# changing order of integration in spherical coordinates

• Jun 27th 2010, 01:56 AM
beowulf
changing order of integration in spherical coordinates
Hi,

I'm teaching myself a bit of calculus and I'm having trouble with this:

Let D be the region bounded below by the plane z=0, above by the sphere x^2 + y^2 + z^2 = 4, and on the sides by the cylinder x^2 + y^2 = 1. Set up the triple integral in spherical coordinates that gives the volume of D using the order of integration .

The solution says that D is:

Attachment 18014

But I thought it should be:

Attachment 18015

Could you please tell me where I'm going wrong?

Many thanks!
• Jun 27th 2010, 06:40 AM
running-gag
Hi

Your book and you are both right but your book's solution is much simpler
It is simpler to express $\rho$ in terms of $\varphi$ than the contrary
• Jun 27th 2010, 06:58 AM
beowulf
Many thanks. I'm a still a bit confused, though, because both solutions express Phi in terms of Rho. (I know it's easier to do it the other way round but I wanted to make sure I understood how to change the order of integration.) The difference between the two solutions is that in mine there's one more 'bit' which seems to be missing from the book's solution. In fact, leaving aside the first integral in both solutions, the remaining integral in the book's solution evaluates to Pi/Sqrt(3), while my remaining two integrals together evaluate to 2*Pi/Sqrt(3) -- which is what I also get if I express Rho in terms of Phi, as you suggested.
• Jun 27th 2010, 10:07 AM
running-gag
I do not agree since $\int_{0}^{2\pi} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \int_{0}^{\frac{1}{\sin \phi}} \rho^2 \sin \phi d\rho d\phi d\theta = \frac{2\pi}{3} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{d\phi}{\sin^2 \phi} = \frac{2\pi}{\sqrt{3}}$

The remaining integral in the book's solution is therefore $\frac{2\pi}{\sqrt{3}}$