Question

Paul's Note - Surface Integral using Divergence Theorem http://tutorial.math.lamar.edu/Problems/CalcIII/DivergenceTheorem.aspx

I am trying to understand (apparently, using Stroke's) how they figured out the spherical limits, in Example 1 solution.
How?

Help greatly appreciated!

Answer 1

that's not Stokes!!

Answer 2

It says so in the "solution" for example 1 ... I dont know. Have not done Strokes yet

Answer 3

On step 2

Answer 4

I guess my issue is parameterizing the sphere surface within the constraints

Answer 5

Step 2

Answer 6

How do they find Sphere Limits?

Answer 7

I feel I am getting closer to understanding... not there yet though... !

Answer 8

Recall the usual change of variables:
\[\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\]which generates the volume element \(\mathrm{d}V=\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z=\rho^2\sin\varphi\,\mathrm{d}\rho\,\mathrm{d}\theta\,\mathrm{d}\varphi\).

Let's fix the angles \(\theta\) and \(\varphi\) for the moment. A sphere of radius \(\rho\) contains all the points \((x,y,z)\) such that the distance of any point in the region bounded by the sphere is no greater than \(\rho\). In this case, the sphere has radius \(\rho=4\), so in order to grab all the points at any fixed values for \(\theta\) and \(\varphi\), \(\rho\) would have to range from the origin to the surface of the sphere. In other words, \(0\le\rho\le4\).

Now let's only fix \(\varphi=\dfrac{\pi}{2}\). This particular value zeroes out the \(z\) component, while leaving behind the \(x\) and \(y\) components. You'll notice that you're left with the usual polar coordinate change of variables that describe a circle in the \(x\)-\(y\) plane for \(0\le\theta\le2\pi\). Your sphere, however, is halved to contain only the points \((x,y)\) that lie in the second and third quadrants of the \(x\)-\(y\) plane. All this means is that \(\theta\) varies to contain the top half of the circle in the \(x\)-\(y\) plane, so \(\pi\le\theta\le2\pi\). (See first attachment)

Now fix \(\theta\), it doesn't matter which value you use. The angle \(\varphi\) is the angle made between a point \((x,y,z)\) and the \(z\)-axis. For an entire sphere, you would have \(0\le\varphi\le\pi\), but the sphere has been quartered to only contain points for which \(\dfrac{\pi}{2}\le\varphi\le\pi\). (See second attachment)

Answer 9

Range of \(\phi\) is
$$
0\le \phi \le \pi
$$
But we are constrained to the lower part of the sphere so
$$
\cfrac{\pi}{2} \le \phi \le \pi
$$
Of course \(\rho \) goes from 0 to 4
and \(\theta \) as drawn normally goes from 0 to \(2\pi\), but we don't need the entire circle, just the second part of the circle, so:
$$
\pi\le \theta\le 2\pi
$$
Does this help?

Answer 10

this is informative
http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-26-spherical-coordinates/

Answer 11

That Prof. Denis Auroux is a great professor - good resource.

Answer 12

Okay. I *think* I got it, regarding the sphere limits in that example.

Although do the order you put it up matter? Is there a strict order you have to triple-integral it in? Or is it the order of polar coordinates P(ρ, φ, θ) ? (in and outwards using integral)

To twist the example a bit (in order to understand it further);
What if instead of having a cut sphere that way; what if we cut it in half twice; so
0 < z < 0.5

A cone-like

Since it wouldn't be a complete "circle" (besides looking from Z axis, in which case it should be a perfect circle), how would you go on limiting this? (This example is closer to an actual task I am working on)