I have this math problem right now, and any way I look at it, it just seems impossible. I don't understand.
So I am given the integral:
$$
\int_0^{2\pi}\int_0^{1}\int_0^{\sqrt{4-r^2}} r^2\,
dz\,dr\,d\theta
$$
The exercise is to convert (no need to calculate) the integral to cartesian and spherical coordinates.
I managed the conversion to cartesian coordinates and I got:
$$
\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_0^{\sqrt{4-x^2-y^2}} \sqrt{x^2+y^2}\,
dz\,dy\,dx
$$
But the conversion to spherical coordinates seems just impossible. I don't understand at all.
I am given a "hint" that I need to give the integral in multiple parts. What really throws me off though, is that the area that I'm integrating is clearly a cylinder, only the top is spherical.
I would appreciate help in figuring out how this is possible.
Thanks
Best Answer
I show here a drawing of your domain, projected in the $x-z$ plane, with $x$ horizontal and $y$ vertical.
Then the radius of integration in spherical coordinates is between the origin an d the black dot. The angle with respect to $z$ axis is $t$ in the figure. As mentioned in the question, you have two domains. While the projection in the horizontal plane is less than $1$, your upper limit is $2$. If the projection is greater than $1$, you only integrate to $1/\sin t$. Why that value? Because than the projection in the plane is $$\frac 1{\sin t}\sin t=1$$
Alternatively, just from a notation point of view, you can put the upper limit for $r$ to be $$\min\left(2,\frac 1{\sin t}\right)$$
In practice, you have domains where the polar angle is below and above $\arcsin\frac12=\frac\pi6$