# Integration bounds when integrating a half sphere

I have to integrate over the northern hemisphere of a sphere. The answer states that the integration bounds for r is 0 to R, for theta it's 0 to pi/2, for phi it's 0 to 2pi.

What I don't understand is why theta doesn't go from 0 to 2pi. If I had to set this up, I would have theta going from 0 to 2pi and phi going from 0 to pi.

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berkeman
Mentor
I'll assume you mean integrating over the volume of the northern hemisphere, since you say r varies from 0 to R, and is not fixed at R (for a surface integration).

Phi is the angle in the plane of the equator, and theta is the angle down from the North Pole. So when you sweep theta pi/2, that's the half of a longitude line that lies in the northern hemisphere. When you sweep that vertical longitude line all the way around for phi = 0 to 2pi to cover the full northern hemisphere.

If instead you sweep theta from 0 to 2pi, that is the full length of the longitude line from North Pole to South Pole.

What happens if you do $$\frac {dx}{dr} of \frac {2}{3} \pi r^3?$$ or the volume of half a sphere, what do you end up with?

$$2 \pi r^2$$

http://en.wikipedia.org/wiki/Sphere

In analytic geometry, a sphere with center (x0, y0, z0) and radius r is the set of all points (x, y, z) such that

$$(x - x_0 )^2 + (y - y_0 )^2 + ( z - z_0 )^2 = r^2 \,.$$

The points on the sphere with radius r can be parametrized via

$$x = x_0 + r \sin \theta \; \cos \phi y = y_0 + r \sin \theta \; \sin \phi \qquad (0 \leq \theta \leq \pi \mbox{ and } -\pi < \phi \leq \pi) \, z = z_0 + r \cos \theta \,$$

A sphere of any radius centered at the origin is described by the following differential equation:

$$x \, dx + y \, dy + z \, dz = 0.$$

This equation reflects the fact that the position and velocity vectors of a point travelling on the sphere are always orthogonal to each other.

The surface area of a sphere of radius r is

$$A = 4 \pi r^2 \,$$

and its enclosed volume is

$$V = \frac{4}{3}\pi r^3.$$

The sphere has the smallest surface area among all surfaces enclosing a given volume and it encloses the largest volume among all closed surfaces with a given surface area. For this reason, the sphere appears in nature: for instance bubbles and small water drops are roughly spherical, because the surface tension locally minimizes surface area.
An image of one of the most accurate spheres ever created by humans, as it refracts the image of Einstein in the background. A fused quartz gyroscope for the Gravity Probe B experiment which differs in shape from a perfect sphere by no more than a mere 40 atoms of thickness. It is thought that only neutron stars are smoother.
An image of one of the most accurate spheres ever created by humans, as it refracts the image of Einstein in the background. A fused quartz gyroscope for the Gravity Probe B experiment which differs in shape from a perfect sphere by no more than a mere 40 atoms of thickness. It is thought that only neutron stars are smoother.

The circumscribed cylinder for a given sphere has a volume which is 3/2 times the volume of the sphere, and also a surface area which is 3/2 times the surface area of the sphere. This fact, along with the volume and surface formulas given above, was already known to Archimedes.

A sphere can also be defined as the surface formed by rotating a circle about any diameter. If the circle is replaced by an ellipse, and rotated about the major axis, the shape becomes a prolate spheroid, rotated about the minor axis, an oblate spheroid.

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HallsofIvy
Homework Helper
In their never ending battle to confuse one another, mathematicians and physicists use opposite conventions for $\theta$ and $\phi$. In most calculus books, you will see $\theta$ used for the "longitude" (measured around the equator) and $\phi$ for the "co-latitude" (measured down from the north pole)- physics text books reverse that. You are using the "physics" convention. In this case, as you have been told, the measurement around the equator, $\phi$, for the entire sphere, goes all the way around- from 0 to $2\pi$. Notice that if the co-latitude, $\theta$, also went from 0 to $2\pi$, points would have two sets of coordinates. The point with $\phi= 3\pi/2$, $\theta= \pi/2$ could also be labeld $\phi= \pi/2$, $\theta= 3\pi/2$. Taking $\theta$ from 0 to $\pi$ only solves that problem. Of course, to get the "northern hemisphere" only, you take $\theta$ from 0 to $\pi/2$.

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In their never ending battle to confuse one another, mathematicians and physicists use opposite conventions for $\theta$ and $\phi$. In most calculus books, you will see $\theta$ used for the "longitude" (measured around the equator) and $\phi$ for the "co-latituded" (measured down from the north pole)- physics text books reverse that. You are using the "physics" convention. In this case, as you have been told, the measurement around the equator, $\phi$, for the entire sphere, goes all the way around- from 0 to $2\pi$. Notice that if the co-latitude, $\theta$, also went from 0 to $2\pi$, points would have two sets of coordinates. The point with $\phi= 3\pi/2$, $\theta= \pi/2$ could also be labeld $\phi= \pi/2$, $\theta= 3\pi/2$. Taking $\theta$ from 0 to $\pi$ only solves that problem. Of course, to get the "northern hemisphere" only, you take $\theta$ from 0 to $\pi/2$.
You'll also notice if you multiply $$\frac {3\pi}{2}. \frac{\pi}{2}= \frac {3}{4} \pi^2$$

Try integrating that with the radius? What do you end up with? yup

$$\frac {4}{3} \pi r^3$$

$$\frac{dy}{dx}\frac {4}{3}\pi r^3=$$

It all makes beautiful sense, and it's where Archimedes got his calculations from, although his integration methods would have been different from the way modern integration worked, and he had no general rules to work from

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berkeman
Mentor
In their never ending battle to confuse one another, mathematicians and physicists use opposite conventions for $\theta$ and $\phi$. In most calculus books, you will see $\theta$ used for the "longitude" (measured around the equator) and $\phi$ for the "co-latituded" (measured down from the north pole)- physics text books reverse that. You are using the "physics" convention. In this case, as you have been told, the measurement around the equator, $\phi$, for the entire sphere, goes all the way around- from 0 to $2\pi$. Notice that if the co-latitude, $\theta$, also went from 0 to $2\pi$, points would have two sets of coordinates. The point with $\phi= 3\pi/2$, $\theta= \pi/2$ could also be labeld $\phi= \pi/2$, $\theta= 3\pi/2$. Taking $\theta$ from 0 to $\pi$ only solves that problem. Of course, to get the "northern hemisphere" only, you take $\theta$ from 0 to $\pi/2$.
Wow, thanks!! That really clears things up. It's funny, when I pulled out a textbook at home last night to double-check the convention so that I could reply to the OP, I pulled out a physics text first, and used that convention (which apparently was the same as the OP's problem was assuming). But as I looked at the convention in the text, I was saying to myself how I thought it was just the opposite! I feel better now.

In their never ending battle to confuse one another, mathematicians and physicists use opposite conventions for $\theta$ and $\phi$. In most calculus books, you will see $\theta$ used for the "longitude" (measured around the equator) and $\phi$ for the "co-latituded" (measured down from the north pole)- physics text books reverse that. You are using the "physics" convention. In this case, as you have been told, the measurement around the equator, $\phi$, for the entire sphere, goes all the way around- from 0 to $2\pi$. Notice that if the co-latitude, $\theta$, also went from 0 to $2\pi$, points would have two sets of coordinates. The point with $\phi= 3\pi/2$, $\theta= \pi/2$ could also be labeld $\phi= \pi/2$, $\theta= 3\pi/2$. Taking $\theta$ from 0 to $\pi$ only solves that problem. Of course, to get the "northern hemisphere" only, you take $\theta$ from 0 to $\pi/2$.
What I don't get is this:

When integrating over an entire sphere, $$\theta$$ would go from 0 to $$2\pi$$. So if I integrate a half sphere, why not go from 0 to $$\pi$$? That would cover the range of $$\theta$$ that is in the northern hemisphere.

EDIT: Never mind, I think I can see it now. So if I was integrating the entire sphere, $$\theta$$ would just go from 0 to $$\pi$$ right?

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berkeman
Mentor
EDIT: Never mind, I think I can see it now. So if I was integrating the entire sphere, $$\theta$$ would just go from 0 to $$\pi$$ right?
Correct. That would pick up the southern hemisphere as well.

What I don't get is this:

When integrating over an entire sphere, $$\theta$$ would go from 0 to $$2\pi$$. So if I integrate a half sphere, why not go from 0 to $$\pi$$? That would cover the range of $$\theta$$ that is in the northern hemisphere.

EDIT: Never mind, I think I can see it now. So if I was integrating the entire sphere, $$\theta$$ would just go from 0 to $$\pi$$ right?
Well I was useless as usual but, I'm used to it. what is 1/2 of. Check your answers with the simple formula that = dy/dx. It's just so obvious if you look at what lies behind it. I guess I tried to convey that and failed.

HallsofIvy
Homework Helper
You'll also notice if you multiply $$\frac {3\pi}{2}. \frac{\pi}{2}= \frac {3}{4} \pi^2$$

Try integrating that with the radius? What do you end up with? yup

$$\frac {4}{3} \pi r^3$$
How did the $\pi^2$ suddenly become $\pi$

$$\frac{dy}{dx}\frac {4}{3}\pi r^3=$$

It all makes beautiful sense, and it's where Archimedes got his calculations from, although his integration methods would have been different from the way modern integration worked, and he had no general rules to work from

How did the $\pi^2$ suddenly become $\pi$
it didn't but I assume if you put r into the equation or relate it, the formula becomes :-

$$\int \frac{3}{4} \pi^2=\frac{4}{3} \pi^3+C$$

Accounting for the radius in relation to a sphere=

$$\frac{4}{3}\pi r^3$$

Am I wrong?

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HallsofIvy
Homework Helper
$$\int \frac{3}{4} \pi^2.=\frac{4}{3} \pi^3+C$$

What does the . represent there? What are you integrating with respect? HOW are you putting r in there?

$$\int \frac{3}{4} \pi^2.=\frac{4}{3} \pi^3+C$$

What does the . represent there? What are you integrating with respect? HOW are you putting r in there?
Oh it's a typo I just mean r in respect to a sphere given that area always integrate to a..

OK I edited it. now theirs no .

It's all covered in the article I quoted anyway.

EDIT:

$$(x - x_0 )^2 + (y - y_0 )^2 + ( z - z_0 )^2 = r^2 \,.$$

$$x = x_0 + r \sin \theta \; \cos \phi y = y_0 + r \sin \theta \; \sin \phi \qquad (0 \leq \theta \leq \pi \mbox{ and } -\pi < \phi \leq \pi) \, z = z_0 + r \cos \theta \,)$$

$$xdx+ydy+zdz=0$$

$$A = 4 \pi r^2 \,$$

$$V = \frac{4}{3}\pi r^3.$$

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