Transforming integral in cylindrical coordinates into cartesian.

Question

I am trying to transform the following integral to an integral in cartesian coordinates. $$\int^{2\pi}_0\int^1_0\int^{\sqrt{1-r^2}}_0r \ dzdrd\theta$$ I cannot really visualise how the region enclosed by the integral borders look like. Could anyone show how to transform it? So far I know that: $$\int^{1}_{-1}\int^{\sqrt{1-x^2}}_{-\sqrt{1-x^2}}\int^{A}_01 \ dzdydx$$ but I have no idea at all how to find $A$.

Answer 1

Since you already have the answer, I'll give a thorough thought description to changing coordinates:

The inner integral gives bounds $z=0$ and $z=\sqrt{1-r^2}$. But $z=\sqrt{1-r^2}$ gives $z^2=1-r^2$, which is $z^2=1-(x^2+y^2)$. Moving the variables to one side gives $x^2+y^2+z^2=1$ -- the unit sphere centered at the origin. So part of the boundary of the region is a sphere. But which part?

We know from the middle integral that $0 \leq r \leq 1$. Notice that $z=\sqrt{1-r^2}$ is smaller the larger $z$ is. So $z$ is smallest when $r=1$, where it is $0$, and largest when $r=0$, where it is $1$. This gives $0 \leq z \leq 1$. But then we must be dealing with some part of the upper half of the sphere. The fact that $0 \leq \theta \leq 2\pi$ tells us that we are looking at the whole upper-half sphere.

Now that we know what we are dealing with, we can set this up any way we'd like (so long as we are able). If we slice in $z$'s first, we notice that the $z$'s 'start' at the plane $z=0$ and 'go up' until they 'hit' the sphere -- $x^2+y^2+z^2=1$. Since this is the upper half this is $z=\sqrt{1-x^2-y^2}$. This gives inner integral: $$ \int_0^{\sqrt{1-x^2-y^2}} 1 \; dz $$ Now that $z$ is gone, what is the boundary of our region in the plane? It's the unit circle! We can slice any which way. For example, slicing in $y$ first gives $$ \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_0^{\sqrt{1-x^2-y^2}} \;dz\;dy\;dx $$ since $y$ always goes from the bottom half of the unit circle -- $-\sqrt{1-x^2}$ -- to the top half -- $\sqrt{1-x^2}$. Then we can choose any $x$ between $-1$ and $1$. Similarly, we could have instead set-up the integral as $$ \int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \int_0^{\sqrt{1-x^2-y^2}} \;dz\;dx\;dy $$ Note that you do not even need to start with $z$! You could have 'easily' made the inner most integral with respect to $x$ or $y$. See if you can set those up, they are a good exercise.

Answer 2

The bound for $z$ implies that $z^2+r^2\leq 1$, which is the same as $x^2+y^2+z^2\leq 1$. Now can you recognize what this region is?