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## Analiza matematyczna funkcji wielu zmiennych

### Kurs: Analiza matematyczna funkcji wielu zmiennych>Rozdział 4

Lekcja 8: Całkowanie we współrzędnych krzywoliniowych: współrzędne biegunowe, współrzędne cylindryczne i współrzędne sferyczne.

# Całkowanie we współrzędnych sferycznych

Wyrażanie całki po danym obszarze lub w zadanych granicach we współrzędnych sferycznych. Tłumaczenie na język polski: Fundacja Edukacja dla Przyszłości.

## Kontekst

Different authors have different conventions on variable names for spherical coordinates. For this article, I will use the following convention. (In each description the "radial line" is the line between the point we are giving coordinates to and the origin).
• r indicates the length of the radial line.
• theta the angle around the z-axis. Specifically, if you project the radial line onto the x, y-plane, theta is the angle that line makes with the x-axis.
• \phi the angle between the radial line and the z-axis.
The following two are not strictly required, but they might help as warm up and practice for this topic.

## Do czego zmierzamy

• When you are performing a triple integral, if you choose to describe the function and the bounds of your region using spherical coordinates, left parenthesis, r, comma, \phi, comma, theta, right parenthesis, the tiny volume d, V should be expanded as follows:
\begin{aligned} &\quad \iiint_R f(r, \phi, \theta)\,dV \\\\ &= \iiint_R f(r, \phi, \theta) \,(\blueE{dr}) (\greenE{r\,d\phi})(\goldE{r\sin(\phi)\,d\theta})\\\\ &= \iiint_R f(r, \phi, \theta) \,\redE{r^2 \sin(\phi)}\,d\theta\,d\phi\,dr \end{aligned}
The key term to remember (or re-derive) is start color #bc2612, r, squared, sine, left parenthesis, \phi, right parenthesis, end color #bc2612
• Converting to spherical coordinates can make triple integrals much easier to work out when the region you are integrating over has some spherical symmetry.

## Dissecting tiny volumes in spherical coordinates

As discussed in the introduction to triple integrals, when you are integrating over a three-dimensional region R, it helps to imagine breaking it up into infinitely many infinitely small pieces, each with volume d, V.
When you were working in cartesian coordinates, these tiny pieces were thought of as rectangular blocks. In spherical coordinates, on the other hand, it helps to think of your tiny pieces as being slightly curved blocks "hugging" a sphere. I'll be drawing a fairly large version of one of these chunks, partly to exagerate its curvature, and partly just so we can see it. For examlpe, here's what one looks like in three dimensions:
The reason for this shape is that each face represents a constant value for one of the spherical coordinates:
• One pair of faces represents constant values of start color #0c7f99, r, end color #0c7f99 (these will be slightly curved, as if hugging a sphere).
• One pair of faces represents constant values of start color #a75a05, \phi, end color #a75a05.
• One pair of faces represents constant values of start color #0d923f, theta, end color #0d923f.
Why is that significant? Because the way multiple integrals work is that each individual integral treats all coordinate as constants, except for one. Therefore, as we consider how the multiple integral as a whole assembles these tiny pieces together, it is more natural to think about pieces whose volume can be expressed in terms of changes to individual coordinates. This will become clearer as you read further.
As the size of these blocks approaches zero, the curve will become so negligible that we can treat them as rectangular prisms. One edge represents a tiny change in the length in the distance from the origin, start color #0c7f99, d, r, end color #0c7f99:
The other two edges are related to the tiny changes in the other two coordinates, start color #0d923f, d, theta, end color #0d923f and start color #a75a05, d, \phi, end color #a75a05 . However, since start color #0d923f, theta, end color #0d923f and start color #a75a05, \phi, end color #a75a05 measure radians, not a unit of length, these values must be multiplied by a unit of length in order to properly reflect the lengths of the edges in our rectangular prism.
For example, the edge representing a change in start color #a75a05, \phi, end color #a75a05 has length start color #a75a05, r, d, \phi, end color #a75a05:
The edge representing a change in start color #0d923f, theta, end color #0d923f is a little trickier. This edge is part of some circle wrapping around the z-axis, and the radius of that circle is not start color #0c7f99, r, end color #0c7f99, but start color #0c7f99, r, end color #0c7f99, sine, left parenthesis, start color #a75a05, \phi, end color #a75a05, right parenthesis. This means the arc length due to a small change in start color #0d923f, theta, end color #0d923f is start color #0c7f99, r, end color #0c7f99, sine, left parenthesis, start color #a75a05, \phi, end color #a75a05, right parenthesis, start color #0d923f, d, theta, end color #0d923f.
That can be confusing at first, so it might be worth a moment of contemplation to ensure you understand how that works.
Putting all this together, we can express the volume of our "rectangular" block in terms of start color #0c7f99, d, r, end color #0c7f99, start color #a75a05, d, \phi, end color #a75a05 and start color #0d923f, d, theta, end color #0d923f by taking the product of all its side lengths.
d, V, equals, left parenthesis, start color #0c7f99, d, r, end color #0c7f99, right parenthesis, left parenthesis, start color #a75a05, r, d, \phi, end color #a75a05, right parenthesis, left parenthesis, start color #0d923f, r, sine, left parenthesis, \phi, right parenthesis, d, theta, end color #0d923f, right parenthesis, equals, start color #bc2612, r, squared, sine, left parenthesis, \phi, right parenthesis, end color #bc2612, d, r, d, \phi, d, theta
In other words, when you have some triple integral,
\begin{aligned} \iiint_R f \,dV \end{aligned}
and you choose to express the bounds and the function using spherical coordiantes, you cannot just replace d, V with d, r, d, \phi, d, theta. You must also remember the start color #bc2612, r, squared, sine, left parenthesis, \phi, right parenthesis, end color #bc2612 term:
\begin{aligned} \iiint_R f(r, \theta, \phi) \: \redE{r^2 \sin(\phi)}\,dr\,d\phi\,d\theta \end{aligned}
Personally, I can never quite remember exactly how to expand the d, V term off the top of my head
"Was it sine, left parenthesis, \phi, right parenthesis or sine, left parenthesis, theta, right parenthesis... and is it r or r, squared...?"
Instead, I think through the process I just illustrated above, asking what the arc lengths resulting from changes to \phi and theta are.

## Example 1: Volume of a sphere revisited

This might be the simplest possible starting example for triple integration in spherical coordinates, but it let's us compute an interesting non-trivial fact: The volume of a sphere.
Question: What is the volume of a sphere with radius R?
Situate the sphere such that its center is on the origin.
If we were doing this integral in cartesian coordinates, we would have that ugly-but-common situation where the bounds of inner integrals are functions of the outer variables. However, because spherical coordinates are so well suited to describing, well, actual spheres, our bounds are all constants.
Concept check: Which of the following sets of bounds on the coordinates r, \phi and theta accurately describes all the points inside a sphere of radius R (without running over the entire sphere multiple times)
Wybierz 1 odpowiedź:

Using these bounds, together with the fact that
d, V, equals, start color #bc2612, r, squared, sine, left parenthesis, \phi, right parenthesis, d, r, d, \phi, d, theta, end color #bc2612
we can start setting up our integral like this:
\begin{aligned} \iiint_{\text{Piłeczka}} dV = \int_0^{2\pi} \int_0^\pi \int_0^R r^2 \sin(\phi) \,dr \,d\phi \,d\theta \end{aligned}
Concept check: Work through this integral, and appreciate just how lovely it is compared with the other nasty triple integrals you may have encountered.
\begin{aligned} \int_0^{2\pi} \int_0^\pi \int_0^R r^2 \sin(\phi) \,dr \,d\phi \,d\theta = \end{aligned}

If you dare, imagine trying to do this integral in cartesian coordinates. It's a nightmare! This gives us an important takeaway:
Key takeaway If you are integrating over a region with some spherical symmetry, passing to spherical coordinates can make the bounds much nicer to deal with.

## Example 2: Integrating a function

Integrate the function
f, left parenthesis, x, comma, y, comma, z, right parenthesis, equals, x, plus, 2, y, plus, 3, z
in the region of the first octant where
x, squared, plus, y, squared, plus, z, squared, is less than or equal to, 3

### Step 1: Express the region in spherical coordinates.

How could you know that we should pass to spherical coordinates? We could do this whole integral in cartesian coordinates, couldn't we? Cylindrical coordinates would work too.
The fact that our boundary includes the condition x, squared, plus, y, squared, plus, z, squared, is less than or equal to, 3 is a description of the distance between points of our region and the origin. Since the spherical coordinate r expresses precisely this idea, we can feel confident that describing the boundary of our region using r will make the bounds of our three integrals simpler than if we did so in terms of x, y and z.
Specifically, this condition becomes
\begin{aligned} x^2 + y^2 + z^2 &\le 3 \\\\ r^2 &\le 3 \\\\ r &\le \sqrt{3\,} \end{aligned}
Concept check: What about theta and \phi? What bounds should we place on these two coordinates to keep our integral within the first octant?
is less than or equal to, theta, is less than or equal to
is less than or equal to, \phi, is less than or equal to

### Step 2: Express the function in spherical coordinates

Next, we convert the function
f, left parenthesis, x, comma, y, comma, z, right parenthesis, equals, x, plus, 2, y, plus, 3, z
into spherical coordinates. To do this, we use the conversions for each individual cartesian coordinate.
• x, equals, r, sine, left parenthesis, \phi, right parenthesis, cosine, left parenthesis, theta, right parenthesis
• y, equals, r, sine, left parenthesis, \phi, right parenthesis, sine, left parenthesis, theta, right parenthesis
• z, equals, r, cosine, left parenthesis, \phi, right parenthesis
Plugging each of these in, we get
\begin{aligned} f(x, y, z) &= x + 2y + 3z \\\\ &= r\sin(\phi)\cos(\theta) + 2r\sin(\phi)\sin(\theta) + 3r\cos(\phi) \\\\ &= r\Big(\sin(\phi)\cos(\theta)+2\sin(\phi)\sin(\theta)+3\cos(\phi)\big) \\\\ \end{aligned}
You might say that this makes things more complicated than they were in cartesian coordinates. And you'd be right! But when it comes to triple integrals, a more complicated function is a relatively small price to pay for getting our bounds to be constants.

### Step 3: Compute the triple integral

Concept check: Putting the previous two steps together, what is the integral that we need to solve?
Wybierz 1 odpowiedź:

Bring it on home: Solve that integral!
Integral from previous question:

## Podsumowanie

• When you are performing a triple integral, if you choose to describe the function and the bounds of your region using spherical coordinates, left parenthesis, r, comma, \phi, comma, theta, right parenthesis, the tiny volume d, V should be expanded as follows:
\begin{aligned} &\quad \iiint_R f(r, \phi, \theta)\,dV \\\\ &= \iiint_R f(r, \phi, \theta) \,(\blueE{dr}) (\greenE{r\,d\phi})(\goldE{r\sin(\phi)\,d\theta})\\\\ &= \iiint_R f(r, \phi, \theta) \,\redE{r^2 \sin(\phi)}\,d\theta\,d\phi\,dr \end{aligned}
The key term to remember (or re-derive) is start color #bc2612, r, squared, sine, left parenthesis, \phi, right parenthesis, end color #bc2612
• Converting to spherical coordinates can make triple integrals much easier to work out when the region you are integrating over has some spherical symmetry.

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