to the sphere limits. So I found the final volume by the following integral sum:
$V=\int_0^{2\pi}\int_0^{\frac{\sqrt{2}}{2}} r^2 dr dt+ \int_0^{2\pi}\int_{\frac{\sqrt{2}}{2}}^1 r\sqrt{1-r^2} dr dt=\frac{\sqrt{2} \pi}{3}$
The answer says $V=\frac{2 \pi}{3}(1-\frac{1}{\sqrt{2}})$. What am I missing??
I'm not sure how you're getting this setup but the boundaries are not correct and the $z$ function is not correct. I am assuming that you are doing this using the double integral method which requires you to input as
$$V=\iint z(r,\theta)rdrd\theta$$
If so, then you need to consider the $z$ for both regions. Remember in single-variable calculus when finding that area between two curves, you need to subtract the upper function from the lower function like
$$A=\int (f(x)-g(x)) dx$$ where $f(x)>g(x)$.
That same goes here. What is the lower function and upper function in the first region. You will find that you actually don't need to break it down into 2 regions, as you are trying to do, although you can. Next, you need to find the smallest value of $r$ and the largest value of $r$. So two things:
$$V=\int_{0}^{2\pi}\int_{r_{min}}^{r_{max}} \left(z_{upper}(r,\theta)-z_{lower}(r,\theta)\right)rdrd\theta$$
Hope this helps!