1. ## Integral...substitution coordinates

The body K is limited by the cone z= -sqrt(x^2+y^2) and the sphere x^2+y^2+z^2=16

Now, the most natural thing to do is to introduce spherical coordinates:

x= rsinωcosφ
y= rsinωsinφ
z= rcosω

In this case, the body K would be described by:

cone z= -sqrt(x^2+y^2)--> -sqrt(r^2(sinω)^2)= - rsinω

and x^2+y^2+z^2=16 --> r^2= 16

z= rcosω = -rsinω --> ω= -pi/4

which means that the limits of the integral is

0<r<sqrt(16)
-pi/4<ω<0
0<φ<2pi

(is this correct?)

However, the task states that I should use cylindrical coordinates to describe the body K:

x= rcosφ
y= rsi
z= z

Then the body K would be described as: the cone = -r and the sphere = r^2 + z=16

But which limits do I set here?

Is this correct, is the transformation to cylindrical spheres this easy when they ask for the body K to be described by cylindrical coordinates?

2. The cone does not have "spherical symmetry" so I would be inclined to agree that cylindrical coordinates is better than spherical coordinates. If you are looking for the volume of the region bounded by the sphere and cone, then your limits of integration of $\omega$ should be from $3\pi/4$ to $\pi$, NOT $-\pi/4$ to 0 (in spherical coordinates, the "co-latitude" goes from 0 to $\pi$).

In cylindrical coordinates, [tex]\theta[tex] will go from 0 to [tex]2\pi[/itex], r from 0 to $\sqrt{16}= 4$ and z from $-sqrt{16- r^2$ to -r.

3. Ok, thanks.

I'm sad to say though that I don't understand why z goes from -sqrt(16) - r^2 to -r in the cylindrical case...

Shouldn't it go from sqrt(16)- r to -r

(there is no minus infront of sqrt(16)- r)

4. That was just my bad LaTex- Since $r^2+ z^2= 16$, $z^2= 16- z^2$ so that $z= \pm\sqrt{16- r^2}$.

Because we are talking about the region between the cone z= -r and the sphere, z is negative: z goes from the sphere, $z= -\sqrt{16- r^2}$ to $z= -r$.