I have two problems which I have not been able to solve using double integrals in polar coordinates.
The first one is to find the volume of the solid above the cone $z = r$ and below the sphere $z^2 = 1-r^2$.
The second question is the volume of the solid bounded by the paraboloids $z = 3r$ and $z = 4-r$.
In both the main problem was finding and defining the limits of integration.
Thanks in advance!
For the first one - You can use rectangular coordinates to get, perhaps, a closer feeling of what's going on:
The cone is $\;z=\sqrt{x^2+y^2}\;$, the sphere is $\;x^2+y^2+z^2=1\;$. You can restrict yourself to the what happens in the first octant and then multiply by four due to symmetry ( draw it if this helps):
$$\int_0^{1/\sqrt2}\int_0^{\sqrt{\frac12-x^2}}\int_{\sqrt{x^2+y^2}}^{\sqrt{1-x^2-y^2}}dzdydx$$
Of course, nobody really expects anyone who isn't a masochist to do the above that way, so we can switch to cylindrical coordiantes, say, and get (don't forget the Jacobian!):
$$\int_0^{1/\sqrt2}\int_0^{\pi/2}\int_r^{\sqrt{1-r^2}}r\,dz\,d\theta\,dr=\frac\pi2\int_0^{1/\sqrt2}\left[r\sqrt{1-r^2}-r^2\right]dr=$$$${}$$
$$=\left.-\frac\pi4\frac23(1-r^2)^{3/2}\right|_0^{1/\sqrt2}-\frac\pi2\left.\frac13r^3\right|_0^{1/\sqrt2}=-\frac\pi6\left(\frac1{2\sqrt2}-1\right)-\frac\pi6\left(\frac1{2\sqrt2}\right)=$$
$$=-\frac\pi{6\sqrt2}+\frac\pi6\implies\text{ the volume wanted is}\;\;\frac{2\pi}3\left(1-\frac1{\sqrt2}\right)$$
Now try a similar reasoning for the other one. Observe that changing to rectangular coordinates isn't really necessary, but sometimes it makes things easier to grasp.
Let us find the projection in the $xy$ plane. Note $z=r$ and $z^2=1-r^2$ our are two surfaces. By setting values of $z$ equal we should find the boundary of the projection in $xy$ plane, draw the region to see this.
Note the first surface implies $z^2=r^2$. This implies,
$$1-r^2=r^2$$
Which implies $r=\frac{\sqrt{2}}{2}$ is the boundary of the projection in the $xy$ plane. Then the volume is,
$$V=\int_{0}^{2\pi} \int_{0}^{\frac{\sqrt{2}}{2}} \int_{r}^{\sqrt{1-r^2}} r dz dr d\theta$$
For the the second integral we repeat the procedure.
$$4-r=3r$$
$$r=1$$
$$V=\int_{0}^{2\pi} \int_{0}^{1} \int_{3r}^{4-r} rdz dr d\theta$$
If you haven't learned triple integrals then what you want to compute is $\iint_{D} (\text{top_Surface}-\text{bottom_Surface}) dA$ to compute the volume. So for example the integral above can be equivalently thought of as the double integral,
$$\int_{0}^{2\pi} \int_{0}^{1} ((4-r)-3r) r dr d\theta$$
The first integral is: $$\int_{0}^{{\sqrt{2}\over{2}}}\int_{0}^{2\pi}\int_{r}^{{\sqrt{1-r^2}}}r \space dzd\theta dr$$ The second is: $$\int_{0}^{1}\int_{0}^{2\pi}\int_{3r}^{4-r}r \space dzd\theta dr$$ You should make a graph and find the intersections to figure out the radius. The angle has no limitations and $z$ is already defined