# Refesher Part Duex - Rectangular -> Cylindrical -> Spherical

1. Apr 12, 2015

### beer

I'm at it again today. Helping the same friend (plus another) study for their final calculus exams; it's a good refresher for me as well. (I'm a senior industrial engineering major so I'm "done" with calculus, and it isn't used much in our upper level classes nor professionally to the best of my knowledge.)

Yesterday I ran in to a "visualization" error on my part. Today, we're (currently) working through the following problem:

∫∫∫ (x2 + y2)1/2 dz dy dx

0 ≤ z ≤ (x^2 + y^2)1/2
0 ≤ y ≤ (25 - x^2)1/2
0 ≤ x ≤ 5

The question doesn't require evaluation of the integral. Rather, it asks to convert the integral in to cylindrical coordinates, and in to spherical coordinates, and to explain and sketch the region. However, solving the integral yields (625π / 8)

Setting up the region, R, yields an eighth of a sphere with radius 5. I cautiously explained that if they set up their boundaries right in any form of integration (spherical, cylindrical, etc) and integrated over a constant (generally 1), it would yield the volume of R... in this case, (1/8) * (4/3 π 53) = 125π/6. I've never explicitly read or been explained that this is a good check of proper boundaries, but it seems to always hold true.

Converting to cylindrical we had:
0 ≤ z ≤ r
0 ≤ r ≤ 5
0 ≤ θ ≤ π/2

Which yielded the volume of R when integrated over a constant. Assuming these were the correct boundaries. And more importantly:

∫∫∫ r z dz dr dθ = (625π / 8)

Check. Things seem to be in order.

Similarly, for spherical, we get that...
0 ≤ θ ≤ π/2
0 ≤ ρ ≤ 5
0 ≤ φ ≤ π/2

These boundaries integrated with a constant yields the the same 125π/6 value of R I assume indicate that the boundaries are indeed correct.

However, when setting up the integral (given the above bounds) I told them it would be:

∫∫∫ρ2 sin(φ) [ρsin(φ)] dρ dθ dφ

There are a few ways to get the [ρsin(φ)] part; direct conversion using the commonly available "rectangular to spherical" formulas, or by noting that r = ρsin(φ) when converting from cylindrical to spherical, etc.

Since the boundaries don't include any variables, the order of integration shouldn't matter. (This is the first red flag in my mind; I recall from habit that most all triple integrals we solved in my Calc III class included variables on the inner bounds)

Solving this integral yields 625π2 / 32.

Again, the problem doesn't actually require solving the integrals, but since the solution is different, I assume we set the spherical version up correctly and I can not identify where things went wrong... either in the boundaries (likely) or in the function itself... (also likely.)

Any help would be appreciated. Sorry for being so long winded.

2. Apr 17, 2015